Topology Notes. A project under construction. Franz Rothe

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1 Topology Notes A project under construction Franz Rothe Department of Mathematics University of North Carolina at Charlotte Charlotte, NC frothe@uncc.edu January 2, FALL/4181topology.tex Contents 1 About functions 3 2 Metric spaces 16 3 Topological spaces 23 4 Limit Points, Accumulation Points 33 5 Continuity 45 6 Products 57 7 The Kuratowski Closure Operator 61 8 Dense Sets and Baire Spaces 69 1

2 9 The Cantor Set and the Devil s Staircase The relative topology Connectedness Pathwise connected spaces The Hilbert curve Compact spaces Compact sets in metric spaces The Bolzano-Weierstrass property 124 References [1] Solomon Feferman Anita B. Feferman, Alfred Tarski: Life and Logic, Cambridge University Press, [2] Herbert B. Enderton, Elements of Set Theory, Academic press, ISBN , [3] David Hilbert, Foundations of Geometry, 2nd English ed., Open Court, La Salle, [4] Bert Mendelson, Introduction to Topology, third edition, Dover publications, Inc, ISBN-13: , [5] Yiannis N. Moschovakis, Notes on Set Theory, Springer, ISBN , [6] James R. Munkres, Topology, 2nd edition., Prentice-Hall, Inc, ISBN-13: , [7] Robert S. Wolf, A Tour through Mathematical Logic, the carus mathematical monographs 30 ed., Mathematical Association of America,

3 1 About functions Definition 1. By a function or mapping f : X Y from the domain X to the tentative target Y 1 is meant any prescription or procedure that assigns to each element of the set X, a unique element y. Following Euler s notation, one writes y = f(x). At the time of Euler and earlier, there was the general opinion that a function is given by an expression, as for example x 2, x 3 5x or sin(3x π/3). That view was first challenged in mathematical physics and differential geometry. Both d Alembert and Euler found, already in the 18-th century, a method to solve the problem of the vibrating string. By Euler s method one obtains the solution as a Fourier series, corresponding to the overtones produced by the string. On the other hand, d Alembert s solution uses the reflection of travelling waves at the endpoints of the string. The two methods both produce the same correct solution of the same physical problem, but nevertheless, one ends up with totally different expressions for the solution. From d Alembert s solution for the vibrating string problem, it became clear that the notion of function is linked much more naturally with graph of a function, and with the use of functions to described physical phenomena, than with the use of formal expressions. One may, for example, want to find the motion of a vibrating string, which is at the time zero deflected by a plug producing a single v-shape dent on the string. There is no easy formal expression which describes such a dent. But such an expression is not at all necessary to consider and solve the physical problem. We see that the natural realistic notion function has to be a more general one. Moreover, in this and similar contexts, the necessity to define a function involving cases became obvious. Nowadays, one also calls this in calculus a split definition. With the work of Fourier on trigonometric series, one discovered that a function given by a split definition, can still be expressed by one single Fourier series. So the link of a function with any expression become even more distant and less obvious or important. Finally, Dirichlet came up with the following example of a function: { 1 for x rational; (1.1) D(x) = 0 for x irrational. It is credited to Dirichlet, to introduce around 1850 the above definition 1 for a notion of function. As a benefit, we have a much more precise and general notion of what a function is. The notion of function becomes detached from the method to produce the function. Whether expressions, calculations, computation, graphing, split cases, mathematical physics, or any other means are involved is left totally open by Dirichlet s 1 For reasons that shall become clear below, there is no generally accepted name for the set Y. 3

4 definition. It even becomes possible that the definition of a function may evolve axioms or methods which were not known at the time of Dirichlet. As examples, one may think of definitions evolving the axiom of choice, or Turing machines. At that stage of affairs, the need for a impaccably precise definition becomes obvious. One has the goal to define the notion of function within the framework of axiomatic set theory. To this end, we remember the possibility to graph a function, and the use of Cartesian coordinates for this purpose. For example graphing a function from the real number to the real numbers is done in the coordinate plane, the points of which are ordered pairs (x, y) where x and y are real numbers. More generally, the graph of any function f : X Y, where X and Y are arbitrary sets is a subset of the Cartesian product X Y. Indeed, we define the graph Γ f of any function f : X Y as follows: (1.2) Γ f := {(x, y) ; x X, y Y and y = f(x)} Recall that it is customary to call any subset R X Y a relation. Clearly not all relations are functions. The reason is that for a function, both existence and uniqueness of the image value f(x) is required, once the point x X is specified. Hence we can state: Proposition 1. Let X and Y be arbitrary sets. Any relation R X Y is the graph of a function f : X Y if and only the following two requirements are satisfied: existence of the image point: For each x X exists y Y such that (x, y) R; uniqueness of the image point: If (x, y 1 ) R and (x, y 2 ) R then y 1 = y 2. In the framework of axiomatic set theory, it is nowadays customary to identify a function with its graph. As a benefit of this view, the elementary properties of function can be discussed, even without introducing any new axioms. 2 Remark. Still, there is widespread feeling in the mathematical community that such an identification is somewhat artificial. Here are some of the reasons for that attitude. Under the premiss of identifying a function f : X Y with its graph, it is possible to reconstruct the domain X from the function. This is a natural and wanted situation. On the other hand, it is not possible to reconstruct the target set Y from the function; any two functions f 1 : X Y 1 and f 2 : X Y 2 with different target sets Y 1 and Y 2 but the same graph turn out to be equal. The function may have been constructed not from its graph, but instead by a totally different independent idea. 2 But remember that the axiom of replacement is needed to define the direct image of a set by a function symbol. 4

5 One may have doubts whether this is a natural and wanted situation. Definition 2 (Domain and range of a function). Given is any function or mapping f : X Y. The set X is called the domain and denoted by dom f. The set f(x) := {y : y B and x [x X and f(x) = y] } is called the range and denoted by ran f. Definition 3 (Direct image, and inverse image). Let f : X Y be any function. Let A X and B be any sets. The image or direct image f(a) of the subset A of the domain is defined to be set f(a) := {y : y Y and x [x X and f(x) = y] } The inverse image f 1 (B) of the any set B is defined to be set f 1 (B) := {x X : f(x) B } Proposition 2. Let f : X Y be any function. Let A, B be any sets, and let {A α } α I be an indexed family. For the inverse image, the following set equalities hold: (1.3) (1.4) (1.5) (1.6) (1.7) (1.8) f 1 ( ) =, f 1 (Y ) = X, f 1 (A B) = f 1 (A) f 1 (B), f 1 (A B) = f 1 (A) f 1 (B), f 1 (A \ B) = f 1 (A) \ f 1 (B), ( ) f 1 A α = f 1 (A α ), α I α I ( ) f 1 A α = f 1 (A α ) for nonempty index set I. α I α I Proposition 3. Let f : X Y be any function. Let A, B X be any subsets of the domain, and let {A α } α I be an indexed family of subsets A α X of the domain. For the direct image, the following set equalities and inclusions hold: (1.9) (1.10) (1.11) (1.12) (1.13) f( ) =, f(x) Y, f(a B) = f(a) f(b), f(a B) f(a) f(b), ( ) f A α = f(a α ), α I α I ( ) f A α f(a α ) for nonempty index set I. α I α I 5

6 10 Problem 1.1. Given is any function f : X Y and any sets A, B X. Give a simple proof that f(a B) = f(a) f(b). f(a B) f(a) f(b) : Take any element y f(a B). Hence there exists x A B such that y = f(x). Hence either x A or x B, and y = f(x) in both cases. Now x A and y = f(x) implies y f(a). Similarly, x B and y = f(x) implies y f(b). Thus we see that either y f(a) or y f(b), and hence y f(a) f(b), as to be shown. f(a B) f(a) f(b) : Take any element y f(a) f(b). Hence either y f(a) or y f(b). In the first case, there exists x A such that y = f(x). In the second case, there exists x B such that y = f(x). Hence in both cases, there exists x A B, such that y = f(x). Thus we see that y f(a B), as to be shown. My second answer using logic. We check that both sets have the same elements: Let y f(a B) be given. We need to check whether y f(a) f(b). Here are the steps: (1.14) (1.15) (1.16) (1.17) (1.18) (1.19) y f(a B) x [y = f(x) and x A B] x [y = f(x) and (x A or x B)] x [(y = f(x) and x A) or (y = f(x) and x B)] x [y = f(x) and x A] or x [y = f(x) and x B] y f(a) or y f(b) y f(a) f(b) Which step uses not so evident quantified logic? Which tautology from quantified logic is used? Answer. Step (4) uses that in quantified logic is a tautology. x [P (x) or Q(x)] x P (x) or x Q(x) Proposition 4. Let f : X Y be any function. Let A X be any subset of the domain, B be any set. Combining direct and inverse images, the following set inclusions and equality hold: (1.20) (1.21) (1.22) f 1 (f(a)) A f(f 1 (B)) B, f(a f 1 (B)) = f(a) B 6

7 10 Problem 1.2. Let f : X Y be any mapping. Convince yourself that f(a f 1 (B)) = f(a) B for all subsets A X and B Y. Proof. Imagine elements x X and y Y. y f(a f 1 (B)) x [y = f(x) and x A f 1 (B)] x[y = f(x) and x A and x f 1 (B)] x[y = f(x) and x A and f(x) B] x[y = f(x) and x A] and y B y f(a) and y B y f(a) B 10 Problem 1.3. Let f : X Y be any mapping and A X be any set. Convince yourself that A f 1 (f(a)) Proof. Take any x A. Assumed is A X. Let y = f(x). Hence y f(a), f(x) f(a) and x f 1 (f(a)). We see that x A x f 1 (f(a)), as to be checked. My second answer using logic. The first line is evidently true. 3 All lines below can be logically deducted x[ z(z = x)] x[x A z(z = x)] x[x A z(z = x and x A)] x[x A z(x X and z = x and z A)] x[x A (x X and z(f(z) = f(x) and z A))] x[x A (x X and f(x) f(a))] x[x A (x f 1 (f(a))] A f 1 (f(a)) Remark. We use the logical axioms and rule of inference for first-order logic taken from Robert S. Wolf, A Tour through Mathematical Logic to prove that x t(x = t) 3 A few days I did depend only on my intuition. A derivation of this assertion from a standard list of logical axioms and rules of inference for first-order logic taken from Robert S. Wolf, A Tour through Mathematical Logic is included below. 7

8 Into the logical axiom (2) about universal specification, we put as formula P(x) simply x t, and let the term t simply be a variable. We get from the logical axiom (2): xp(x) P(t) x(x t) t t One has to use the generalization t to obtain in a similar way: t[ xp(x) P(t)] t[ x(x t) t t] The next steps are using a tautology P Q Q P, and the definition of the existential quantifier (5) and logical axiom (3): t[t = t x(x t)] t[t = t x(x = t)] t[t = t] t x(x = t) Because of axiom (6) t = t holds, and hence the generalization t[t = t] holds. Hence the modus ponens implies as claimed. t x(x = t) 10 Problem 1.4. Let f : X Y be any mapping and X be nonempty. Convince yourself that A f 1 (B) f(a) B for all subsets A X and B Y. Proof of A f 1 (B) f(a) B. Assume that A f 1 (B). Hence by problem 1.2 f(a) f(f 1 (B)) = f(x f 1 (B)) = f(x) B B Proof of f(a) B A f 1 (B). Assume that f(a) B. Hence by problem 1.3 A f 1 (f(a)) f 1 (B) Definition 4 (One-to-one, injective, surjective, bijective). Let f : X Y be any function. The function is called one-to-one or injective if and only if (1.23) f(x 1 ) = f(x 2 ) x 1 = x 2 holds for all x 1, x 2 X. The function is called onto Y if and only if for all y Y there exists x X such that f(x) = y. A function f : X Y is called a bijection from X to Y if and only if it is both one-to-one and a mapping onto Y. 8

9 10 Problem 1.5. Let f : X Y be any function. Convince yourself of the following: The function is called injective if and only if for all x 1, x 2 X holds (1.24) x 1 x 2 f(x 1 ) f(x 2 ) The function is onto Y if and only if the set equation f(x) = Y holds. 10 Problem 1.6. Given are two functions (i) f : {1, 2} {1, 2} such that f(1) = 1 and f(2) = 2; (ii) g : {1, 2} N such that g(1) = 1 and g(2) = 2. Are these two functions equal or not? Answer. Indeed, these two functions are equal. In general, two functions f and g are equal if and only if their domains are equal: dom f = dom g; f(x) = g(x) holds for all x in this common domain. Remark. Because of this matter of affairs we need to be aware: Two functions are equal if and only if their graphs are equal. A function f : X Y does uniquely specify its domain X but not its target set Y. One can only say a function is onto some set but not simply onto. The notion surjective makes only sense for a function f : X Y, in the context where the target set Y is specified additionally to the function. To specify a bijection f : X Y, one needs, in addition to the function with its domain X, to specify the target set Y, and to prove that the function is both one-to-one and onto Y. Lemma 1 (Not enough for existence of inverse). Given are mappings f : X Y and g : Y X such that g f = id X. Then the mapping f is injective and the mapping g is surjective, onto X. Proof. The assumption g f = id X implies that the function f is injective. Indeed, assume f(x) = f(x ). Applying function g on both sides, one gets x = g(f(x)) = g(f(x )) = x Hence the function f is injective. Too, this same assumption g f = id X implies that g is mapping onto X. Let any x X be given. We put y := f(x) and check that g(y) = g(f(x)) = x. Hence the function g assumes any arbitrary x X as a value. 9

10 Remark. For arbitrary mappings f : X Y and g : Y X, the assumption g f = id X does neither imply the mapping f is surjective, nor need the mapping g to be injective. The following two problems 1.7 and 1.8 give elementary examples. 10 Problem 1.7. Define the mappings f : N N and g : N N by setting (a) Check that g f = id N. f(n) = n + 1 for all naturals n 1; g(1) = 1 and g(n) = n 1 for all naturals n 2. (b) Check that the other composite mapping f g satisfies [f g](1) = 2 and [f g](n) = n for all naturals n 2. Answer. (a) [g f](n) = g(f(n)) = g(n + 1) = n = n for all natural n 1, and hence g f = id N holds. (b) [f g](n) = f(g(n)) = f(n 1) = n = n for all natural n 2, but [f g](1) = f(g(1)) = f(1) = Problem 1.8. Define the mappings f : R 2 R 3 and g : R 3 R 2 by setting f(x 1, x 2 ) = (x 1, x 2, 0) for all (x 1, x 2 ) R 2 ; g(x 1, x 2, x 3 ) = (x 1, x 2 ) for all (x 1, x 2, x 3 ) R 3. (a) Check that g f = id R 2. In other words [g f](x 1, x 2 ) = (x 1, x 2 ) for all (x 1, x 2 ) R 2. (b) Check that the other composite mapping f g is [f g](x 1, x 2, x 3 ) = (x 1, x 2, 0) for all naturals (x 1, x 2, x 3 ) R 3. Remark. Written using the matrices, the linear mappings from problem 1.8 are 1 0 [ ] F = and G = The composite mapping g f corresponds to the matrix GF = I 2, whereas the composite mapping f g corresponds to the matrix F G = P. These two matrix products are [ ] GF = and F G = One sees that the matrix GF = I 2 is the identity matrix in two dimensions. But the other product F G = P is a projection, and not I 3. 10

11 Proposition 5 (Minimal assumptions giving the inverse). For any mappings f : X Y and g : Y X, the following statements are equivalent: (a) g f = id X and f is surjective; (b) g f = id X and g is injective; (a2) f g = id Y (b2) f g = id Y and g is surjective; and f is injective; (c) g f = id X and f g = id Y ; (d) f and g are both bijections, and (c) holds. Reason for (a) (c). The assumptions g f = id X being surjective together imply and the function f : X Y f g f = f id X = f (f g)(f(x))) = f(x) for all x X, (f g)(y)) = y for all y Y, f g = id Y and hence item (c) holds. Reason for (a2) (c). The assumptions f g = id Y being surjective together imply and the function g : Y X g f g = g id Y = g (g f)(g(y))) = g(y) for all y Y, (g f)(x)) = x for all x X, g f = id X and hence item (c) holds. Reason for (b) (c). The assumptions g f = id X and the function g being injective together imply and hence item (c) holds. g f g = id X g = g g([f g](y)) = g(y) for all y Y, [f g](y)) = y for all y Y, f g = id Y 11

12 and the function f being injec- Reason for (b2) (c). The assumptions f g = id Y tive together imply f g f = id Y f = f f([g f](x)) = f(x) for all x X, [g f](x) = x for all x X, g f = id X and hence item (c) holds. 4 Reason for (c) (d). This part cannot be left to the reader. By lemma 1, the assumption g f = id X implies that the function f is injective and the function g is surjective. Switching the roles of f and g, similarly, the assumption f g = id Y implies that the function g is injective and the function f is surjective. Hence both functions are bijections. Reason for (d) (a),(b),(c). This implication is obvious. Corollary 1. Let f : X Y and g : Y X be any two mappings. It is assumed that f and g are both injective and g f = id X holds. Then both mappings are bijections, and they are inverse to each other. Corollary 2. Let f : X Y and g : Y X be any two mappings. It is assumed that f and g are both surjective and g f = id X holds. Then both mappings are bijections, and they are inverse to each other. Corollary 3. Let f : X Y and g : Y X be any two mappings. It is assumed that either f or g is bijective, and that g f = id X holds. Then both mappings are bijections, and they are inverse to each other. 10 Problem 1.9. For any maps f : X Y and g : f(x) X, equivalent are (a) g f = id X ; (b) The map f is injective and f g = id f(x). (c) f is a bijection from X to f(x), and g is its inverse. Reason for (a) (b). Assume f(x) = f(x ). Apply function g on both sides and get x = g(f(x)) = g(f(x )) = x Hence the function f is injective. Moreover, the assumption g f = id X implies f g f = f id X = f. One concludes (f g)(f(x)) = f(x) for all x X, and hence (f g)(y)) = y for all y f(x). 4 No, it can t. 12

13 Reason for (b) (a). The assumption f g = id f(x) implies f g f = id f(x) f = f. Hence f((g f)(x)) = f(x) for all x X. Since f is injective one concludes (g f)(x) = x for all x X. Hence g f = id X, as to be shown. 10 Problem We define three functions a : R 2 R 2, b : R 2 R 2 and c : R 2 R by the formulas ( x + y a(x, y) =, x y ) 2 2 b(u, v) = ( u 2, v 2) c(w, z) = w z Calculate the composite function c b a Answer. ( x + y c b a(x, y) = c b, x y ) 2 2 (x + y)2 (x y)2 = 4 4 ( (x + y) 2 = c, 4 = xy ) (x y) Problem Let X, Y be any sets and f : X Y be any function. We define the relation R = {(x, x ) X X : f(x) = f(x )} (i) Check that R is an equivalence relation. (ii) Show that the mapping F : X/R ranf, obtained by setting F ([a]) = f(a) for all a X, is well-defined. (iii) We define the projection π : X X/R which assigns to each a X its equivalence class [a]. Let I : ranf Y denote the inclusion. Prove that f = I F π. In other words, the diagram below is commutative. X π X/R f Y I F ran f (iv) The mapping F : X/R ranf is a bijection. Especially, the sets X/R and ranf have the same number of elements. Answer. (i) The relation R is an equivalence relation since it is 13

14 reflexive: symmetric: xrx holds since f(x) = f(x) is a logical axiom. xry yrx holds since f(x) = f(y) implies f(y) = f(x). transitive: xry and yrz xrz holds since f(x) = f(y) and f(y) = f(z) imply f(x) = f(z). (ii) The mapping F : X/R ranf by setting F ([a]) = f(a) for all a X is welldefined. Indeed, f(a) = f(b) implies [a] = [b]. Hence f(a) depends only on the equivalence class [a]. (iii) We define the projection π : X X/R which assigns to each a X its equivalence class [a]. Let I : ranf Y denote the inclusion. Hence I F π(a) = I(F (π(a))) = I(F ([a])) = F ([a]) = f(a) f = I F π for all a X (iv) To check whether the mapping F : X/R ranf is injective, suppose that F ([a]) = F ([b]). Hence f(a) = f(b), moreover arb and finally [a] = [b]. To check whether the mapping F : X/R ranf is mapping onto ran f, take any element y of this set. Hence there exists a X such that f(a) = y. Moreover F ([a]) = f(a) = y and hence y is in the range of the induced mapping F. 10 Problem Let F : X X be any injective function. For n 0, we define the iteration f [0] = I X, f [1] = f, f [n+1] = f f [n] Prove by induction that all iterates f [n] are injective. 10 Problem Continuing the previous problem, we define the relation R X X such that arb k N 0 b = f [k] (a) or a = f [k] (b) Prove that R is an equivalence relation. Answer. The relation R is an equivalence relation since it is reflexive: ara holds since a = f [0] (a) by definition. symmetric: arb bra holds both b = f [k] (a) and a = f [k] (b) imply bra. transitive: We check that arb and brc arc. Assuming arb and brc we get several cases: (i) b = f [k] (a) and c = f [l] (b). Hence c = f [k+l] (a) and arc holds. (ii) a = f [k] (b) and c = f [l] (b) with l k. Hence c = f [l k] (f [k] (b)) = f [l k] (a) and arc holds. 14

15 (iii) a = f [k] (b) and c = f [l] (b) with l < k. Hence a = f [k l] (f [l] (b)) = f [k l] (c) and arc holds. Three more similar cases are left to the reader. 15

16 2 Metric spaces Definition 5 (Metric space, pseudo-metric space). A metric space is a pair (X, d) where X is a nonempty set and d : X X [0, ) is a function called metric with the properties (i) d(x, y) 0; (ii a) d(x, x) = 0; (ii b) d(x, y) = 0 implies x = y; (iii) d(y, x) = d(x, y); (iv) d(x, z) d(x, y) + d(y, z). holding for all x, y, z X. For a pseudo-metric space, only items (i), (ii a), (iii) and (iv) are required. Definition 6 (Normed space, pseudo-normed space). A normed space is a pair (X, ) where X is a linear space over the real or complex field and : X [0, ) is a function called norm with the properties (i) x 0; (ii) x = 0 implies x = 0; (iii) λx = λ x for all λ in the field; (iv) x + y x + y. holding for all x, y X. For a pseudo-normed space, only items (i), (iii) and (iv) are required. Lemma 2. Every normed space is a metric space. 10 Problem 2.1. Explain why every normed space is a metric space. Proof. We define d(x, y) = x y and check that it is a distance function. (i) d(x, y) = x y 0; (ii a) d(x, x) = x x = 0 0 = 0 0 = 0; (ii b) d(x, y) = 0 implies x y = 0 and thus x y = 0 and x = y; (iii) d(y, x) = y x = ( 1)(x y) = 1 x y = d(x, y); (iv) d(x, z) = x z = (x y) + (y z) x y + y z = d(x, y) + d(y, z). 16

17 Definition 7 (Open ball). Let (X, d) be any metric space, and x X, r 0. The open ball around x of radius r is the set B(x; r) := {y X : d(x, y) < r} Definition 8 (The standard topology of a metric space). A subset O of a metric space (X, d) is called open, if and only if for every point x O, there exists δ > 0 such that B(x; δ) O. 10 Problem 2.2. Check for the standard topology of a metric space (X, d) that the postulates for the open sets are valid: (O1) and X are open sets. (O2) If U, V are open, their intersection U V is open. (O3) If U α is a family of open sets, with indexes α I, their union α I U α is open. Answer. (O1) open is vacuously true. The entire space X is open since we can take any δ > 0 and point x X and get B(x; δ) X. (O2) Take any open sets U and V. To check whether their intersection is open, take any point x U V. There exist δ U > 0 and δ V > 0 such that B(x; δ U ) U and B(x; δ V ) V. Let δ := min{δ U, δ V }, which is clear still positive. B(x; δ) = B(x; δ U ) B(x; δ V ) U V We see the intersection U V contain a ball of positive radius around each one of its points, and hence is open, as to be checked. (O3) Let U α be a nonempty family of open sets, with indexes α I. To check whether their union U := α I U α is open, take any point x U. There exist β I and δ > 0 such that B(x; δ) U β. Hence B(x; δ) U β α I U α We see the union U contains a ball of positive radius around each one of its points, and hence is open, as to be checked. 10 Problem 2.3. Let (X, d) be any metric space, and x X, r > 0. Prove that a ball B(x; r) is an open set. (a) Assume d(x, y) < r and check that B(y; r d(x, y)) B(x; r). 17

18 (b) Check that the open ball B(x; r) is indeed open. Answer. (a) Assume d(x, y) < r and let z B(y; r d(x, y)). Hence d(x, z) < d(x, y) + d(y, z) < d(x, y) + (r d(x, y)) = r implies z B(x; r). Thus we have confirmed that B(y; r d(x, y)) B(x; r). (b) We check that the set B(x; r) is open. Indeed, for any y B(x; r), by definition d(x, y) < r. Item (a) yields that B(y; r d(x, y)) B(x; r). Hence B(y; r d(x, y)) is an open ball around y contained in the ball B(x; r). Hence the set B(x; r) is open. Lemma 3. A subset O of a metric space (X, d) is open, if and only if it is the union of open balls: O = α I B(x α ; δ α ) 10 Problem 2.4. Prove Lemma 3. Definition 9 (Neighborhood of a point). A subset N of a metric space X is called a neighborhood of point x, if and only if there exists a real number δ > 0 such that x B(x, δ) N. The set of all neighborhoods of point x is denoted by N x. Definition 10 (Closed set). A subset F of a topological or metric space is called closed if and only if its complement X \ F is open. 10 Problem 2.5. Let (X, d) be any metric space, and x X, r 0. Prove that a disk D(x; r) := {y X : d(x, y) r} is a closed set. (a) Assume d(x, y) > r and check that B(y; d(x, y) r) X \ D(x; r). (b) Check that the set X \ D(x; r) is open. Answer. (a) Assume d(x, y) > r and let z B(y; d(x, y) r). Hence d(y, z) < d(x, y) r and d(x, z) d(x, y) d(y, z) > r imply z X \D(x; r). Thus we have confirmed that B(y; d(x, y) r) X \ D(x; r). (b) We check that the set X \ D(x; r) is open. Indeed, let y X \ D(x; r). Hence d(x, y) > r. Item (a) yields that B(y; d(x, y) r) X \ D(x; r) is an open ball around y contained in the complement X \ D(x; r). Hence the set X \ D(x; r) is open. Thus we see that the X \ D(x; r) is open and hence the set D(x; r) is closed. Definition 11 (Closure in a metric space). Let (X, d) be a metric space and F X be any subset. A point b X lies in the closure F if and only if for all δ > 0 there exists a point c such that d(b, c) < δ and c F. 18

19 Definition 12 (Closure of a set). Let (X, d) be a topological or metric space and F X be any subset. The closure of F is denoted by F. A point b X lies in the closure F if and only if every neighborhood of b contains a point c of the set F. Remark. It does not matter whether c = b or c b. Remark. The inclusion F F holds always. Remark. Formulas defining b F and the negation b / F : b F V N b c [c V F ] b / F V N b [V F = ] In words once more: A point b lies in the closure F if and only if every neighborhood of b contains a point c F. A point b does not lie in the closure F if and only if there exists a neighborhood V of b which is disjoint from the set F. Proposition 6. A subset F of a topological or metric space is closed if and only if F = F. Proof. F is closed X \ F is open b[b / F V N b [V X \ F ]] b[b / F V N b [V F = ]] b[b / F b / F ] F F F = F Definition 13. A sequence of elements of any set S is mapping from N S. The number or index n N is mapped to the element a n. Definition 14. A sequence of elements a n of a metric space (X, d) is convergent to a if and only if for all δ > 0, there exists a natural number N such that n > N d(a n, a) < δ Proposition 7. A subset F of a metric space (X, d) is closed if and only if for any convergent sequence of elements a n F, its limit lim n a n F lies in the set F. 19

20 Proof. We assume that F is closed, and check whether for any convergent sequence of elements a n F its limit a = lim n a n is an element a F. Given is a convergent sequence with elements a n F. We distinguish the two cases: (i) There exists n such that a n = a; (ii) a n a for all n N. In the first case, a F does hold. Now assume case (ii) occurs. By the assumed convergence of the sequence, for all δ > 0, there exists a natural number N such that n > N a n B(a; δ) and a n a Hence for all neighborhoods V of a, there exists δ > 0 such that a N+1 B(a; δ) V and a N+1 a Hence a F. By assumption, the set F is closed and hence F F. Hence a F holds, as to be checked. For the proof of the converse we need the Archimedean principle 15 and the Principle of Countable Choice 17, which are explained below. Proof. We assume that any convergent sequence of elements a n F has its limit lim n a n F, and check whether F is closed. We have to check whether F F. To this end, take any b F. Hence for all neighborhood V of b, there exists c b such that c V F. We choose the neighborhoods V := B(b; 1 ) for any n N. Hence n n N a [ ] a b and d(a, b) < 1 n It is quite common to write in the last formula a n instead of simply a. The exact point here is : by the axiom of choice ACN, there exists the sequence n a n. We now check that this sequence has the limit b. Let any number δ > 0 we given. By the Archimedean principle 15, there exists N such that Nδ > 1. Hence n > N implies d(a n, b) < 1 n < δ and finally lim n a n = b Our assumption above implies b F. Thus we have checked with the last paragraph that F F, and hence F is closed. Definition 15 (The Archimedean principle). For any reals a, b > 0 there exists a natural number n N such that na > b. Definition 16 (Upper bound). The number M is called an upper bound of the set S R iff (2.1) x M for all x S Lemma 4. The Archimedean principle has the following equivalent formulations: 20

21 (a) For every real number x R, there exists a natural number n N such that x < n. In other word, the set of natural numbers has no upper bound. (b) For any ɛ > 0 there exists a natural number n N such that 1 n < ɛ. (c) For any ɛ > 0 there exists a natural number r N such that 2 r < ɛ. Items (b) and (c) can be expressed by the formulas 1 lim n n 1 = 0 and lim r 2 = 0 r Definition 17 (Countable Principle of Choice, ACN ). For the function H on the domain N, we assume that H(i) for all i N. Then there exists a function f with domain N such that f(i) H(i) for all i N. In other words, the countable Cartesian product i N H(i) of nonempty sets H(i) is nonempty. Indeed f i N H(i). The function d : (x, y) X X [0, ) used to define the metric is to some rather large degree arbitrary. We investigate now some continuous functions f : [0, ) [0, ) for which the composition f d turns out to be a metric, too. Let f : [0, ) [0, ) be a continuous function for which (f1) f(0) = 0 and f(x) > 0 for x > 0; (f2) 0 < x < y implies f(x) < f(y); (f3) f(x + y) f(x) + f(y) for all x, y > 0. It is a bid tedious, but straightforward, to check that the functions f(x) = x 1 + x and g(x) = x 5 have the above properties. Too, one can check that any differentiable function for which f(0) = 0, f (x) > 0 for all x 0 and f is weakly decreasing has the properties (f1) through (f3). 10 Problem 2.6. Let (X, d) be a metric space and assume the continuous function f has the properties (f1) through (f3). Check the following assertions: (a) The function d (x, y) := f(d(x, y)) for all x, y X defines a metric. (b) An open ball B(x; ɛ) with x X and ɛ > 0 for the metric d is a ball for the metric d with radius ɛ = f(ɛ). 5 From a purist s point of view, one may wish to restrict attention to these examples. 21

22 (c) Any open ball B (x; ɛ ) with x X and ɛ > 0 for the metric d is either a ball for the metric d or the entire space X. (d) The metrics d and d are topologically equivalent. Answer. (a) The function d (x, y) := f(d(x, y)) is a metric. Indeed d (x, y) 0 and d (x, y) = 0 d(x, y) = 0 x = y; d (y, x) = f(d(y, x)) = f(d(x, y)) = d (x, y); d (x, z) = f(d(x, z)) f(d(x, y)+d(y, z)) f(d(x, y))+f(d(y, z)) = d (x, y)+ d (y, z). (b) Given is any open ball B(x; ɛ) with center x X and radius ɛ > 0 for the metric d. Put ɛ := f(ɛ) and get B (x; ɛ ) = {y X : d (x, y) < ɛ } = {y X : f(d(x, y)) < f(ɛ)} = {y X : d(x, y) < ɛ} = B(x; ɛ) (c) With center x X and radius ɛ > 0 for the metric d, an open ball B (x; ɛ ) = {y X : d (x, y) < ɛ } is given. In case that ɛ is so large that it is not in the range of the function f, the inequality f(d(x, y)) < ɛ holds for all x, y X, and hence B (x; ɛ ) = X. In the other case, there exists a unique ɛ such that f(ɛ) = ɛ and one gets B (x; ɛ ) = B(x; ɛ) as above. (d) Both metrics d and d define the same family of open sets. Hence they are is topologically equivalent. 22

23 3 Topological spaces Definition 18 (Topological space). A topological space is a pair (X, T ), where X is a nonempty set and the family T is any family of subsets of the space X for which the requirements below hold (O1) T, X T (O2) U, V T U V T (O3) U α T for all α I implies α I U α T Definition 19 (Neighborhood of a point). A subset N of a topological space X is called a neighborhood of point x, if and only if there exists an open set O such that x O N. The set of all neighborhoods of point x is denoted by N x. Lemma 5. A subset O of a topological space X is open if and only if it is a neighborhood for each of its points. Proof. By putting V = O, we see that an open set is a neighborhood of each of its points. Conversely, assume that set O has the property to be a neighborhood for each of its points: For every point a O, there exists a neighborhood V a such that a V a O. We take from these inclusions the union over all a O to obtain O = a O {a} a O V a a O O = O and hence O = a O V a Thus the set O is a union of open sets. By axiom (O3), the union set O is open. Remark. We have used here the axiom scheme of replacement to produce from the set O the family {V a : a O}, and finally the union set axiom to get the set {Va : a O} =. Definition 10 (Closed set) A subset F of a metric or topological space X is called closed, if and only if its complement X \ F is open. 10 Problem 3.1. For an arbitrary topological space, the family T of all open sets is postulated have the properties (O1)(O2)(O3). Which properties follow for the family F of all closed sets of any topological space. Answer. (F1) F, X F (F2) F, G F F G F 23 a O V a

24 (F3) F α F for all α I and I imply α I F α F Proposition 8. Let (X, d) be any metric or topological space and A X be any subset. The closure A is given by the formula (3.1) A := {F X : A F and F is closed } Proof. Recall that by definition 12, any point b lies in the closure A if and only if every neighborhood of b contains a point c A: b A V N b c [c V A] We need the negated statement. A point b does not lie in the closure A if and only if there exists a neighborhood V of b which is disjoint from the set A: b / A V N b [V A = ] V [b V and V X is open and A X \ V ] Making use of the union set, we get the set equation b / A b {V X : V is open and A X \ V } X \ A = {V X : V is open and A X \ V } We let F = X \ V be the complement of any open set. The complement of the last formula yields the formula (3.1) as claimed. A = {X \ V X : V is open and A X \ V } = {F X : F is closed and A F } Definition 20 (Basis for the neighborhoods). Let X be a topological space and a X any point. A basis for the neighborhoods at a is a collection B a of sets such that (B1) a B for all B B a ; (B2) For all B B a there exists an open neighborhood V such that a V B; (B3) For all neighborhoods V a there exists B B a such that a B V. In other words, (B1)(B2) tell that the members of the collection B a are neighborhoods of a. Item (B3) tells that the collection B a contains enough small neighborhoods. 10 Problem 3.2. Let (X, d) be any metric space and x X. Check that the collection of disks D(x; 1 ) with any natural number n is a basis of the neighborhood at n x. Make clear for which item the Archimedean axiom is needed. 24

25 Answer. We check the requirements for a neighborhood basis from definition 20. (B1) x D(x; 1 ) holds for all natural numbers n; n (B2) For any natural number n, we put V := B(x; 1 ). Since B(x; 1 ) D(x; 1 ), the n n n set V is an open neighborhood such that x V D(x; 1 ); n (B3) For all neighborhoods V x there exists ɛ > 0 such that B(x; ɛ) V. By the Archimedean principle 15, there exists a natural number n such that nɛ > 1 and hence 1 n < ɛ and x D ( x; 1 n) B(x; ɛ) V Thus D(x; 1 n ) B x is the smaller neighborhood from the neighborhood basis, as required. Definition 21 (Basis of a topology). Let X be any nonempty set. A basis for a topology of X is a collection B of sets of subsets of X such that (BT1) for each x X there exists at least one B B such that x B; (BT2) if x belongs to the intersection of two basis elements B 1 and B 2, there exists a basis element B 3 such that x B 3 B 1 B 2. Proposition 9. Let X be any nonempty set. Let B be a collection of subsets of X which satisfies (BT1) and (BT2). There exists a unique topology T on X such that for all x X B x := {B B : x B} is a basis for the neighborhoods of x. Lemma 6. The topology from proposition 9 is obtained by requiring a subset C X to be open if and only if for all x C exists B B such that x B C. (3.2) C T x C B B [x B and B C] Definition 22 (Topology generated by a basis). Let X be any nonempty set. The topology T generated by B is obtained by requiring that a subset C X is open if and only if for all x C exists B B such that x B C. Lemma 7. The topology T generated by B is the collection of the unions S of all the subsets of B: (3.3) T = { S : S B} = {C X : S B [C = S]} 25

26 Lemma 8. We assume that F is a closed set in a topological space. For any convergent sequence of elements a n F, the limit a = lim n a n is an element a F, too. Proof. Given is a convergent sequence with elements a n F. By the assumed convergence, for all neighborhoods V N a, there exists a natural number N such that n > N a n V Hence all neighborhoods V of a contain a point of F, for example a N+1 V F. Hence a F. By assumption, the set F is closed and hence F F. Hence a F holds, as to be checked. The proof of the converse for lemma 8 is more tricky, basically because of the following issues. Do there exist enough convergent sequences as those occurring in lemma 8? Are the sequences long enough that their limits can reach all points of the closure? In order to get a positive answer to the first question, we shall use the Principle of Countable Choice 17. For a positive answer to the second question, we need to assume the space to be first countable, which one may think of as a generalization of the Archimedean principle 15. Definition 23 (First countability axiom). A space X is said to have a countable basis at point x if and only if there exists a countable collection B N x of neighborhoods of x such that each neighborhood of x contains one of the neighborhoods from family B. A space that has a countable basis at each of its points is said to satisfy the first countability axiom, or to be first-countable. Lemma 9. We assume (a) the principle 17 of countable choice ACN; (b) the topological space X to be first-countable, as explained by the first countability axiom 67 above. Let F X be any subset of the topological space X with the property that (c) any convergent sequence of elements a n F has its limit lim n a n = a F. Under these assumptions, the set F is closed. 26

27 Proof. We have to check whether F F. To this end, take any b F. By the definition of the closure, for all neighborhood V of b, there exists a point c V F. We choose the countable sequence of neighborhoods V := B n for any n N which are a countable basis at point b. Hence n N a [a B n F ] It is quite common to write in the last formula a n instead of simply a. The exact point here is: by the axiom of choice ACN, there exists the sequence n a n. Hence n N [a n B n F ] We now check that the sequence n a n has the limit b. Let any neighborhood V N b we given. There exists n N and a neighborhood B n from the countable neighborhood basis at point b such that B n V : Hence which confirms V N b n N [B n V ] V N b n N [a n B n F V F ] lim a n = b n Our assumption (c) above implies b F. Thus we have checked by the entire argument that F F, and hence F is closed. Together, the lemmas 8 and 9 confirm the following result, which is analogous to the proposition 7 for metric spaces. Proposition 10. We assume (a) the principle 17 of countable choice ACN to hold; (b) the topological space X to be D:firstcountablefirst-countable, as explained by the first countability axiom 67 above. Any subset F X is closed if and only if be any subset of the topological space X with the property that for any convergent sequence of elements a n F has its limit lim n a n F lying in the set F. Definition 24 (Clopen set). A subset A X of a topological space X is called clopen if and only if the set A is both open and closed 10 Problem 3.3. For an arbitrary topological space, the family T of all open sets is postulated have the properties (O1)(O2)(O3). Which properties follow for the family F T of all clopen sets of any topological space. 27

28 Answer. If the sets A and B are clopen, then the sets X \ A, A B, A B, A \ B are clopen. Definition 25 (Hausdorff space). A topological space for which any two distinct points have disjoint open neighborhoods is called a Hausdorff topological space. 10 Problem 3.4. Prove that every metric space is a Hausdorff space. Here is a list of further separation properties: (T0) For any point b, the derived set {b} is closed. (T1) Any one-point set is closed. (T2) Any two distinct points have disjoint open neighborhoods. Hausdorff property. (T2) is called the (T3) Any one-point set is closed. For any closed set F and point b / F exist two disjoint open sets U V = such that F U and b V. (T4) Any one-point set is closed. For any two disjoint closed sets F G = exist two disjoint open sets U V = such that F U and G V. 10 Problem 3.5. Convince yourself that any topological space is (T1) if and only if for any any point b, the derived set {b} is empty. Hence any (T1)-space is a (T0) space, too. 10 Problem 3.6. Convince yourself that any topological space is (T1) if and only if for any two distinct points a 1 a 2 there exists a neighborhood U of a 1 such that a 2 / U. Answer. Assume the (T1) property holds. Take any two distinct points a 1 a 2. The one-point set {a 2 } is closed, and the complement set X \ {a 2 } is open. Hence there exists a neighborhood U of a 1 such that U X \ {a 2 } and hence a 2 / U. The converse is as easy to check. Definition 26. Any set A is called finite if and only if either it is empty, or there exists a natural number n N and a bijection from the set A to the interval [1, 2,..., n]. 10 Problem 3.7. Prove that for a (T1) topological space, every finite set is closed. Proof. Given is any finite set A. Either A =, or there exists a natural number k such that A = {a 1, a 2,..., a k }. We prove by induction on k that such a set is closed. Induction start for k = 1: By the (T1)-property, any one-point set {a 1 } is closed. 28

29 Induction step k k + 1 : Let A = {a 1, a 2,..., a k, a k+1 }. Clearly A = {a 1, a 2,..., a k } {a k+1 } By the induction assumption, the set {a 1, a 2,..., a k } is closed. By the argument from the induction start, the one-point set {a k+1 } is closed. By axiom (O2), the intersection of any two open sets is open. Hence the union any two closed sets is closed. Hence we see that the set A is closed. Definition 27 (discrete topology). Let X be any set. For the discrete topology, all subsets of X are open. In other words T = P(X). 10 Problem 3.8. Let X be any set. Find a metric, which generates the discrete topology. Answer. We put (3.4) d(x, y) = { 0 for all x = y X ; 1 for all x y X. This function defines a metric, and generates the discrete topology. 10 Problem 3.9. Let X be an arbitrary set. Let T be the collection of all subsets O X either O =, or the complement X \ O is finite. (a) Describe the collection F consisting of the complements of the sets in T. (b) Check that the collection F satisfies the properties postulated for the closed sets. Hence we conclude that (X, T ) is a topological space. (c) Assume that X is finite. Convince yourself that (X, T ) is the discrete topology. Answer. (a) The collection F consists of all subsets A X either A = X, or the set A is finite. (b) The collection F contains the sets and X. If A F and B F, their union is either a union of two finite sets, and hence finite, or the union is X. In both cases A B F. Let A α be any collection of closed sets. There are two cases: (i) in case that A α = X for all α I, their intersection is A β = X which is closed. (ii) in case that there exists α I for which A α X, this A α is a finite set. Hence the intersection is A β A α is a finite set and hence closed. 29

30 (c) Assume that X is finite. The empty set and all one-point sets {x} with arbitrary x X are closed. Since the union of any finite number of closed sets is closed, and X is assumed to be finite, we conclude that all subsets of X are closed. Hence all subsets of X are open, and (X, T ) is a discrete topological space. 10 Problem Let X be an infinite set. Let T be the collection of all subsets O X either O =, or the complement X \ O is finite. We have already shown in problem 4.3 that these are the open sets for some topology. Check that this topology has the (T1) separation property, but is not Hausdorff. Answer. We have already seen in problem 4.3 that the finite sets, and hence the onepoint sets are closed. Thus the (T1) separation property holds. To check that the space is not Hausdorff, take any two distinct points a 1 a 2 and let U a 1 and V a 2 be any open neighborhoods of them. Their complements are closed and not equal to the space X. Hence the sets X \ U and X \ V are both finite. This implies their union (X \ U) (X \ V ) to be finite. Hence the complement X \ [(X \ U) (X \ V )] = U V is infinite and hence cannot be empty. Thus we have checked that any two neighborhoods of two distinct points intersect, contradicting the Hausdorff property. 10 Problem Convince yourself that property (T3) holds if and only if any one-point set is closed and each open neighborhood V and each point a V contains a closed neighborhood of point a. Definition 28. A sequence of elements a n of a topological space (X, T ) is convergent to a if and only if for all neighborhoods V N a, there exists a natural number K such that n > K a n V. 10 Problem Let N = N { } be the set of natural numbers, to which one element, called, is adjoint. Let J be the collection of all subsets O X for which holds O K [n > K n O] In other words, either / O, or O and there exists some natural number K for which {n N : n > K} O. (a) Describe the collection F consisting of the complements of the sets in T. (b) Check that the collection F satisfies the properties postulated for the closed sets. Hence we conclude once more that (N, J ) is a topological space. Answer. (a) The collection F consists all subsets A N for which holds / A K [n > K n / A] In other words, either A, or / A and A {n N : n K} for some natural K. The latter are the finite subsets of the naturals N. 30

31 (b) F and N F. Assume A, B F. There are two cases: (i) In case both sets are finite sets of naturals, there union A B is a finite set of naturals and hence A B F. (ii) In the case that A or A, we see that A B. Hence again A B F holds. In both cases, we conclude A B F. Let A α F be any collection of closed sets. There are two cases: (i) in case that A α = N for all α I, their intersection is A β = N which is closed. (ii) In case that there exists α I for which A α N, this set is a finite set of naturals. Hence the intersection is again a finite set of naturals, and hence member of the collection F. Definition 29 (Weaker and stronger topology). Let (X, T ) and (X, J ) be two topologies on the same set. The topology (X, T ) is called weaker (or coarser) than (X, J ) iff T J. Thus the weaker or coarser topology (X, T ) has less open (and less closed) sets. The stronger or finer topology (X, J ) has more open (and more closed) sets. Clearly, the topology were the only open sets are the empty set and the set X is the weakest topology. On the other hand, the discrete topology is the strongest topology. The identity map id : (X, J ) (X, T ) is continuous iff the topology (X, J ) is stronger than (X, T ). Thus strong convergence implies weak convergence, but the converse is not true. On the real line X = R is defined an unusual topology. We say that a set O R is "open" 6 if either O = or O = X, or there exists a real number a such that either O = [a, ) or O = (a, ). 10 Problem Convince yourself that we have indeed defined a topology. Answer. We check the postulates for the family T of the open subsets of the space X: (O1) T, X T are valid for the example; (O2) We check whether U, V T U V T. Clearly the postulate holds if one of the sets U or V is either or R. In the other cases U = (a, ) or U = [a, ) and V = (b, ) or V = [b, ), this implies either U V = (max[a, b], ) or U V = [max[a, b], ); which are both "open". 6 I use this typesetting to distinguish the new topology here defined, from the usual topology. 31

32 (O3) We check whether U α I for all α I implies α I U α I. Clearly the postulate holds if one of the sets U α equals R. Too in the case that all sets U α are empty. Otherwise, we may cancel the empty sets from the collection. Thus U α = (a α, ) or U α = [a α, ), and the union is either (m, ) or [m, ) with m = inf{a α : α I} [, ). In each case the union α I U α I is "open". 10 Problem Continuing the last problem, convince yourself that (i) For any a R, the set [a, ) is a neighborhood; (ii) a basis B a for the neighborhoods at a is given by just this single neighborhood; (iii) the Hausdorff separation property does not hold; (iv) for any "open" set O = {[x, x + 1] : x O}. Answer. (i) For any a R, the set [a, ) is a neighborhood since this is an "open" set containing a. (ii) A basis of neighborhoods at a is a collection B a of neighborhoods of a such that (B3) For all neighborhoods V a there exists B B a such that a B V. All other neighborhoods V a contain a set (b, ) for some b < a. Hence [a, ) (b, ) V. and postulate (B3) holds for the collection containing just the single neighborhood [a, ). (iii) Indeed for any two distinct a b and all neighborhoods a U and b V, the intersection U V [max[a, b], ) is not empty. Hence the Hausdorff separation property does not hold. (iv) For any "open" set x O [x, ) O [x, x + 1] O. Hence O = {[x, x + 1] : x O}. 32

33 4 Limit Points, Accumulation Points Definition 30 (Limit points, derived set). Let (X, d) be a topological or metric space and F X be any subset. The set of limit points of F is denoted by F and called the derived set. A point b is called a limit point of F if and only if every neighborhood of b contains a point of the set F different from b. Formulas defining b F and the negation b / F : b F V N b c [c V F and c b] b / F V N b [V F = or V F = {b}] In words once more: A point b is a limit point of F if and only if every neighborhood of b contains a point c F different from b. A point b is not a limit point of F if and only if there exists a neighborhood V of b containing no further point of F than possibly b. Definition 31 (Isolated point). Let (X, d) be a topological or metric space and F X be any subset. A point b F is called an isolated point if and only if there exists a neighborhood V of b which does not contain any second point of F. Let us denote F iso the set of its isolated points. Clearly F iso F = holds always. Here is definition 31 as a logical formula: b F iso V N b [V F = {b}] Proposition 11. A subset F of a topological or metric space is closed if and only if F F. Proof. F is closed X \ F is open b[b / F V N b [V X \ F ]] b[b / F V N b [V F = ]] b[b / F V N b [V F {b}]] b[b / F b / F ] F F Proposition 12. In a (T1) topological space and hence in any Hausdorff space, the derived set F of any set F X satisfies F F. Hence the derived set F is closed. 33

34 Proof of Proposition (12). Let F X be for any set. To show that F F holds, we assume that there exists a point b F but b / F and derive a contradiction. As shown in the lemma 10 below, there exists a point c b such that (4.2) U N c [b U] In other words, c {b}. On the other hand, the (T1) property tells that any one-point {b} is closed and hence {b} =. Because of this contradiction F F holds. By proposition 11 the set F is closed. Lemma 10. Let X be any topological space, and F X be any subset. Assume a point b F \ F exists. There exists a neighborhood V N b such that (4.1) V F = {b} In other words, the point b is an isolated point of the set F. Moreover, there exists a point c b such that (4.2) U N c [b U] and hence c {b}. Hence the derived set {b} is nonempty. Proof of the lemma 10. Let F X be for any set, and assume b F but b / F. Since b / F, there exists an open neighborhood V N b for which (4.3) V F = or V F = {b} Since b F, all neighborhoods W N b contain a point y b such that y W F : (4.4) W N b y [y W F and y b] Especially, we may put W := V into equation (4.4), and claim existence of a point c b such that c V F. Since V is open, the set V is a neighborhood of c. For all neighborhoods U of c, the intersection U V is a neighborhood of c, too. Now c F implies (4.5) U N c z [z c and z U V F ] We may choose U := V and conclude that in equation (4.3), only the second alternative is possible. Hence (4.1) V F = {b} Because of (4.1), we see that z claimed to exist in (4.5) can only be z = b. Moreover c b and hence (4.5) implies the claim (4.2) U N c [b U] and hence c {b}, as to be shown. 34

35 Proposition 13. In a (T0) topological space, the derived set F of any set F X satisfies F F. Hence the derived set F is closed. Corollary 4. The derived set F is closed for all subsets F X of a topological space if and only if the derived set is closed for all one-point sets {b}. Lemma 11. Let X be any topological space, and F X be any subset. Assume a point b F \ F exists. Let V N b be any open neighborhood of b such that (4.1) V F = {b} In lemma 10, such a neighborhood was shown to exist. For any such neighborhood, there exists a point c V \ {b} such that (4.6) U N c [b U] and hence c V {b}. and moreover holds (4.7) (4.8) V F {b} F c V F = V {b} Proof of the lemma 11. Since b F, all neighborhoods W N b contain a point y b such that y W F : (4.4) W N b y [y W F and y b] Let V N b be any open neighborhood of b such that (4.1) V F = {b} Especially, we may put W := V into equation (4.4), and claim existence of a point c b such that c V F. Since V is open, the set V is a neighborhood of c. For all neighborhoods U of c, the intersection U V is a neighborhood of c, too. Now c F implies (4.9) U N c z [z c and z U V F ] Because of (4.1), we see that z claimed to exist in (4.9) can only be z = b. Moreover c b and (4.6) U N c [b U] and hence c V {b}. To check the inclusions (4.7) take any y V F \ {b}. There hold U N y z [z y and z U F ] U N y z [z y and z U V F ] U N y z [z y and z U {b}] U N y [b U] y V {b} 35

36 We have obtained the inclusion V F \ {b} V {b} We remark that b / {b} since X is an open neighborhood of b but does not contain a second point x b for which x X {b}. Thus one gets the first inclusion of (4.7). To get the second one is obtained from {b} F since A B A B holds universally. Now V F = V {b} is immediate and (4.8) holds, too. Proof of Proposition (13). Let F X be for any set. To show that F F holds, we assume that there exists a point b F but b / F and derive a contradiction. By Lemma 10, there exists a neighborhood V N b such that V F = {b}. We remark that b / {b} since X is an open neighborhood of b but does not contain a second point x b for which x X {b}. By the (T0) property, the set {b} is closed. Hence V 1 := V \ {b} is an open neighborhood of b and V 1 F = {b}. We use lemma 11 for the neighborhood V 1. As shown in the lemma, there exists a point c b such that c V 1 {b}. On the other hand V 1 {b} = (V \ {b} ) {b} = Because of this contradiction F F holds. By proposition 11 the set F is closed. Proposition 14 (The exaggerating proposition). Let X be any topological space, and F X be any subset. For any point b F \ F holds b {b} \ {b}. Corollary 5. Assume that b / {b} holds for all points b X of the topological space X. Then the derived set F is closed for all subsets F X. Proof. Let b F \ F. Assume towards a contradiction that b F \ F but b / {b} \ {b}. By Lemma 10, there exists a neighborhood V N b such that V F = {b}. We remark that b / {b} since X is an open neighborhood of b but does not contain a second point x b for which x X {b}. Hence b / {b} {b}, and thus V 1 := V \ {b} = V \ ({b} {b} ) is an open neighborhood of b, moreover V 1 F = {b}. We use lemma 11 for the neighborhood V 1. The equation (4.8) yields correspondingly V 1 F = V 1 {b} = V \ ({b} {b} ) {b} = F X \ V 1 F F = F X \ V 1 V 1 F = Now b V 1 and b F cannot both hold, a contradiction arises. Thus b {b} \ {b} is confirmed. 36

37 10 Problem 4.1. Prove that for a (T1) topological space, every finite set consists only of isolated points. Proof. Given is any finite set A. Either A =, or there exists a natural number k such that A = {a 1, a 2,..., a k }. We prove by induction on k that such a set consists only of isolated points. Induction start for k = 1: Any one-point set {a 1 } consists only of isolated points. Indeed, the space X is an open neighborhood of a 1 and contains no other points of the set {a 1 }. Induction step k k + 1 : Let A = {a 1, a 2,..., a k, a k+1 } = {a 1, a 2,..., a k } {a k+1 }. By the induction assumption, the subset {a 1, a 2,..., a k } consists only of isolated points. In other words, for i = 1... k there exist open neighborhoods W i a i such that W i {a 1, a 2,..., a k } = {a i } By the (T1)-property their exist open neighborhoods V i a i such that a k+1 / V i. The intersection U i := W i V i is an open neighborhood of a i such that U i {a 1, a 2,..., a k, a k+1 } = {a i } By the (T1)-property the point a k+1 has neighborhoods W 1, W 2,..., W k such that a 1 / W 1, a 2 / W 2,..., a k / W k. Their intersection U k+1 := W 1 W 2 W k is an open neighborhood of a k+1 disjoint from the set {a 1, a 2,..., a k } The neighborhoods U 1, U 2,..., U k, U k+1 witness that the set A consists only of isolated points. Definition 32 (Accumulation point). Let (X, d) be a topological or metric space and A X be any subset. A point a is called an accumulation point of set A if and only if every neighborhood of a contains infinitely many points from the set A. Recall that a point b is called a limit point of set A if and only if every neighborhood of b contains a point from the set A different from b. Proposition 15. Let X be a topological or metric space and A X be any subset. Clearly every accumulation point is a limit point of set A. For a Hausdorff space (see definition 25 below) every limit point is an accumulation point. Moreover, assuming the axiom of choice, the following statement is valid: For any limit point a of any set A and any neighborhood N N a, there exists a sequence k a k of points a k A N such that a n a m for all natural n m; a n a for all natural n; for a metric space, additionally lim k a k = a. 37

38 The proof of proposition 15 needs the axiom of dependent choices DC*, which is indeed a bid stronger than the principle of countable choices AC. The formulation DC* is more convenient in analysis. It is only seemingly stronger than a more common axiom DC, not stated here. Some information is contained in my notes on discrete mathematics, and more completely in the book [5] Notes of Set Theory by Y. Moschovakis. Definition 33 (The set of strings (or words) from an alphabet). For any set A, which one may think of as being an alphabet, the set A is defined to be the set of strings or words from this alphabet. The set A consists of the empty word A, and all words w with n letters, where n N is any natural number. The latter are the mappings w : [1... n] A, and may denoted by (w(1),..., w(n)), too. Definition 34 (Axiom of Dependent Choices, DC* ). Let A be any nonempty set, and P A A a relation between the strings from A and the set A, which has the property to assign to each string in A at least one member of A: (4.10) A and u A v A [u P v] Choose any member a A such that P a holds. Then there exists a function f : N A such that f(1) = a and 7 (4.11) P f(1) and (f(1),..., f(n)) P f(n + 1) for all n N 0. Proof. Let point a be a limit point of set A. Take any neighborhood N N a, for example N := X. Since a is a limit point, the intersection N A contains a point a 1 a. By the Hausdorff property there exist disjoint neighborhoods V N a1 and W N a. Put A 1 := V N N a1 and V 1 := W N N a. Inductively, we construct for any natural k: distinct points a 1, a 2,..., a k a in A; disjoint neighborhoods A 1, A 2,..., A k and V k for the points a 1, a 2,..., a k and a such that A i N for i = 1, 2,..., k and V k V k 1. Here is the induction step k k + 1 : Since a is a limit point, the neighborhood V k N a contains a point a k+1 V k A different from a. By the Hausdorff property there exist disjoint neighborhoods: V, W [V N ak+1 and W N a and V W = ] Put A k+1 := V V k N ak+1 to get a neighborhood of a k+1, which is disjoint from the neighborhoods A 1, A 2,..., A k. Put V k+1 := W V k N a. In the case of a metric space, we put V k+1 := W V k B(a, 2 k 1 ) N a. Thus we get a small disjoint neighborhood of a. 7 (f(1),..., f(n)) P f(n + 1) for n = 0 just means P f(1). 38

39 By the principle of induction, the above claims hold for all natural numbers k. By the axiom 34 of dependent choices DC, there exists the sequence k a k. Hence there exist infinitely many points in any neighborhood of the accumulation point. Indeed, for any finite number of such points, the induction step yields an additional point. In the case of a metric space, using the Archimedean axiom 15, we get additionally lim k a k = a. Proposition 16. The following statements are equivalent for any subset F of any metric space (X, d). (i) b F ; (ii) there exists a sequence of elements a n F such that a n b for all natural n and lim n a n = b; (iii) there exists a sequence of elements c n F such that c n b for all natural n; the sequence k d(c k, b) is strictly decreasing; lim k c k = b; c n c m for all natural n m; (i) implies (ii). Take any point b F. By definition 30 of the derived set F, for all neighborhood V of b, there exists c b such that c V F. We choose the neighborhoods V := B(b; 1 ) for any n N. Hence n [ ] n N a n an b and d(a n, b) < 1 n By the axiom 17 of countable choice ACN, there exists the sequence n a n. We now check that this sequence has the limit b. Let any number δ > 0 we given. By the Archimedean axiom 15, there exists N such that Nδ > 1. Hence n > N implies d(a n, b) < 1 < δ and finally n (4.12) lim n a n = b as to be shown. (ii) implies (iii). The sequence k c k is a subsequence c k := a nk of the sequence a n. The numbers n k and c k are defined recursively as follows. Let n 1 := 1 and c 1 := a 1. Assume that the numbers n 1, n 2,..., n k 1 have already been defined. Since a nk 1 b by assumption (ii), the distance d(a nk 1, b) > 0 is positive. By the convergence (4.12), there exists a natural number N such that n > N implies d(a n, b) < d(a nk 1, b). Let n k be the smallest natural number such that n k > n k 1 and d(a nk, b) < d(a nk 1, b). Let c k := a nk. Since d(c k, b) < d(c k 1, b) < d(c k 2, b) < < d(c 1, b) the number c k is different from the numbers c 1, c 2,..., c k 1. 39

40 (iii) implies (i). To check whether b F, take any neighborhood V N b. By definition of the standard topology for a metric space, there exists δ > 0 and an open ball such that B(b; δ) V. Because of the assumption lim k c k = b, there exists as natural number K such that finally k > K d(c k, b) < δ c k B(b; δ) c k V By definition 30 for the derived set, we conclude b F, as to be shown. Remark. The Archimedean axiom 15 is needed to show the implication (i) implies (ii). The axiom 17 of countable choice ACN, is needed to show the implication (i) implies (ii) and (ii) implies (iii). The axiom 34 of dependent choices DC*, is needed to show the implication (ii) implies (iii). Corollary 6. For any finite set F of a metric space (X, d), the derived set F = is empty. Definition 35 (discrete topology). Let X be any set. For the discrete topology, all subsets of X are open. In other words T = P(X). 10 Problem 4.2. Let X be any set. Find a metric, which generates the discrete topology. Answer. We put (4.13) d(x, y) = { 0 for all x = y X ; 1 for all x y X. This function defines a metric, and generates the discrete topology. 10 Problem 4.3. Let X be an arbitrary set. Let T be the collection of all subsets O X either O =, or the complement X \ O is finite. (a) Describe the collection F consisting of the complements of the sets in T. (b) Check that the collection F satisfies the properties postulated for the closed sets. Hence we conclude that (X, T ) is a topological space. (c) Assume that X is finite. Convince yourself that (X, T ) is the discrete topology. Answer. (a) The collection F consists of all subsets A X either A = X, or the set A is finite. (b) The collection F contains the sets and X. If A F and B F, their union is either a union of two finite sets, and hence finite, or the union is X. In both cases A B F. 40

41 Let A α be any collection of closed sets. There are two cases: (i) in case that A α = X for all α I, their intersection is A β = X which is closed. (ii) in case that there exists α I for which A α X, this A α is a finite set. Hence the intersection is A β A α is a finite set and hence closed. (c) Assume that X is finite. The empty set and all one-point sets {x} with arbitrary x X are closed. Since the union of any finite number of closed sets is closed, and X is assumed to be finite, we conclude that all subsets of X are closed. Hence all subsets of X are open, and (X, T ) is a discrete topological space. 10 Problem 4.4. Let X be an infinite set. Let T be the collection of all subsets O X either O =, or the complement X \ O is finite. We have already shown in problem 4.3 that these are the open sets for some topology. Check that this topology has the (T1) separation property, but is not Hausdorff. Answer. We have already seen in problem 4.3 that the finite sets, and hence the onepoint sets are closed. Thus the (T1) separation property holds. To check that the space is not Hausdorff, take any two distinct points a 1 a 2 and let U a 1 and V a 2 be any open neighborhoods of them. Their complements are closed and not equal to the space X. Hence the sets X \ U and X \ V are both finite. This implies their union (X \ U) (X \ V ) to be finite. Hence the complement X \ [(X \ U) (X \ V )] = U V is infinite and hence cannot be empty. Thus we have checked that any two neighborhoods of two distinct points intersect, contradicting the Hausdorff property. 10 Problem 4.5. Convince yourself that property (T3) holds if and only if any one-point set is closed and each open neighborhood V and each point a V contains a closed neighborhood of point a. Definition 36. A sequence of elements a n of a topological space (X, T ) is convergent to a if and only if for all neighborhoods V N a, there exists a natural number K such that n > K a n V. 10 Problem 4.6. Let N = N { } be the set of natural numbers, to which one element, called, is adjoint. Let J be the collection of all subsets O X for which holds O K [n > K n O] In other words, either / O, or O and there exists some natural number K for which {n N : n > K} O. (a) Describe the collection F consisting of the complements of the sets in T. 41

42 (b) Check that the collection F satisfies the properties postulated for the closed sets. Hence we conclude once more that (N, J ) is a topological space. Answer. (a) The collection F consists all subsets A N for which holds / A K [n > K n / A] In other words, either A, or / A and A {n N : n K} for some natural K. The latter are the finite subsets of the naturals N. (b) F and N F. Assume A, B F. There are two cases: (i) In case both sets are finite sets of naturals, there union A B is a finite set of naturals and hence A B F. (ii) In the case that A or A, we see that A B. Hence again A B F holds. In both cases, we conclude A B F. Let A α F be any collection of closed sets. There are two cases: (i) in case that A α = N for all α I, their intersection is A β = N which is closed. (ii) In case that there exists α I for which A α N, this set is a finite set of naturals. Hence the intersection is again a finite set of naturals, and hence member of the collection F. Proposition 17. Let N = N { } be the set of natural numbers, to which one element, called, is adjoint. Let J be the collection of all subsets O X for which holds O K [n > K n O] A sequence of elements a n of a topological space (X, T ) is convergent to a if and only if the mapping a : N X obtained by setting a(n) = a n for all natural n and a( ) = a is continuous. Proof. For each natural n, the one-element set {n} is open. Hence the mapping a is always continuous at the points n. The mapping a is continuous at the point if and only if for all open neighborhoods V a the preimage a 1 (V ) is open in N. This holds if and only if and hence if and only if lim n a n = a. V N a K N [n > K a n V ] 42

43 Definition 37 (Weaker and stronger topology). Let (X, T ) and (X, J ) be two topologies on the same set. The topology (X, T ) is called weaker (or coarser) than (X, J ) iff T J. Thus the weaker or coarser topology (X, T ) has less open (and less closed) sets. The stronger or finer topology (X, J ) has more open (and more closed) sets. Clearly, the topology were the only open sets are the empty set and the set X is the weakest topology. On the other hand, the discrete topology is the strongest topology. The identity map id : (X, J ) (X, T ) is continuous iff the topology (X, J ) is stronger than (X, T ). Thus strong convergence implies weak convergence, but the converse is not true. On the real line X = R is defined an unusual topology. We say that a set O R is "open" 8 if either O = or O = X, or there exists a real number a such that either O = [a, ) or O = (a, ). 10 Problem 4.7. Convince yourself that we have indeed defined a topology. Answer. We check the postulates for the family T of the open subsets of the space X: (O1) T, X T are valid for the example; (O2) We check whether U, V T U V T. Clearly the postulate holds if one of the sets U or V is either or R. In the other cases U = (a, ) or U = [a, ) and V = (b, ) or V = [b, ), this implies either U V = (max[a, b], ) or U V = [max[a, b], ); which are both "open". (O3) We check whether U α I for all α I implies α I U α I. Clearly the postulate holds if one of the sets U α equals R. Too in the case that all sets U α are empty. Otherwise, we may cancel the empty sets from the collection. Thus U α = (a α, ) or U α = [a α, ), and the union is either (m, ) or [m, ) with m = inf{a α : α I} [, ). In each case the union α I U α I is "open". 10 Problem 4.8. Continuing the last problem, convince yourself that (i) For any a R, the set [a, ) is a neighborhood; (ii) a basis B a for the neighborhoods at a is given by just this single neighborhood; (iii) the Hausdorff separation property does not hold; (iv) for any "open" set O = {[x, x + 1] : x O}. 8 I use this typesetting to distinguish the new topology here defined, from the usual topology. 43

44 Answer. (i) For any a R, the set [a, ) is a neighborhood since this is an "open" set containing a. (ii) A basis of neighborhoods at a is a collection B a of neighborhoods of a such that (B3) For all neighborhoods V a there exists B B a such that a B V. All other neighborhoods V a contain a set (b, ) for some b < a. Hence [a, ) (b, ) V. and postulate (B3) holds for the collection containing just the single neighborhood [a, ). (iii) Indeed for any two distinct a b and all neighborhoods a U and b V, the intersection U V [max[a, b], ) is not empty. Hence the Hausdorff separation property does not hold. (iv) For any "open" set x O [x, ) O [x, x + 1] O. Hence O = {[x, x + 1] : x O}. 44

45 5 Continuity Definition 38. Let (X, d) and (Y, d) be metric spaces. A function f : X Y is called continuous at point a X if and only if forall ɛ > 0 exists δ > 0 such that d(x, a) < δ d(f(x), f(a)) < ɛ A function f is called continuous if and only if it is continuous for all a X. Remark. It is instructive to use quantified logic. Note that a third forall x quantifier does appear. A function f : X Y is called continuous at point a X if and only if ɛ > 0 δ > 0 x X [d(x, a) < δ d(f(x), f(a)) < ɛ] That a function f is not continuous at point a means that the ɛ, δ property for continuity at some point a does not hold: ɛ > 0 δ > 0 x X [d(x, a) < δ and d(f(x), f(a)) ɛ] We see that a function f is not continuous at point a means existence a (special) ɛ > 0 such that for all δ > 0 exists a (bad) x for which d(x, a) < δ but yet d(f(x), f(a)) ɛ. 10 Problem 5.1. Prove directly from the ɛ, δ definition that the function y = x 2 is continuous at any a R. Answer. Given is ɛ > 0. We need to give an upper bound for δ, take for example δ 1. Hence x a < δ implies x a + 1 and f(x) f(a) = x 2 a 2 = x + a x a ( x + a ) x a (2 a + 1) x a Under the assumption x a < δ, we have to make this difference less than ɛ. achieve We may now choose x 2 a 2 (2 a + 1) x a < (2 a + 1) δ! ɛ [ ] ɛ δ = min 1, 2 a + 1 To 10 Problem 5.2. Let (X, d), (Y, d) and (Z, d) be any metric spaces and f : X Y and g : Y Z be any functions. Assume that the function f is continuous at the point a X, and the function g is continuous at the point f(a) Y. Give the reason why the composition function g f is continuous at the point a X. Answer. Given is any ɛ > 0. Since the function g is continuous at the point f(a), there exists a δ > 0 such that d(y, f(a)) < δ d(g(y), g(f(a))) < ɛ 45

46 Since the function f is continuous at the point a, there exists a η > 0 such that d(x, a) < η d(f(x), f(a)) < δ We may put y = f(x) and get from both implications together d(x, a) < η d(f(x), f(a)) < δ d(g(f(x)), g(f(a))) < ɛ Since d(g(f(x)), g(f(a))) = d(g f(x), g f(a)), we see that the composite function g f is continuous at the point a X. Recall definition 19 for neighborhood and definition 20 for a basis of neighborhoods. Proposition 18. Let (X, d) and (Y, d) be metric spaces, let f : X Y be any function and let a X be any point. Equivalent statements are: (i) the function f : X Y is continuous at point a; (ii) for all ɛ > 0 there exists δ > 0 such that for all x X, indeed d(x, a) < δ implies d(f(x), f(a)) < ɛ; (iii) for any nonempty open ball B(f(a); ɛ), there exists a nonempty open ball B(a; δ) such that f(b(a; δ)) B(f(a); ɛ); (iv) for any neighborhood M of the image point f(a), there exists a neighborhood N of point a such that f(n) M. 10 Problem 5.3. The set of all neighborhoods of point a is denoted by N a. Write down the statements (ii) and (iv) from proposition 18 in terms of symbolic logic. Answer. (ii) ɛ > 0 δ > 0 x X [d(x, a) < δ d(f(x), f(a)) < ɛ]; (iv) M N f(a) N N a [f(n) M]. (iv) implies (iii). We assume the neighborhood condition for continuity, and have to check the ɛ δ condition. Given is any ɛ > 0. Let M := B(f(a); ɛ). Since this is a neighborhood of image point f(a), by assumption (iv) there exists a neighborhood N of point a such that f(n) M. From the defining property of a neighborhood, there exists a nonempty ball B(a; δ) N. Clearly B(a; δ) implies δ > 0. Hence we have obtained the number δ > 0 for which as required. f(b(a; δ)) f(n) M = B(f(a); ɛ) 46

47 (iii) implies (iv). We assume the ɛ δ condition for continuity and have to check the neighborhood condition. Given is any neighborhood M of the image point f(a). From the defining property of a neighborhood, there exists a nonempty ball B(f(a); ɛ) M. Clearly B(f(a); ɛ) implies ɛ > 0. By the ɛ δ condition for continuity, there exists δ > 0 such that f(b(a; δ)) B(f(a); ɛ). Let N := B(a; δ). Since this is a neighborhood of point a, we have obtained neighborhood N of point a such that as required. f(n) = f(b(a; δ)) B(f(a); ɛ) M Definition 39 (Continuity at a point). Let (X, O) and (Y, U) be topological spaces and f : X Y be any mapping. The function f is called continuous at point a X if and only if for any neighborhood M of the image point f(a), there exists a neighborhood N of point a such that f(n) M. Proposition 19. Let (X, O) and (Y, U) be topological spaces and f : X Y be any mapping. Equivalent are: (1) the function f is continuous at all points a X; (2) for each open set O U, the preimage f 1 (O) O is open. Proof of (1) (2). Take any open set O U. To check whether the preimage f 1 (O) is open, take any point a f 1 (O). Since the set O is open and f(a) O, it is a neighborhood M := O of the image point f(a). Since continuity at the point a is assumed, there exists a neighborhood N of point a such that f(n) M. By the defining property of neighborhoods, there exists an open set such that a P a N. From the inclusions (5.1) {a} P a N f 1 (M) = f 1 (O) we see that the set f 1 (O) contains a neighboorhood of each of its points. By Lemma 5 we conclude that the set f 1 (O) is open. For convenience of the reader, the reasoning is included. We have to infer that f 1 (O) is open, which is done as follows. By the argument above, for all a f 1 (O), there exists an open set P a such that inclusion (5.1) holds. We take of these inclusions the union over all a f 1 (O) to obtain f 1 (O) = {a} P a f 1 (O) = f 1 (O) hence a f 1 (O) f 1 (O) = a f 1 (O) a f 1 (O) P a a f 1 (O) Thus the set f 1 (O) is a union of open sets, and hence by axiom (O3) it is open. 47

48 Proof of (2) (1). We assume the preimage f 1 (O) of any open set O U to be open. To check the neighborhood condition for continuity at any a X, take any neighborhood M of the image point f(a). By the defining property of neighborhoods, there exists an open set U such that f(a) U M. Hence a f 1 (U) f 1 (M). By the assumption (2), the set f 1 (U) is open, and hence it is a neighborhood of point a. We put N := f 1 (U) and conclude f(n) = f(f 1 (U)) = U f(x) U M Thus we have checked that for any neighborhood M of the image point f(a), there exists a neighborhood N of point a such that f(n) M, which is the defining condition for continuity at point a. 10 Problem 5.4. Let (X, d) and (Y, d) be metric spaces. We prove that a function f : X Y is continuous if and only if the preimage f 1 (V ) of any open set V Y is open. Complete the proof below. If the function f : X Y is continuous, then the preimage of any open set is open. Given is any open set V Y. Let a f 1 (V ). Hence f(a) V. Since V is assumed to be open, it contains a neighborhood of a. Hence there exists an ɛ > 0 such that B(f(a); ɛ) V. Since the function f is continuous at a, there exists δ > 0 such that d(x, a) < δ d(f(x), f(a)) < ɛ. In other words, x B(a; δ) implies f(x) B(f(a); ɛ). Hence B(a; δ) f 1 (B(f(a); ɛ)) f 1 (V ) With the last paragraph, we have checked that f 1 (V ) is open, as to be shown. If the preimage f 1 (V ) of any open set V Y is open, the ɛ, δ property holds. Let a X and ɛ > 0 be given. The set B(f(a); ɛ) is open. Hence its preimage f 1 (B(f(a); ɛ)) is open. Hence there exists δ > 0 such that B(a; δ) f 1 (B(f(a); ɛ)) In other words, x B(a; δ) implies f(x) B(f(a); ɛ). Hence the ɛ δ requirement for continuity d(x, a) < δ d(f(x), f(a)) < ɛ holds. Lemma 12. Let (X, O) and (Y, U) be topological spaces and f : X Y be any mapping. Let the collections B a and B f(a) be two bases for the neighborhoods at points a and f(a). Equivalent are: (1) the function f is continuous at point a X; (2) for any neighborhood C B f(a), there exists a neighborhood B B a such that f(b) C. 48

49 Figure 1: About continuity. Remark. In the figure on page 49 I have illustrated lemma 12. The spaces X and Y are depicted as two rectangles, and get the usual topology. The arbitrary neighborhoods of a respectively f(a) are illustrated by patches around these points of irregular shapes. As an example for bases for the neighborhoods at points a and f(a) is chosen the set of balls around these points: B a = {B(a, δ) X : δ > 0} B f(a) = {B(f(a), ɛ) Y : ɛ > 0} Thus item (1) means the continuity using arbitrary neighborhoods, whereas item (2) 49

50 means the ɛ, δ definition of continuity. (1) M N f(a) N N a [f(n) M] (2) ɛ > 0 δ > 0 [d(a, x) < δ d(f(a), f(x)) < ɛ] 10 Problem 5.5. Prove that item (1) implies item (2). Explain the figure illustrating that item (1) implies item (2). Proof of (1) (2). We have assumed item (1), the definition 39 for continuity of the function f at point a X using neighborhoods. Given is any neighborhood C B f(a). In the upper row of the figure on page 49, it is depicted as a blue circle. By item (1) there exists a neighborhood N of point a such that f(n) C. By the assumption (B3) about a basis, there exists a neighborhood B B a from the basis such that B N. In the upper row of the figure on page 49, it is depicted as a red circle. We have obtained the inclusion f(b) f(n) C as required. 10 Problem 5.6. Prove that item (2) implies item (1). Proof of (2) (1). We have assumed item (2), and have to check item (1). Given is any neighborhood M of the image point f(a). In the lower row of the figure on page 49, it is depicted as a blue polygon. By the assumption (B3) about a basis, there exists a neighborhood C B f(a) from the basis such that C M. By the assumption (2), there exists a neighborhood B B a such that f(b) C. Let N := B. In the lower row of the figure on page 49, circle B it is depicted in red. Since this is a neighborhood of point a, we have obtained neighborhood N of point a such that as required. f(n) = f(b) C M Proposition 20. Let (X, O) and (Y, U) be topological spaces and f : X Y be any mapping. Moreover, we assume at point a X, the topology of X allows for a countable basis B a N a. Too, we use the axiom of choice. Equivalent are: (i) the function f is continuous at point a; (ii) for any convergent sequence of elements x n X with limit lim n x n = a, the mapped sequence f(x n ) is convergent, too, and lim f(x n) = f( lim x n ) = f(a) n n 50

51 (i) implies (ii), which is the easier part. We have assume the function f to be continuous at point a. In terms of symbolic logic, this statement can be written as M N f(a) N N a [f(n) M] Given is any convergent sequence of elements x n X with limit lim n x n = a. In terms of symbolic logic, this statement can be written as V N a K N [n > K x n V ] We have to check whether the mapped sequence f(x n ) is convergent to f(a). To this end, let any neighborhood M N f(a) of point f(a) be given. By the continuity of the function, there exists a neighborhood V N a such that f(v ) M. Because of lim n x n = a, there exists a natural number K such that n > K x n V. Hence n > K x n V f(x n ) f(v ) M The entire paragraph confirms that as to be shown. lim f(x n) = f(a) n Remark. We see that this part requires neither a countable basis B a N a, nor the axiom of choice. not (i) implies not (ii). We assume the function f not to be continuous at point a. Thus the condition from definition 39 is violated. In terms of symbolic logic, this statement can be written in the equivalent forms: [ M N f(a) N N a [N f 1 (M)] ] M N f(a) [ N N a [N f 1 (M)] ] M N f(a) N N a [ N f 1 (M) ] M N f(a) N N a x [ x N x f 1 (M) ] M N f(a) N N a x [ x N x f 1 (M) ] M N f(a) N N a x [ x N and x / f 1 (M) ] M N f(a) N N a x [x N and f(x) / M] In other words, there exists a (critical) neighborhood M of point f(a) such that all neighborhoods N of point a contain a point x N such that f(x) / M. 9 We have assumed the existence of a countable basis of neighborhoods at a. We may call them V n with n N, and may assume that the inclusions (5.2) n > k V n V k 9 It may perhaps be easier for you to convince yourself of this statement immediately. 51

52 hold for all n, k N. We now put V := V n and get existence of a neighborhood M at point f(a) such that n N x [x V n and f(x) / M] By the axiom of (countable) choice, there exists the sequence n x n. Since we have assumed that the set {V n : n N} is a basis of neighborhoods at a, we can deduce (5.3) lim n x n = a Indeed definition 20 tells the assumption (B3) to hold for a basis of neighborhoods. Thus for all neighborhoods V N a, there exist K N and V K B a such that a V K V. Hence by the inclusions (5.2), we see that n > K x n V n V K V confirming the convergence (5.3). On the other hand, the sequence n f(x n ) of images does not converge to f(a) since f(x n ) / M for all n N. Thus there exists a convergent sequence x n a for which f(x n ) is not convergent to f(a). (Whether the latter sequence converges to another limit, or not at all, is not important at this point.) Thus the negation of item (ii) is confirmed. 10 Problem 5.7. Let (X, d) and (Y, d) be metric spaces. Suppose that the mapping f : X Y has the property that for any convergent sequence of elements x n X, the mapped sequence f(x n ) is convergent, too, and Prove that f is continuous. lim f(x n) = f( lim x n ) n n Towards a contradiction, assume that the ɛ, δ property for continuity at some point a does not hold. Why would there exist a convergent sequence x n a for which f(x n ) is not convergent to f(a)? Answer. Towards a contradiction, we assume that the ɛ, δ property for continuity at some point a does not hold. Hence ɛ > 0 δ > 0 x [d(x, a) < δ and d(f(x), f(a)) ɛ] We may put δ = Hence there exists ɛ > 0 such that n n N x [d(x, a) < 1 n and d(f(x), f(a)) ɛ] By the axiom of (countable) choice, there exists the sequence n x n. We now check that this sequence has the limit a. Let any number δ > 0 we given. By the Archimedean 10 Since he is standing in the desert of Arizona, the purist logician here does write simply x, instead of the more intuitive x n. 52

53 axiom, there exists N such that Nδ > 1. Hence n > N implies d(x n, a) < 1 < δ, and n finally lim n x n = a. Since d(f(x n ), f(a)) ɛ for all natural n, we see that the sequence f(x n ) cannot converge to f(a). Thus there exists a convergent sequence x n a for which f(x n ) is not convergent to f(a). (Whether this sequence converges to another limit, or not at all, is not important at this point.) Proposition 21. Let N = N { } be the set of natural numbers, to which one element, called, is adjoint. Let J be the collection of all subsets O X for which holds O K [n > K n O] A sequence of elements a n of a topological space (X, T ) is convergent to a if and only if the mapping a : N X obtained by setting a(n) = a n for all natural n and a( ) = a is continuous. Proof. For each natural n, the one-element set {n} is open. Hence the mapping a is always continuous at the points n. The mapping a is continuous at the point if and only if for all open neighborhoods V a the preimage a 1 (V ) is open in N. This holds if and only if and hence if and only if lim n a n = a. V N a K N [n > K a n V ] 10 Problem 5.8. Let X = N { } be the set N of natural numbers, and an extra element called. Let U be the collection of all subsets O X for which (i) either / O; (ii) or O and there exists some natural m such that n m n O. (a) Check that this collection O satisfies the properties postulated for the open sets. (b) Given is any function f : X R such that f( ) = lim n f(n), where the limit is assumed to exist. Convince yourself that f is continuous. Check that the preimage f 1 (U) of any open set U R is open. (c) Convince yourself that lim n n = in the space X. (d) We assume that the function f : X R is continuous. Convince yourself that f( ) = lim n f(n), and the limit exists. Answer. (a) The collection O satisfies the properties postulated for the open sets. 53

54 Indeed, the empty set is open by item (i). The set X is open by item (ii), using m = 1. Let U and V be two open sets. If both satisfy item (ii), their intersection U V satisfies item (ii) with m = max[m U, m V ]. If either U or V satisfies item (i), the intersection U V satisfies item (i). Let U α be any collection of open sets. If all U α satisfy item (i), their union Uα satisfies item (i). If there exists α for which U α satisfies item (ii), their union U α satisfies item (ii) with m = m(u α ). (b) Given any function f : X R such that f( ) = lim n f(n) where and the limit is assumed to exist. Given is any open set U R. We distinguish two cases: either (i) f( ) / U; or (ii) f( ) U. In the case (i), the assumption means that / f 1 (U).We see that the set f 1 (U) satisfies item (i). In the case (ii), there exists ɛ > 0 such that (f( ) ɛ, f( ) + ɛ) U. By the definition of convergence, there exists a natural N such that n > N f(n) f( ) < ɛ n N + 1 f(n) (f( ) ɛ, f( ) + ɛ) U n f 1 (U) We see that the set f 1 (U) satisfies item (ii) with m := N + 1. In either case f 1 (U) turns out to be open. Since the preimage of any open set is open, the function is continuous. (c) Convince yourself that lim n n = in the space X. For any open neighborhood V, by item (ii) there exists some natural m such that n m n V. This statement is rewritten as lim n n =. (d) We assume that the function f : X R is continuous at the point. Since lim n n =, the continuity of f implies lim n f(n) = f( ), and existence of the limit. 10 Problem 5.9. Let X and Y be topological spaces and f : X Y a continuous function. In the product X Y, the topology is defined the usual way. Moreover, we assume the space Y has the Hausdorff property 25. Prove that the graph G f := {(x, y) : x X and y = f(x)} is closed. An easy approach is to check that the complement of G f is open. Proof. We check that the complement X Y \ G f := {(a, b) : a X and b f(a)} 54

55 is open. Take any point (a, b) X Y \ G f. Since b f(a), by the Hausdorff property (T2) there exist disjoint neighborhoods and U b and V f(a). Since the function f is continuous at point a, there exists an open neighborhood W a such that f(w ) V. The product W U is an open neighborhood of the point (a, b) which is entirely contained in the complement X Y \ G f : Take any (x, y) W U, hence x W and y U, f(w ) V hence f(x) V U V = hence y / V hence y f(x), hence (x, y) / G f Thus we have checked that W U X Y \ G f. Hence X Y \ G f is open and G f is closed. Proposition 22 (The composition of two continuous functions is continuous). Let f : X Y and g : Y Z be continuous mappings between the topological spaces (X, O), (Y, U) and (Z, V). Then their composition g f : X Z is continuous. Proof. We may use item (2) from proposition 27 to check continuity. Let O V be any open set O Z. Since the function g is assumed to be continuous, the preimage g 1 (O) U is open in Y. Since the function f is assumed to be continuous, the preimage f 1 (g 1 (O)) U is open in X. Hence each open set O Z has an open preimage (g f) 1 (O) = f 1 (g 1 (O)) U By item (2) from proposition 27, the composite function g f is continuous. 10 Problem Let (X, d) and (Y, d) be metric spaces and f : X Y a continuous function. Use sequences to show more easily that f(a) f(a) holds for all subsets A X. Answer. Take any subset A X and y f(a). Hence there exists x such that y = f(x) and x A. Hence there exists a convergent sequence such that lim a n = x and a n A n for all natural n. By the continuity of the function f, we conclude lim f(a n) = f(x) = y n Since f(a n ) f(a) for all natural n, we got a convergent sequence in the set f(a) with limit y. Hence y f(a). Thus we have indeed confirmed that f(a) f(a) holds. 10 Problem Let X = R, Y = R and f(x) = 1 1+x 2. This is indeed a continuous function between the given metric spaces. Find an example where f(a) f(a). 55

56 Answer. We choose the set A = [0, ) which is closed. Obviously f(a) = (0, 1] since the function f does not assume the value zero. Hence (0, 1] = f(a) = f(a) but f(a) = [0, 1] 10 Problem 5.12 (Another example where f(a) f(a)). Let X = R, Y = R 2 and f(x) = (cos x, sin x). This is indeed a continuous function between the given metric spaces. We choose the set A := {2πn + 1 n : n N} (i) Convince yourself that the set A is closed. (ii) Calculate the image set f(a). (iii) Convince yourself that f(a) is not closed; (iv) and calculate the closure. Which point is in the difference f(a) \ f(a). Answer. (i) Indeed all points of the set A are isolated. Hence any convergent sequence of points in A is eventually constant, and has its limit in A. Therefore A is closed. (ii) f(a) = {(cos 1 n, sin 1 n ) : n N}. 1 (iii) We know that lim n = 0. Hence by continuity and easy check n ( lim cos 1 n n, sin 1 ) = (1, 0) n ( cos 1 n, sin 1 ) (1, 0) for all natural n. n (iv) The closure is f(a) = {(cos 1 n, sin 1 ) : n N} {(1, 0)} n The point (1, 0) f(a) \ f(a) is in the closure but not in f(a). 56

57 6 Products 10 Problem 6.1. Given are finitely many metric spaces (X i, d i ) for i = 1... N. (i) The projections are the mappings p i : X X i with p i (x) = x i. Convince yourself that these are continuous mappings. (ii) Given is any continuous function f : Y X, where Y is any metric space. Convince yourself that all compositions p i f are continuous. = Y Y f X = 1 i N X i p i f p i X i (iii) Suppose that the compositions p i f for 1 i N are continuous. Check by the ɛ, δ-property that f is continuous at any point b Y. Answer. (i) d i (p i (x), p i (x )) = d i (x i, x i) d(x, x ) for any x = (x i ) and y = (y i ). Hence p i : X X i is continuous. (ii) The compositions p i f are continuous since p i and f are continuous. (iii) Given is any point b Y and ɛ > 0. By assumption, the compositions p i f for 1 i N are continuous. Hence there exist δ i > 0 such that d(y, b) < δ i d i (p i (f(y)), p i (f(b))) < ɛ. Hence d(y, b) < min 1 i N δ i d(y, b) < δ i d i (p i (f(y)), p i (f(b))) < ɛ for 1 i N d(y, b) < δ max 1 i N d i(p i (f(y)), p i (f(b))) = d(f(y), f(b)) < ɛ were we have put δ := min 1 i N δ i > 0 which is positive. Thus, given any point b Y and ɛ > 0, we have δ > 0 such that d(y, b) < δ d(f(y), f(b)) < ɛ. Hence f is continuous, as to shown. 10 Problem 6.2. Suppose O 1, U 1 X 1 and O 2, U 2 X 2. Convince yourself that (i) (O 1 O 2 ) (U 1 U 2 ) = (O 1 U 1 ) (O 2 U 2 ) (ii) X 1 X 2 \ O 1 O 2 = [(X 1 \ O 1 ) X 2 ] [X 1 (X 2 \ O 2 )] You may use either pictures with some explanation, or logic. 57

58 Reason using logic. (i) Use (x 1, x 2 ) = (y 1, y 2 ) x 1 = y 1 and x 2 = y 2, in the third line. (ii) x (O 1 O 2 ) (U 1 U 2 ) x O 1 O 2 and x U 1 U 2 x = (x 1, x 2 ) and x 1 O 1 and x 2 O 2 and x = (y 1, y 2 ) and y 1 U 1 and y 2 U 2 x = (x 1, x 2 ) and x 1 O 1 and x 2 O 2 and x 1 U 1 and x 2 U 2 x = (x 1, x 2 ) and x 1 O 1 U 1 and x 2 O 2 U 2 x (O 1 U 1 ) (O 2 U 2 ) x X 1 X 2 \ O 1 O 2 x X 1 X 2 and x / O 1 O 2 x = (x 1, x 2 ) and x 1 X 1 and x 2 X 2 and (x 1, x 2 ) / O 1 O 2 x = (x 1, x 2 ) and x 1 X 1 and x 2 X 2 and [x 1 / O 1 or x 2 / O 2 ] x = (x 1, x 2 ) and ([x 1 X 1 and x 2 X 2 and x 1 / O 1 ] or [x 1 X 1 and x 2 X 2 and x 2 / O 2 ]) x = (x 1, x 2 ) and ([x 1 X 1 \ O 1 and x 2 X 2 ] or [x 1 X 1 and x 2 X 2 \ O 2 ]) x = (x 1, x 2 ) and [(x 1, x 2 ) (X 1 \ O 1 ) X 2 or (x 1, x 2 ) X 1 (X 2 \ O 2 )] x = (x 1, x 2 ) and (x 1, x 2 ) [(X 1 \ O 1 ) X 2 ] [X 1 (X 2 \ O 2 )] x (X 1 \ O 1 ) X 2 X 1 (X 2 \ O 2 ) Reason using figures. Let X 1 and X 2 be any topological spaces. The product topology for X := X 1 X 2 is now defined in the usual way: For any point (x 1, x 2 ) X 1 X 2, the products of open sets O 1 O 2 with x 1 O 1 and x 2 O 2 are a basis for the neighborhoods of (x 1, x 2 ). Hence the products of open sets O 1 O 2 are a basis for the open sets. Thus any open set O is defined to be a (possibly infinite) union of such products: O = {O 1 (α) O 2 (α) : α I} 10 Problem 6.3. Convince yourself that with the above definition of open set, the postulates for a topology hold. One needs to be a bid more careful to prove that any two open sets O = {O 1 (α) O 2 (α) : α I} and U = {U 1 (β) U 2 (β) : β J} have an open intersection O U. Answer. Given are any two open sets O and U. Their intersection is O U = {O 1 (α) O 2 (α) : α I} {U 1 (β) U 2 (β) : β J} = {O 1 (α) O 2 (α) U 1 (β) U 2 (β) : α I and β J} = {(O 1 (α) U 1 (β)) (O 2 (α) U 2 (β)) : α I and β J} 58

59 The intersections O 1 (α) U 1 (β) and O 2 (α) U 2 (β) are open in X 1 and X 2, respectively. Hence we see that the intersection O U is a union of products of open sets in X 1, X 2, and hence open. It is left to the reader to check the remaining postulates, too. 10 Problem 6.4. We continue the previous problem. (i) Convince yourself that the sets (X 1 \ O 1 ) X 2 and X 1 (X 2 \ O 2 ) are closed. (ii) Convince yourself that in the product space X 1 X 2, each closed set is a (possibly infinite) intersection of finite unions of sets F 1 F 2, where F 1, F 2 are closed sets in X 1, X 2. Answer. (i) (X 1 \ O 1 ) X 2 = X 1 X 2 \ (O 1 X 2 ). The set O 1 X 2 is open, hence its complement is closed. Similarly, we see that X 1 (X 2 \ O 2 ) is a closed set. (ii) Given any closed set F X 1 X 2, its complement X \ F is open and hence X \ F = O 1 O 2 F = X \ O 1 O 2 = X 1 X 2 \ O 1 O 2 = (X 1 \ O 1 ) X 2 X 1 (X 2 \ O 2 ) = F 1 F 2 G 1 G 2 with F 1 = X 1 \ O 1, G 1 = X 1 and F 2 = X 2, G 2 = X 2 \ O 2 which are closed sets in X 1, X 2, respectively Definition 40 (Arbitrary product). Let I be a nonempty index set. For all α I, an arbitrary set X α is given. The Cartesian product α I X α consists of all functions x : I α I X α for which x(α) X α holds for all α I. Proposition 23. Let the index set I be nonempty, and assume the sets X α are nonempty for all α I. The axiom of choice implies that the the Cartesian product α I X α is nonempty. In other words, there exists at least one function x : I α I X α which chooses an x(α) X α for all α I. 59

60 Proof. Proposition?? item (8b) gives the following formulation of the axiom of choice: Given is function H on a nonempty domain I, and we assume that H(i) for all i I. Then there exists a function f with domain I such that f(i) H(i) for all i I. We may choose the function H to be defined by setting H(α) = X α for all α I. Hence we see that the Cartesian product α I X α is nonempty, provided that the sets X α are nonempty for all α I. Definition 41 (Product topology). Let the index set I be nonempty, and assume the topological spaces X α are nonempty for all α I. The topology on the Cartesian product is obtained from the following basis B x for the neighborhoods of any point x α I X α in the product. Let J I be any nonempty finite subset of indices. For each index α J, any open neighborhood V (α) N x(α) is given. Then the set B x := {y α I X α : y(α) V (α) for all α J} is a basic neighborhood of the point x. Proposition 24. Given is the nonempty index set I and a topological spaces X α for all α I. Let X = α I X α be the product space. (i) The projection mappings p α : β I X β X α are defined by p α (x) = x(α). They are continuous. (ii) Given is any continuous function f : Y X, where Y is any topological space. All compositions p α f are continuous. = Y Y f X pα p α f X α (iii) Suppose that the compositions p α f are continuous for all α I. function f is continuous at any point b Y. Then the 60

61 7 The Kuratowski Closure Operator Definition 42 (Kuratowski closure properties). Given is any nonempty set X. A mapping assigning to all subsets A X its closure A X is a closure operation iff it satisfies the following postulates: (CL1,2) = and X = X; (CL3) A A; (CL4) A B = A B; (CL5) A = A; for all subsets A, B X. A closure space is a set X together with a closure operation. Lemma 13. Any closure operation satisfies the monotonicity (CL4 ) A B implies A B; (CL*) A B and B = B imply A B for all subsets A, B X. Proof. Given are any subsets A B X. We conclude A B = B and hence item (CL4) implies A A B = A B = B confirming the monotonicity item (CL4 ). Now item (CL*) follows immediately. Definition 43 (quasi-closure properties). Given is any nonempty set X. A mapping assigning to all subsets A X its closure cl(a) X is a quasi-closure operation iff the postulates (CL1),(CL2),(CL3), (CL4 ) and (CL5) are satisfied. A quasi-closure space is a set X together with a quasi-closure operation. Lemma 14. Any closure or quasi-operation satisfies (CL) A = {B X : A B and B = B} for all subsets A X. Proof. Let C := {B X : A B and B = B} We put B := A and get from the items (CL3) and (CL5) that A B X and B = B. Hence the inclusion C A holds. To confirm the inclusion A C, we assume x A. Given any set B such that A B X and B = B, we conclude from the assumed monotonicity (CL4 ) x A B = B hence x B. Finally x is in the intersection of all sets B for which A B X and B = B. Hence x C. 61

62 Proposition 25 (A (quasi-)topological space yields a (quasi-)closure space). Given is any set X and a family F of subsets satisfying the postulates (F1) F, X F; (F3) Any nonempty collection A α F has an intersection α I A α F. The closure operation A A with (7.1) A := {B X : A B and B is closed } satisfies the postulates (CL1,2)(CL3) and (CL5), and the monotonicity (CL4 ). Moreover, a set A is closed if and only if A = A. If additionally the postulate (F2) A, B F A B F and hence all postulates for a topological space hold, the closure operation satisfies all postulates (CL1) through (CL5). Proof. Directly from the definition 7.1 and postulate (F1) follow (CL1,2) = and X = X; (CL3) A A; (CL*) D C and C closed imply D C for all subsets A, B, C, D X. Moreover the postulate (F3) implies that the set B is closed for all B X. By putting C := B and D := B into (CL*) one gets Together with item (CL3) we obtain (CL5) B = B for all B X. B B To check item (CL4 ), we assume A B X. By item (CL3) we conclude A B B. Since B is closed, item (CL*) with D := A and C := B implies A B. Thus we have confirmed the item (CL4 ) A B A B for all A, B X. We check that a set A is closed if and only if A = A. For a closed set, the choice B := A in the defining formula (7.1) is admissible. Hence the intersection satisfies A A. The converse inclusion A A holds by (CL3). Hence the equality A = A holds. Conversely, we assume A = A. The set A is defined as an intersection of closed set, and hence it is closed by postulate (T3). Hence A = A implies that A is closed. 62

63 Finally we assume additionally the postulate (F2) holds, and check item (CL4). Let A, B X be any sets. Using item (CL4 ), the inclusions A A B and B A B imply A A B and B A B and finally A B A B We know that the sets A and B are both closed, and hence by postulate (F2), the union A B is closed. By putting C := A B and D := A B into (CL*) one obtains A B A B A B A B By item (CL3) the assumption true, and we get the conclusion as wanted. Both inclusions together imply A B = A B. Thus we have confirmed item (CL4) A B = A B for all A, B X. Proposition 26 (A (quasi-)closure space yields a (quasi-)topological space). Given is any closure or quasi-closure space with closure operation A A. We define the collection F of closed sets by the requirement A F A = A From any closure space, a topological space is obtained. From any quasi-closure space, a space satisfying the items (F1) and (F3), but not necessarily (F2), is obtained. In both cases the closure operation (7.1) cl(a) := {B X : A B and B is closed } turns out to be the originally given closure operation: cl(a) = A. Proof. It has been assumed that A F A = A defines the collection F of closed sets. For this collection, we need to verify the postulates for closed sets: (F1) F, X F is obtained from item (CL1,2). (F2) Take any sets A, B F. Hence A = A and B = B and item (CL4) imply and hence A B F as required. A B = A B = A B 63

64 (F3) Take any nonempty collection A α F for all α I. Let B := α I A α. By monotonicity property (CL4 ) this set satisfies B A α for all α I; B A α = A α for all α I; B α I A α = B The converse inclusion B B is given by item (CL3). B = α I A α F as to be shown. Hence B = B, and We now compare the (new) closure operation A cl(a) as defined by equation (7.1) with the originally given operation A A. By the definition of closed set, the new closure operation is cl(a) = {B X : A B and B is closed} = {B X : A B and B = B} By item (CL) from lemma 14 A = {B X : A B and B = B} Hence cl(a) = A holds for all subsets A X. Proposition 27. Let (X, O) and (Y, U) be topological spaces and f : X Y be any mapping. Equivalent are: (1) The function f is continuous. (2) For each open set O U, the preimage f 1 (O) O is open. (3) For each closed set B Y, the preimage f 1 (B) X is closed. (4) f(a) f(a) holds for all sets A X. (5) f 1 (C) f 1 (C) holds for each set C Y. Proof of (3) (4). f(a) f(a) by closure property (CL3). A f 1 f(a) by previous problem 1.4. A f 1 f(a) by item closure property (CL4 ), and since the set f 1 f(a) is closed by assumption (3). f(a) f(f 1 f(a)) f(a) by previous problem

65 Proof of (4) (3). Let B Y be any closed set. We need to check whether f 1 (B) is closed and put A := f 1 (B). Hence f(a) = f(f 1 (B)) = f(f 1 (B) X) = B f(x) B Now we use the closure property (CL4 ) and item (4) and that B is closed: By the previous problem 1.3, conclude f(a) f(a) B = B A f 1 f(a) f 1 (B) = A The converse inclusion A A holds by closure property (CL3). Hence A = A = f 1 (B) is closed, as to be shown. Proof of (3) (5). Let C Y be any set. Hence C is closed and by assumption (3), the preimage f 1 (C) is closed. C C by closure property (CL3). f 1 (C) f 1 (C) f 1 (C) f 1 (C) = f 1 (C) by closure property (CL4 ), and since the set f 1 (C) is closed. Proof of (5) (3). Let B Y be any closed set. Using the assumption (b), we get f 1 (B) f 1 (B) f 1 (B) = f 1 (B) Hence f 1 (B) = f 1 (B), and f 1 (B) is closed, as claimed. Definition 44 (Interior, boundary). The interior int(a) and the boundary A of any subset A of a topological space X are int(a) := X \ X \ A Lemma 15. The interior of a set A X is A := A X \ A int(a) = {C X : C A and C is open } A point lies in the interior of a set if and only if there exists a neighborhood of the point contained in the set. 65

66 Proof. From the equation (7.1) about the closure, and the definition of the interior we conclude, by means of demorgan s formula X \ A = {B X : X \ A B and B is closed } int(a) = X \ X \ A = X \ {B X : X \ A B and B is closed } = {X \ B X : X \ B A and B is closed } = {C X : C A and C is open } We have used that X \ A B X \ B A and have put C = X \ B, which is open. From this result we see that x int(a) C[x C and C A and C is open ] Thus point x lies in the interior int(a) of the set A if and only if there exists an open neighborhood C x which, by C A, is contained in the set A. Lemma 16. A point is in the closure of a set if and only if all its neighborhoods intersects the set. Proof. Use X \ int(a) = X \ A and put B = X \ A. By formal logic x int(a) C[x C and C A and C is open ] x X \ int(a) x X and C [x C and C A and C is open ] x X \ A x X and C[x C and C is open C A] x X \ A x X and C[x C and C is open (X \ A) C ] x B V [ V is open and x V V B ] We see that a point x lies in the closure B of a set B if and only if all its open neighborhoods V x intersects the set, because of V B. Proposition 28. Given is any topological space X and any subset A X. (i) A point x lies in the closure of a set A if and only if each neighborhood of point x intersects the set A. (ii) A point x lies in the boundary of a set A if and only if each neighborhood of point x contains both a point lying inside the set A and a point lying outside the set A. (iii) A point x lies in the interior of a set A if and only if there exists a neighborhood of point x contained in the set A. (iv) A point x is a limit point of a set if and only if each neighborhood of point x contains another point of the set different from point x. 66

67 10 Problem 7.1. Let X be any topological space and A be any subset. Let O x denote the family of open neighborhoods of any point x. Match the definitions for the closure A, boundary A, interior int(a), and limit set A. x int(a) V O x V A x A V O x c [c V A and c x] x A V O x V A x A V O x [V A and V \ A ] 10 Problem 7.2. Let X = R. Find the closure, interior and boundary for the following sets. A A Int A A [0, 2) [0, 2] (0, 2) {0, 2} [0, ) [0, ) (0, ) {0} R \ {0} R R \ {0} {0} R R R [0, 1) (1, 2) [0, 2] (0, 1) (1, 2) {0, 1, 2} [0, 2) ( 2, 2] [0, 2] (0, 2) ( 2, 2) {0, 2, 2} { p 2 n : p Z and n N} R R (0, ) Q [0, ) [0, ) A = { n+1 n : n N} {1} A {1} A 10 Problem 7.3. Let X = R. Find the closure, interior and boundary for the following sets. 67

68 A A Int A A [0, 1) [0, 1] (0, 1) {0, 1} [0, ) [0, ) (0, ) {0} (0, ) [0, ) (0, ) {0} R R R [0, 1) (1, 2) [0, 2] (0, 1) (1, 2) {0, 1, 2} [0, 2) ( 2, 2] [0, 2] (0, 2) ( 2, 2) {0, 2, 2} [0, ) \ Q [0, ) [0, ) (0, ) Q [0, ) [0, ) { 1 : n N} n {0} A {0} A 68

69 8 Dense Sets and Baire Spaces 10 Problem 8.1. Give an example where A B A B. It is enough to consider the example X = R and take for A and B suitable intervals. Answer. Take for example A = (0, 1) and B = (1, 2). Hence A B = =, but A B = [0, 1] [1, 2] = {1} which are different. 10 Problem 8.2. Which one of the inclusions A B A B, or A B A B does always hold? Give a proof. Answer. The first inclusion is always true. Indeed, since the closure operation is monotone, A B A, implies A B A. Similarly, we get A B B, and hence A B A B. Proposition 29. Any real interval (a, b) with a < b contains a rational number. 10 Problem 8.3. Prove Proposition 29, assuming additionally that a > 0. Make clear which two basic principles are needed. Answer. Since b a > 0, the Archimedean axiom implies existence of a natural number q such that q(b a) > 1. Moreover, the Archimedean axiom implies existence of a natural number p such that p > qa. The principle of complete induction implies that a nonempty set A N has a least element. Hence there exists a least natural number p such that p > qa, and hence p > qa p 1. We rearrange to get qb > qa + 1 p > qa b > p q > a Thus p q is a rational number lying in the interval (a, b). Definition 45 (Dense set). A subset A X of a topological space X is called dense iff A = X. 10 Problem 8.4. Let A X be any set in the topological space X. Prove that A A = A. Answer. A A = A (A X \ A) = (A A) (A X \ A) = (A A) X = A 10 Problem 8.5. Use the previous problem to check that any A X of a topological space X is dense if and only if A X \ A. Answer. A X \ A A A = X A = X 69

70 Proposition 30. Let X be any topological space and A X be any set. The following are equivalent assumptions: (i) the set A is dense, which means A = X; (ii) the complement has empty interior: int(x \ A) = ; (iii) any nonempty open set contains some point of A: O O A ; (iv) for every point x X, each neighborhood V N x contains some point of A; (v) for every open set O O A; (vi) for every open set O = O A; (vii) A = X \ A. (viii) A X \ A. Proof. Equivalent are A is dense in X A = X int(x \ A) = X \ A = [O X \ A and O is open in X O = ] [O A = and O is open in X O = ] [O and O is open in X O A ] Hence (i),(ii) and (iii) are equivalent. To check that item (iii) implies item (iv), take any x X and neighborhood V of x. By definition, the neighborhood contains an open neighborhood: x O V. By item (iii), the intersection O A is nonempty. Hence V A is nonempty, too. To check that item (iv) implies item (v), let any open set O and point x O be given. Take any open neighborhood V N x. Hence V O N x is an open neighborhood of point x, too. By item (iv) we see that V O A. Since V is an arbitrary the neighborhood of point x, the definition of the closure yields x O A. Thus we have checked that O O A. To check that item (v) implies item (vi), note that assuming item (v) O O A O and hence the equality (vi) holds. Taking O := X in (vi), we conclude (i). To check that item (i) implies item (vii), note that A = X implies Clearly, item (vii) implies item (viii). A = A X \ A = X \ A X \ A 70

71 To check that item (viii) implies item (i), note that item (viii) implies A = A X \ A = X \ A. Hence A X \ A X \ A and Thus A = X as to be checked. X A A X \ A = X 10 Problem 8.6. Let X = R have the standard topology. Prove that both the set of rational Q and its complement, the set R \ Q of all irrational real numbers, are dense in the real numbers. Answer. It is enough to realize that every nonempty open interval contains a rational number. This fact has been established above by proposition 29. Hence every nonempty open set contains a rational number. Hence by item (iii) of proposition 30 the set of rationals is dense in the set of real numbers. Similarly, we proceed for the set R \ Q of all irrational numbers. Given any interval (a, b), we need to find an irrational number in that interval. We already know there exists a rational number p (a, b). By the Archimedean axiom, there exists a natural q number r such that 2 < b p. On checks that the number r q y = p q + 2 r lies in the interval (a, b) and is irrational. Hence every nonempty open set contains an irrational number. By item (iii) of proposition 30 the set of irrational numbers is dense in the set of real numbers. 10 Problem 8.7. Let X, Y be topological spaces, A X be a dense subset, and f : X Y a continuous function onto Y. Prove that f(a) is dense in Y. Answer. We know from proposition 27 that f(a) f(a). Since A is dense and the function f maps onto Y, we know A = X and f(x) = Y. Hence Y = f(x) = f(a) f(a) f(x) = Y implies f(a) = Y and hence f(a) is dense in Y. 10 Problem 8.8. Let X, Y be topological spaces, A X be a dense subset, and f : X Y a continuous function onto Y. Complete the proof below showing that f(a) is dense in Y. Proof. It is enough to check that every nonempty open subset of Y intersects the image f(a). Given is a nonempty open set C Y. Take any y C. Since the function f is surjective, there exists a preimage x f 1 (C) with f(x) = y and f 1 (C) is nonempty. 71

72 Since the function f is continuous, the preimage f 1 (C) is open. Since the set A is dense and f 1 (C) is open and nonempty, the intersection A f 1 (C) is nonempty. Take any a A f 1 (C). Thus f(a) f(a) C, and we see that the image f(a) intersects each nonempty open set C Y. We know by item (iii) of proposition 30 that a set intersecting each open nonempty set is dense. Hence we have confirmed that the set f(a) is dense in Y. Let X and Y be any topological spaces. The product topology for X Y is defined in the usual way: For any point (x, y) X Y, the products U V of any open neighborhoods U x and V y are a basis for the neighborhoods of (x, y). Proposition 31 (The product of two dense sets is dense). Let A X and B Y be any dense sets in the topological spaces X and Y. Then the product A B is dense in the product space X Y. Proof of proposition 31. We check item (iii) from proposition 30 holds for the set A B. Let C be any nonempty open set in the product space X Y, and let (x, y) C be any point. By the definition of the product topology, there exist open sets U and V such that x U X, y V Y and (x, y) U V C Since A X and B Y be any dense sets, item (iii) from proposition 30 implies that the intersections A U and B V are both nonempty. Hence their product is nonempty. Because of (A U) (B V ) = (A B) (U V ) (A B) C we see that the product A B intersect any arbitrary nonempty open set C. By item (iii) from proposition 30 we conclude that the set A B is dense. Corollary 7. The product of two, and hence finitely many separable spaces is separable. Proof. For any separable spaces X and Y, by definition there exist countable dense sets A X and B Y. By proposition 31, the A B is dense in the product space X Y. Too, the product A B is countable. Hence the product space X Y is separable. 10 Problem 8.9. Let A X be any subset of a topological space X. Check that A is open and dense if and only if A = X \ A. Answer. Proof of: A open dense A = X \ A. We have assumed A is open and dense, thus A = X and X \ A = X \ A. Hence by the definition of the boundary as to be shown. A = A X \ A = X (X \ A) = X \ A 72

73 Proof of: A = X \ A A open dense. The boundary is A = A X \ A, and hence always closed. We assume that A = X \ A. Hence X \ A is closed and A is open. Moreover X \ A X \ A = A = A X \ A A implies X \ A A and X = (X \ A) A = A. Hence the set A is dense, as to be shown. 10 Problem Let the subset A X of a topological space X be open and dense, and the subset B X be dense. Prove that the intersection A B is dense. Proof. Let O be any open nonempty set. Since the set A is dense, by item (iii) of the proposition 30, we conclude A O. Moreover, A O is an open set. Using item (i) once more, now for the dense set B, we conclude that the intersection B (A O) is nonempty. Obviously, B (A O) = (A B) O. The entire paragraph implies, again via item (iii) of the proposition 30 that the set A B is dense. 10 Problem Prove that the finite intersection of open dense sets is open and dense. Answer. For two open dense sets A and B the assertion follows from the previous problem: the intersection A B is dense by the previous problem, and the intersection of two open sets is always open. For all natural n 1, we assume the sets A n are open dense. The general case is proved by induction. Here is the induction step n n + 1. We have already shown that the intersection 1 i n A i is open dense. Too, the set A n+1 is open dense. Hence the first paragraph implies that A i A n+1 = is open and dense. 1 i n 1 i n+1 Definition 46 (Nowhere dense set). A subset A X of a topological space X is called nowhere dense iff int(a) =. (8.1) (8.2) 10 Problem Prove that for any finite collections of sets A n and B n int(b n ) = X for all 1 n N implies int( int(a n ) = for all 1 n N implies A i 1 n N int( 1 n N B n ) = X A n ) = 73

74 Proof. The sets int(b n ) with 1 n N are a finite collection of open dense sets. By the result of the previous problem 8.11, their intersection is open dense. Hence as to be shown int( B n ) = int(b n ) = X 1 n N 1 n N For any sets A n, let B n := X \ A n be their complements. One checks that (8.3) (8.4) X \ int(a n ) = int(x \ A n ) int(a n ) = int(b n ) = X Let be given a finite collection nowhere dense sets A n. We conclude int(a n ) = for all 1 n N, hence int( int( 1 n N A n ) = int( 1 n N 1 n N int(b n ) = X for all 1 n N, hence B n ) = X X \ B n ) = X \ int( 1 n N B n ) = Proposition 32. The following are equivalent characterizations for a Baire space X. (a) For any countable collection of nowhere dense sets A n, the union has empty interior: (8.5) int(a n ) = for all natural n implies int( n N A n ) = (b) For any countable collection of sets B n, each one of which has dense interior, their intersection is dense: (8.6) int(b) n = X for all natural n implies B n = X (c) For any countable collection of open dense sets O n, their intersection is dense: (8.7) O n = X and O n open for all natural n implies O n = X Proof of (a) (b). Given is a countable collection of sets B n, and the interior of each of them is dense. In other words, int(b n ) = X holds for all natural n. Let A n := X \B n. The equivalence (8.3) yields int(a n ) =. Hence item (a) yields int( n N A n ) = n N B n = X \ int( n N A n ) = X n N n N 74

75 Proof of (b) (a). This proof can be left to the reader. Proof of (b) (c). Indeed item (c) is just a special case of (b) with the open sets B n := O n for which B n = int(b n ). Proof of (c) (b). Given is any countable collection of sets B n, each one of which has dense interior. Let O n := int(b n ). By item (c), the intersection is dense: O n = X and hence a fortiori n N X n N B n n N O n = X Proposition 33. Let X be any topological space. The following are equivalent properties of any subset A X: (i) the interior of set A is dense; (ii) int(a) = X; (iii) the closure of the complement has empty interior: int(x \ A) = ; (iv) any nonempty open set contains some interior point of A: O O int(a) ; (v) for every point x X, each neighborhood V N x contains some point of int(a); (vi) given any point x X and any neighborhood V N x, it contains a point y V such that there exists a neighborhood W N y for which W V A; Proof. I leave it to the reader to check that items (i) through (v) are equivalent. Here is the check that item (v) implies item (vi), Let any point x and any (open) neighborhood V N x be given. By item (v), there exists a point y V int(a). Since the latter is an open set, there exists an open neighborhood W N y such that as claimed y W V int(a) V A Conversely, this point y lies in V int(a). Hence we see that item (vi) implies item (v). 10 Problem Let X be a topological space and A X be any subset. For this problem, I find it convenient to denote the interior by int(a) and the closure by cl(a). Prove the inclusions int(a) int(cl(int(a))) int(cl(int(cl(a)))) cl(int(cl(a))) cl(a) cl(int(cl(int(a)))) 75

76 Answer. Let B := cl(a). Since int(b) B, the closure property (CL4 ) implies cl(int(b)) cl(b) = B, which yields the inclusion cl(int(cl(a))) cl(a). Taking complements yields int(a) int(cl(int(a))). The remaining parts are even easier to check. 10 Problem Show that by applying the operations of interior and inclusion successively to one set, no more than the six sets as in problem 8.13 can be obtained. Especially cl(int(cl(int(cl(a))))) = cl(int(cl(a))) int(cl(int(cl(int(a))))) = int(cl(int(a))) and hold for all subsets A X. Answer. cl(int(cl(b))) cl(b) cl(int(cl(int(cl(a))))) cl(int(cl(a))) int(c) int(cl(int(c))) int(cl(a)) int(cl(int(cl(a)))) with B:=int(cl(A)) implies cl(int(cl(a))) cl(int(cl(int(cl(a))))) cl(int(cl(a))) = cl(int(cl(int(cl(a))))) The remaining parts are even easier to check. with C:=cl(A) implies both inclusions imply 10 Problem Show that by applying the operations of interior and inclusion successively to an appropriate set S, indeed the six sets as in problem 8.13 turn out to be all distinct. Let A := { 1 q : q N and q 2 } and let S be the union of the four sets in the first column. The disjoint unions of the remaining columns give six different sets. S int(s) int(cl(int(s))) int(cl(int(cl(s)))) [0, 1] (0, 1) (0, 1) (0, 1) (2, 3) Q (2, 3) (4 + A) [6, 7] \ (6 + A) (6, 7) \ (6 + A) (6, 7) (6, 7) 76

77 S cl(int(cl(int(s)))) cl(int(cl(s))) cl(s) [0, 1] [0, 1] [0, 1] [0, 1] (2, 3) Q [2, 3] [2, 3] (4 + A) {4} (4 + A) [6, 7] \ (6 + A) [6, 7] [6, 7] [6, 7] 77

78 9 The Cantor Set and the Devil s Staircase To give a theoretical discussion of real-valued functions requires not only to just consider specific examples, but one has to go on and consider all possible realvalued functions. No reasonable limit can be set on the means which might be, now or in future, available for defining these functions. Under this premisses, the notion of a possible real-valued function cannot be tied to, or restricted by any particular conception of the means for defining functions. On the other hand, at any given historical period, such ties and restrictions will de facto be the case. The same could be thought to go for sets or classes. Any universally applicable mathematical theory cannot be tied to any specific means for defining or determining real function, sets or classes. The fundamental issue is then how the mathematician reasons about all possibilities prior to their specification. This makes the realism-antirealism issue difficult to resolve head on. After Mary Tiles The Philosophy of Set Theory 1989 And there are other things too, governed by that incommensurability of experience and understanding. But it is always the case that what we experience in one moment, whole and unquestioning, becomes incomprehensible and confused when we seek to bind it to our enduring ownership with the chains of thought. And what looks big and unhuman while our words reach it from afar, becomes simple and ceases to be unsettling the moment it enters our s life s field of action. Robert Musil The Confusions of Young Törless 1906 Lemma 17. Given are sequences with a n, b n {0, 2} for all n N, but no a n, b n = 1 occurring. For any natural number N 1, the inequality b n a n (9.1) 3 n < 2 3 N implies a n = b n for 1 n N. 1 n N Proof. We use induction by N. The assertion is valid for N = 1. For the induction step, we assume, the assertion to be proved for all N < N, and check it for N. Assume now the inequality (9.1) to hold, as written. Hence 1 n N 1 b n a n 3 n < b N a N 3 N < 4 N 3 < 2 N 3 N 1 78

79 By the induction assumption, this inequality implies a n = b n for all 1 n N 1. Hence the inequality (9.1) implies b N a N 3 N < 2 3 N but this can hold only for a N Lemma is proved. = b N. Thus the induction step is completed, and the Definition 47 (Cantor set). The subset of the real interval [0, 1] consisting of the real numbers a which can be represented by a triadic fraction (9.2) a = n 1 with only digits a n {0, 2} occurring, is called the Cantor set. For each natural number n 1, let X n = {0, 2} and X = n 1 X n be the product. We define a map f : X [0, 1] by setting f(x) = n 1 and a map g : f(x) X mapping a g(a) := (a n ). In other words, the map g assigns to a real number a which can be represented, as in equation (9.2) above, the sequence (a n ) of its triadic digits a n {0, 2}. a n 3 n x n 3 n 10 Problem 9.1. (i) Prove that the map f is injective. (ii) Convince yourself that the map g is well-defined and f g = id f(x). (iii) Convince yourself once more that g f = id X. (Of course this part follows in general from the previous problem) Hence there exists a bijection mapping the product X onto the Cantor set f(x). Proof. (i) To check whether f is injective, we assume f(x) = f(x ). Hence holds 1 n N n>n n 1 x n 3 = x n n 3 n n 1 which implies that for all N N x n x n 3 n = x n x n 3 n x n x n 3 n n>n n N = 1 n 3 < 2 N 3 N By the Lemma 9.1, this inequality implies x n x n for all 1 n N. Since this assertion holds for all natural N, one finally concludes x = x as expected. 79

80 (ii) Part (i) implies that for any real number a f(x), the sequence of its triadic digits a n {0, 2} is unique. Hence the map g : f(x) X is well defined. The definition of g assures that f(g(a)) = a n 3 = a n n 1 for all a f(x), and hence f g = id f(x). (iii) To convince yourself once more that g f = id X, take any sequence x X. Thus f(x) is the corresponding triadic number. Put a := f(x). Since the triadic digits a n {0, 2} are uniquely determined by the number a, they are given by the original sequence (x n ) = x. Thus g(a) = x and, in other words, finally g(f(x)) = x for all x X means g f = id X is valid. Let each X n = {0, 2} have the discrete topology, and the product X = n 1 X n have the standard weak topology. Hence a basis for the neighborhoods of (x n ) X consists of the (countably many) sets B N (x) := {y X : y = (y n ) and y n = x n for 1 n N} with N N. Let the image f(x) R have the relative topology induced by the usual topology for the real numbers. 10 Problem 9.2 (Homeomorphism of the Cantor set to a discrete countable product). We continue the previous problem. (i) Prove that the map f is continuous; (ii) prove that the map g is continuous. Hence the set X with the product topology and the Cantor set f(x) with the relative topology are homeomorphic. Proof. (i) To check the continuity of the map f, say at point x X, let ɛ > 0 be given. By the Archimedean axiom, there exists a natural number N such that 3 N < ɛ. For any sequence x B N (x) in this basic neighborhood, x n = x n holds for 1 n N. Hence f(x ) f(x) = n N+1 x n x n 3 n x n x n 3 n n N+1 n N = 1 n 3 < ɛ N 80

81 (ii) To check the continuity of the map g say at point a f(x), let a basic neighborhood B N (x) of the image point g(a) = x be given. We choose δ := 1. Now assume 3 N b a < δ. Indeed, we conclude b n 3 a n n 3 n < δ n 1 n 1 b n a n 3 n < δ + b n a n 3 n δ = δ + 1 n 3 = 2 N 3 N 1 n N n>n n>n By the Lemma 9.1, this inequality implies a n = b n holds for all 1 n N. The calculation has checked that b a < δ implies b B N (g(a)). Thus the map g is continuous. Define a function h : f(x) [0, 1] by the requirement a = n 1 a n 3 h(a) = a n /2 n 2 n n 1 Question. In which cases do we get a < b but h(a) = h(b)? Answer. This happens if and only if h(a) = p 2 N with some N 1 and 1 p < 2 N. In other words if and only if h(a) happens to be a finite binary fraction. Indeed, the finite binary fractions are exactly those which have two distinct presentations: h(a) = 1 n<n a n /2 2 n + n>n 1 2 = h(b) = n 1 n<n In other words, the sequences (a n ) X and (b n ) X are b n = a n for 1 n < N, a N = 0, b N = 2, a n = 2, b n = 0 for n > N a n /2 2 n N = p 2 N with some N 1. For any N 1, there exist 2 N 1 such pairs. The triadic values a and b are (steps) a = We have used n>n 1 n<n 2 = 1. 3 n 3 N a n n 3 < b = N 1 n<n a n n 3 N Proposition 34. There exists a unique extension of the function h to a monotone function s : [0, 1] [0, 1]. 81

82 Definition 48 (Devil s staircase). The unique extension of the function h to a monotone function s : [0, 1] [0, 1] is called Devil s staircase. Question. What are the horizontal (open) steps of Devil s staircase? What is there open union? Answer. The horizontal steps are (a, b) given by equations (??) with any N 1 and 1 p < 2 N. The open union of these steps is [0, 1] \ f(x). Definition 49. The open union [0, 1] \ f(x) is called Cantor s middle third set. 10 Problem 9.3. Prove that the Devil s staircase is a continuous function. Proof. For any number a in Cantor s middle third set, there exists a neighborhood in which the Devil s stair case is horizontal. In other words, the function s is constant in this neighborhood. Hence the function s is continuous on Cantor s middle third set. We need to check continuity at any given point a f(x) in the Cantor set. Let ɛ > 0 be given. There exists a natural number N such that 2 N < ɛ. We choose δ := 3 N. Now assume b a < δ and b f(x). As shown in Problem 9.2, we conclude a n = b n holds for all 1 n N. s(b) s(a) = n 1 b n a n 2 n+1 < n>n 1 2 = 1 n 2 < ɛ N If b / f(x), there exists c f(x) such that s(c) = s(b) and c a b a < δ. Continuity of Devil s staircase has been checked! Question. Is h 1 a function? Is s 1 a function? Answer. Neither h 1 nor s 1 are functions. There restrictions to the set S := [0, 1] \ {p 2 N : N 1 and 1 p < 2 N } are a function, and indeed the same function h 1 S to [0, 1]. Definition 50 (Devil s rock). The monotone right-continuous extension of h 1 S to the domain [0, 1] is unique. I call it the Devil s rock and denote it by r. Question. What are the jump discontinuities of the Devil s rock. What are the heights of these jumps, and of their sum. Answer. The jump discontinuities of the Devil s rock occur at the values of set s S since h(a) = h(b) = s. The jump at s = p 2 N = 1 n<n a n /2 2 n N 82

83 goes from given by the equations (steps) a = r(s 0) = a to 1 n<n r(s) = r(s + 0) = b a n n 3 < b = N 1 n<n a n n 3 N The height of this jump is 3 N. There occur 2 N 1 jumps of this height. Hence the total height of all jumps is 2 N 1 3 = 1 N N 1 So we see that the Devil s rock function r is constant on [0, 1] \ S, which means on each component of [0, 1] \ S. But this means nothing, since Lemma 18. With the topology induced by real topology, all components of the set [0, 1]\S are single points. Hence one calls this set totally disconnected. Question. In which sense are the devil s staircase s and the devil s rock r inverse functions? Answer. Indeed s r = id [0,1]. Let X 0 X be the subset where it does not happen that x n = 2 for all n > N. Note that r[0, 1] = f(x 0 ) is really a proper subset f(x 0 ) f(x) of the Cantor set f(x). All the lower (left) endpoints of the gaps are deleted. The above Problem 1.9 can be used with f := r and g := s r[0,1] h. Indeed the devil s rock is a bijection r : [0, 1] r[0, 1], and the restriction s : r[0, 1] [0, 1] to domain r[0, 1] is its inverse. Question. Determine the function r s. Answer. The restriction of the function r s to the subset f(x 0 ) f(x) of the Cantor set is the identity: r s f(x0 ) = id f(x0 ). The restriction to the complement [0, 1] \ f(x) of the Cantor set is piecewise constant. The function r s is right-continuous. 10 Problem 9.4. For N = 1, 2, 3 draw the piecewise linear approximations of the devil s staircase s, the devil s rock r, and the composition r s. 10 Problem 9.5. Let { } B := ( p q 2 2q, p + q 2 2q ) : p, q N and 1 p < q and q 2 (0, 1) (i) Convince yourself that the set B is open dense in the space [0, 1]. (ii) Convince yourself that the Lebesgues measure of B is less than 1. Hence the complement [0, 1] \ B is a closed, nonempty set. 83

84 (iii) Convince yourself that int([0, 1] \ B) =. Answer. (i) The set B is open since it is the union of open intervals, and contains all rational numbers in the interval (0, 1). But these are dense and hence [0, 1] B (0, 1) Q = [0, 1] implies B = [0, 1]. Thus the set B is open dense in the space [0, 1]. (ii) The Lebesgues measure of B is less than B < q 2 (q 1)2 1 2q < q q = 1 since q < 2 q. Hence the complement [0, 1] \ B is a closed, nonempty set. (iii) The set [0, 1]\B does not contain any rational numbers and hence int([0, 1]\B) =. 10 Problem 9.6. Continuing the previous problem, complete the table below S int(s) int(cl(int(s))) int(cl(int(cl(s)))) B B (0, 1) (0, 1) [0, 1] \ B S cl(int(cl(int(s)))) cl(int(cl(s))) cl(s) B [0, 1] [0, 1] [0, 1] [0, 1] \ B [0, 1] \ B 84

85 10 The relative topology Definition 51 (Relative topology). Given any topological space X and any nonempty subset Y X, the relative topology induced on Y is defined by requiring that any set A Y is (relatively) open if and only if there exists a set B X, which is open in X and A = B Y. Lemma 19. The above requirement indeed defines a topology on the set Y. 10 Problem To confirm the lemma 19, convince yourself that the axioms (O1), (O2) and (O3) from definition 18 of a topological space, are indeed valid for the relative topology R. Lemma 20. For a Hausdorff space and any subset Y X, the relative topology yields again a Hausdorff space. Proof. Take any two distinct points a b, both a Y and b Y. Since X is assumed to be Hausdorff, by definition 25 there exist disjoint open neighborhoods N a and M b. Now the intersections N Y and M Y are relatively open and disjoint neighborhoods of points a Y and b Y. The reasoning confirms that, according to definition 25, the subspace Y is Hausdorff. Lemma 21. For any metric space (X, d) and nonempty subset Y X, the restriction of the metric to Y is a metric space, too. The standard topology of a metric space (Y, d) is the relative topology of the standard topology of a metric space (X, d). Proof. Indeed for any set A Y : A is open in the standard topology of (Y, d) x A δ > 0 [(d(x, y) < δ and y Y ) y A] x A δ > 0 B(x; δ) Y A B = {B(x; δ(x)) : x A} and B Y A with balls B(x; δ(x)) = {y X : d(x, y) < δ(x)} and their radiuses δ(x) depending on x. But the set B X is open in the standard topology of the space (X, d). This follows directly from the definition 8 of the standard topology of a metric space, and of course once more from lemma 5. Moreover A B Y and hence A = B Y. Hence any set A which is open in the standard topology of (Y, d), is of the form A = B Y with B open in X and hence open in the relative topology of Y. The converse is as obvious: Any set A Y of the form A = B Y with B open in X, is open in the standard topology of (Y, d). 10 Problem Given is any topological space X and subset Y X. Check that the definition cl Y (A) := A Y gives a closure operation on the subsets of Y, for which the Kuratowski closure properties hold. It needs a bid of care to check that cl Y (cl Y (A)) = cl Y (A) 85

86 holds for any subset A Y. Proof. Given are any subsets A, B Y. (CL1,2) cl Y ( ) = Y = and cl Y (Y ) = Y Y = Y ; (CL3) A A Y = cl Y (A) since A Y and A A; (CL4) Check cl Y (A B) = cl Y (A) cl Y (B): Answer. cl Y (A B) = A B Y = (A B) Y = (A Y ) (B Y ) = cl Y (A) cl Y (B) (CL5) We have already proved earlier that (CL4) implies the monotonicity. (CL3) implies cl Y (A) cl Y (cl Y (A)). To obtain the converse inclusion, use A B A B from a previous problem: Answer. Hence cl Y (cl Y (A)) = cl Y (A) Y = A Y Y (A Y ) Y = A Y = cl Y (A) since A = A and Y Y = Y. The converse inclusion cl Y (A) cl Y (cl Y (A)) has already been shown. Both inclusions together give cl Y (cl Y (A)) = cl Y (A), as to be shown. Lemma 22. Given is any topological space X and subset Y X. The definition cl Y (C) := C Y gives a closure operation on the subsets of Y with the Kuratowski closure properties being valid. On the subsets of Y, from this closure operation is obtained the relative topology. Proof. Given any subset C Y. Assume that C is closed in the topology generated by the closure operation cl Y (). Hence C = cl Y (C) = C Y. But the set C is closed in X, and hence C Y is relatively closed in Y. Conversely, assume that the set C is relatively closed in Y. Hence there exists a closed set B X such that C = B Y. Hence C cl Y (C) = C Y = B Y Y B Y = B Y = C There is everywhere equality and hence C = cl Y (C) which means that the set C is closed in the topology generated by the closure operation cl Y (). 86

87 Lemma 23. Given are topological spaces X and Z and any function f : X Z. Take any set W such that f(y ) W Z and let W have the relative topology inherited from the topological space Z. Let the function g : X W be defined by putting g(x) = f(x) for all x X. 11 The function f is continuous if and only if the function g is continuous. Proof. We assume that the function f is continuous and check whether the function g is continuous. To check this claim, take any open set V W. By the definition of the relative topology, there exists a set P, open in Z, for which V = P W. Because the function f is continuous, the preimage f 1 (P ) is open in X. Since f(x) W has been assumed, f 1 (W ) = X holds. But we need the preimage g 1 (V ) = g 1 (P W ) = f 1 (P W ) = f 1 (P ) f 1 (W ) = f 1 (P ) and can see it is open in X. We have checked that the preimage g 1 (V ) of any open set V W is open. Hence the function g is indeed continuous. Conversely, assume that the function g is continuous. To check whether the function f is continuous, take any open set P Z. Hence P W is relatively open, and by the continuity of g, the preimage g 1 (P W ) is open. By the assumption f(x) W, the preimage is f 1 (P ) = f 1 (P f(x)) = f 1 (P W ) = g 1 (P W ) and we see it is open in X. Hence the function f is indeed continuous. Definition 52. For any function f : X Z and any set Y X, the restriction of the function f to the set Y is the function h : Y Z such that h(x) = f(x) for all x Y. We denote the restriction by f Y. Lemma 24. Assume the function f : X Z is continuous. Let the subset Y X have the relative topology inherited from the topological space X. For any open set V Z, the preimage (f Y ) 1 (V ) is relatively open in Y. Hence the restricted function f Y is continuous. Proof. To check this claim, take any open set V Z. Because the function f is continuous, the preimage f 1 (V ) is open in X. But we need the preimage (f Y ) 1 (V ) = Y (f Y ) 1 (V ) and can see it is relatively open in Y. We have checked that the preimage (f Y ) 1 (V ) of any open set V is open. Hence the function f Y is indeed continuous. 11 I do not enter the discussion whether the functions f and g are equal, anyway. 87

88 Lemma 25. Assume the function f : X Z is continuous. Let the subset Y X have the relative topology inherited from the topological space X. Let the direct image f(y ) Z have the relative topology inherited from the topological space Z. Define the double-restricted function f : Y f(y ) by setting f(x) = f(x) for all x Y. For any relatively open set V f(y ), the preimage f 1 (V ) is relatively open in Y. Hence the double-restricted function f is continuous. Of course, lemma 25 is a consequence of the two lemmas 23 and 24. But, because repetition makes the master, I give an independent proof. Independent proof of the lemma 25. To check this claim, take any open set V f(y ). By the definition of the relative topology, there exists a set P, open in Z, for which V = f(y ) P. Because the function f is continuous, the preimage f 1 (P ) is open in X. But we need the preimage f 1 (V ) = Y f 1 (V ) = Y f 1 (f(y ) P ) = Y f 1 (P ) and can see it is relatively open in Y. We have checked that the preimage f 1 (V ) of any open set V f(y ) is open. Hence the function f is indeed continuous. 11 Connectedness Definition 53 (Connectedness). A topological space X is called connected if and only if the only clopen sets are and X. A subset Y X is called connected if and only if the space Y is connected in the relative topology. Definition 54 (Separation). A pair A, B is called a separation of a subset Y X of the topological space X iff (11.1) A B = Y, A, B, A B =, A B = 10 Problem Convince yourself that A B = and A B = A B (A B) = Hence the pair A, B is a separation of a subset Y X iff (11.2) A B = Y, A, B, A B Y = Proposition 35. Equivalent are (i) The topological space X is connected. (ii) The space X has no separation. 88

89 (iii) The space X is not the union of two disjoint nonempty open sets. (iv) The space X is not the union of two disjoint nonempty closed sets. Proof. 10 Problem Convince yourself that this proposition is valid. Proposition 36. Equivalent are (i) The subset Y X of the topological space X is connected. (ii) The subset Y has no separation. (iii) It is impossible that there exist open sets P and Q such that (11.3) Y P Q, P Y, Q Y, P Q Y = (iv) It is impossible that there exist closed sets F and G such that Y F G, F Y, G Y, F G Y = Proof of (i) (ii). The closure operation for the subspace topology is given by cl Y (A) = A Y. The subspace Y has a separation A, B in the relative topology iff A B = Y, A, B, cl Y (A) B =, A cl Y (B) = but these conditions are equivalent to the original conditions (11.1). The subspace Y is connected if and only if the subspace Y has no separation A, B in the relative topology, this happens if and only if the subset Y has no separation. Definition 55 (Surrounded Separation). A pair P, Q is called a surrounded separation 12 of a subset Y X of the topological space X iff the equation (11.3) holds. Proof of not (ii) not (iii). Suppose the subset Y has a separation A, B. We define the open sets P := X \ B and Q := X \ A. Since Y = A B we get P Q = (X \ B) (X \ A) = X \ (B A) = X \ (A B) = X \ Y X \ Y and conclude P Q Y = as required. Moreover A B = yields and similarly A B = yields Finally P Y = (Y \ B) = (A B) \ B = (A \ B) = A Q Y = (Y \ A) = (A B) \ A = (B \ A) = B (P Q) Y = (P Y ) (Q Y ) = A B = Y implies Y P Q. Hence ther pair P, Q is a surrounded separation of Y. 12 That is my own choice of name. 89

90 Proof of not (iii) not (ii). Suppose that the condition of (11.3) hold for the surrounded separation P, Q of Y. We put A := P Y and B := Q Y and verify that the pair A, B is a separation. By assumption A and B. The assumption Y P Q implies A B = Y. From A = P Y and P Q Y =, we get A = P Y X \ Q and since Q is open, we conclude even A X \ Q. Hence A B (X \ Q) Q = We conclude that A B = and similarly get A B =. Hence the pair A, B is a separation and item (ii) does not hold, as claimed. 10 Problem Check that any A X of a topological space X has empty boundary if and only if the set is both open and closed. (Also call clopen): Proof of A clopen A =. A is open and closed A = Answer. The boundary is defined as A = A B with B = X \ A. Since both A and B are closed, we conclude Hence the boundary is empty. A = A B = A B = A (X \ A) = Proof of A = A is clopen. We assume that the boundary is empty, hence A = A B = with B = X \ A. Hence A A X \ B X \ B = A Hence A = A and A is closed. In the same way B B X \ A X \ A = B Hence B = B and B is closed and A is open. 10 Problem Let X, Y be topological or metric spaces and let f : X Y be a continuous function. Prove or disprove (a) If C Y is connected, then f 1 (C) is connected. It is enough to consider a counterexample where X = Y = R. 90

91 Answer. We get a counterexample where X = Y = R and f(x) = x 2. Let C = [1, 2], which is connected. Hence f 1 (C) = [ 2, 1] [1, 2] which is disconnected. Proposition 37 (The continuous image of a connected set is connected). Let X, Z be topological spaces and let f : X Z be a continuous function. If the set Y X is connected, its image f(y ) is connected in Z. My most simple proof for proposition 37. We prove: if the set f(y ) is disconnected, then the set Y is disconnected. Thus it is assumed that f(y ) can be partitioned into two nonempty open sets. In other words A B = f(y ), A, B, A B =, and both A and B are relatively open in f(y ). We put à := f 1 (A) Y and B := f 1 (B) Y and check that the partition {Ã, B} witnesses that the set Y is disconnected. à B = (f 1 (A) f 1 (B)) Y = f 1 (A B) Y = f 1 (f(y )) Y = Y f(ã) and hence à f( B) and hence B à B = f 1 (A) f 1 (B) = f 1 (A B) = f 1 ( ) = Moreover the sets A and B are relatively open in f(y ). By lemma 25 their preimages f 1 (A) and f 1 (B) are relatively open in Y. Thus the pair Ã, B witnesses that the set Y is disconnected. By the contrapositive we see: if the set Y is connected, then the set f(y ) is connected. A second proof for proposition 37. We prove: if the set f(y ) has a separation, then the set Y has a separation. Thus it is assumed that (11.4) A B = f(y ), A, B, A B =, A B = We put à := f 1 (A) Y and B := f 1 (B) Y and check that Ã, B is a separation of Y. à B = (f 1 (A) f 1 (B)) Y = f 1 (A B) Y = f 1 (f(y )) Y = Y f(ã) = f(f 1 (A) Y ) = A f(y ) = A and hence à f( B) = f(f 1 (B) Y ) = B f(y ) = B and hence B à B f 1 (A) f 1 (B)) f 1 (A) f 1 (B) = f 1 (A B) = f 1 ( ) = à B f 1 (A) f 1 (B)) f 1 (A) f 1 (B) = f 1 (A B) = f 1 ( ) = and hence à B = à B = We have used f 1 (A) f 1 (A) from item (5) of Proposition 27. Hence the set Y has a separation. By the contrapositive: if the set Y is connected, then the set f(y ) is connected. 91

92 Lemma 26. Let the subset Y X of the topological space X have the relative topology. The two point set {0, 1} is given the discrete topology. The set Y is connected if and only if the only continuous functions f : Y {0, 1} are the two constants. Proof. Suppose there exists a nonconstant continuous function f : Y {0, 1}. The sets f 1 (0) and f 1 (1) are both nonempty and open in the relative topology of Y. Clearly f 1 (0) f 1 (1) = Y. Hence these two sets are clopen, and thus the set Y is disconnected. Conversely, suppose that the set Y is disconnected. Hence there exist two nonempty disjoint relatively open set A and B such that A B = Y. We define a function f : Y {0, 1} by setting (11.5) f(x) = { 1 for x A; 0 for x B. For this function, the preimage of any set is either, A, B or Y which are all relatively open. Hence the function f is continuous and nonconstant. A third proof of proposition 37. We check that the only continuous functions g : f(y ) {0, 1} are the two constants. Assume that the function g : f(y ) {0, 1} is continuous. Since the function f : X Z is assumed to be continuous, by lemma 25, its restriction f : Y f(y ) is continuous, too. The composite function g f : Y {0, 1} is continuous. Since the set Y is assumed to be connected, the above lemma 26 yields that the function g f is a constant. Take the case g f 0. Thus for all points z f(y ), one gets g(z) = 0. In both cases g f 0 and g f 1 one checks that the function g is constant. By the lemma 26, we conclude that the set f(y ) is connected. Lemma 27 (Any connected set is subset of one part of any separation). Let space Z be disconnected and have the separation A, B. Assume the set Y Z to be connected and nonempty. There are two mutually exclusive cases: (i) either Y A and Y B = ; (ii) or Y B and Y A =. Proof. We assume that Y A and Y B, put à := A Y and B := B Y and check: à B = Y à B A B = à B A B = Since Y is assumed to be connected, Ã, B cannot be a separation of Y. Hence either (i) B Y = B = or (ii) A Y = à =. From Y Z = A B we conclude in the first 92

93 case (i) Y = (A Y ) (B Y ) A, and in the second case (ii) Y = (A Y ) (B Y ) = B Y B, as claimed. Proposition 38 ( spider theorem ). Let Z α be a nonempty family of connected nonempty sets with index α I. Assume these sets have a common point z Z α for all α I. Then the union Z := {Z α : α I} is connected. Proof. Assume there exists a separation A, B of the union Z and show that either A or B are empty. Thus there exists no separation, and Z is connected. By Lemma 27, for each index α, either (i) Z α A or (ii) Z α B holds. But both cannot hold since Z α is nonempty. Hence for all α I exactly one of the cases (i) or (ii) hold. But it is also impossible that there exist two distinct indexes γ for which Z γ A and β for which Z β B. This assumption leads to the contradiction z Z γ Z β A B =. There are left the two following possibilities: (a) Z α A holds for all α I; (b) there exists an index β for which Z β B. In this case we conclude Z β B holds for all β I; Case (a) implies Z A, hence B =. Case (b) implies Z B, hence A =. In neither case is the pair A, B a separation of Z. Hence Z is connected. Corollary 8. If two connected sets Z and W have a nonempty intersection, their union Z W is connected. 10 Problem Prove or disprove: (b) The intersection of any two connected sets is connected. It is enough to consider a counterexample using a circle. Answer. In the Cartesian plane R 2, we use the semicircles A = {(x, y) R 2 : x 2 + y 2 = 1 and y 0} B = {(x, y) R 2 : x 2 + y 2 = 1 and y 0} Both A and B are connected, but the intersection A B = {(1, 0), ( 1, 0)} is disconnected. Proposition 39. Let the subset Y be connected in the space X. Thus Y X. Let Z be any set such that Y Z Y. Then the set Z is connected. Especially, any maximal connected set is closed. 93

94 Proof. We assume that Z = A B has a separation A, B and derive a contradiction. By assumption, the set Y is connected and Y Z. Hence Lemma 27 tells that (i) either Y A and Y B = ; (ii) or Y B and Y A =. It is enough to consider the first case (i). Monotonicity of closure, and A B = imply Z Y A B = Z B A B = Hence A, B is not a separation of Z. This contradiction shows that Z is connected. Corollary 9. If the space X has a dense connected set, it is connected. Lemma 28. The Cartesian product of two connected spaces is connected. First proof of lemma 28. Assume the topological spaces X and Y are connected. Fix the points a X and b Y. Indeed the function f : X X {b} mapping point x X to f(x) = (x, b) is easily seen to be continuous. Hence the sets X {b} and {a} Y are connected in the topology of the product X Y. Since these two sets intersect, their union (X {b}) ({a} Y ) is connected by corollary 8. Take the family of sets S y = (X {y}) ({a} Y ) with indexes y Y. All members of this family are connected, and they intersect in the common point (a, b). Hence by the spider theorem 38, their union is connected. But their union is the entire Cartesian product since S y = X {y} {a} Y = X Y y Y y Y Second proof of lemma 28. Assume the topological spaces X and Y are connected. We check that the only continuous functions g : X Y {0, 1} are the two constants. Assume that the function g : X Y {0, 1} is continuous. Fix any point (a, b) X Y. The restrictions g X {b} and g {a} Y are continuous. Since both functions have a connected domain, lemma 26 tells that these functions are constant. Since their two domains intersect at point (a, b), the function g X {b} {a} Y is a constant. For any second point (c, d) X Y, the function g X {d} {c} Y 94

95 is a constant. Since (X {b} {a} Y ) (X {d} {c} Y ) {(a, d), (c, b)} the function g is constant on the union (X {b} {a} Y ) (X {d} {c} Y ) and hence g(a, b) = g(c, d) holds for any points (a, b) and (c, d) X Y. Thus we see that the function g is a constant. By lemma 26, we see that the product X Y is connected. Corollary 10. The product of finitely many connected spaces is connected. Definition 56 (Component). A subset A X of a topological space X is called a component if and only if (i) the set A is connected; (ii) if any set B such that A B X is connected, then B = A. In other words, a component is a maximal connected subset. Theorem 1. The relation is defined among the elements of a topological space by a b if and only if there exists a connected set A such that a A and b A. This is an equivalence relation. The equivalence classes are the components. They are a partition for the space X. All components are closed. Proof. The relation is clearly reflexive and symmetric. To check the transitivity, we assume that a b and b c. Thus there exists a connected set A such that a A and b A, and a connected set B such that b B and c B. The union A B of two intersecting connected sets is connected, according to corollary 8. Since a A B and c A B, we see that a c holds, as to be shown. The equivalence class [a] of an element a X is the set [a] = {A X : a A and A is connected.} By proposition 38 ( the spider theorem ), this set is connected. It is indeed a maximal connected subset, and hence a component. By proposition 39 any maximal connected set is closed. By the general theorem about equivalence, the equivalence classes, and hence the components, are a partition for the space X. Corollary 11. If a connected space has only finitely many components, these components are all clopen. 10 Problem Use Theorem 1 to prove Corollary

96 Proof of Corollary 11. Let Q 1, Q 2,..., Q N be the components of X. For any i = 1... N holds X \ Q i = k i Q k Hence the complement of any component Q i is a finite union of components. By Theorem 1, we know that any component is closed. Since a finite union of closed sets is closed, we see that both Q i and its complement are closed. Hence the components are clopen, as to be shown. Proposition 40. Given is the nonempty index set I and the connected topological spaces X α for all α I. Then the product space X = α I X α is connected, too. Lemma 29. Let J I be any nonempty finite subset of indices. Let z X be any point. The two point set {0, 1} is given the discrete topology. Any continuous functions f : X {0, 1} is constant on the set (11.6) S := {y X : y (α) = z(α) for α I \ J} Proof. It is easy to see that S is homeomorphic to the finite product α J X α. By Corollary 10 this finite product is connected. By lemma 26 the restriction f S of function f is constant. First proof of theorem 40. We use lemma 26 to confirm that X is connected. The two point set {0, 1} is given the discrete topology. The space X is connected if and only if the only continuous functions f : X {0, 1} are the two constants. Suppose towards a contradiction that there would exist a nonconstant continuous function f : X {0, 1}. Take any points x X such that f(x) = 0 and z X such that f(z) = 1. Since f is continuous, there exists a basic neighborhood B x on which f is identically zero. By the definition 41 of the product topology, the basic neighborhood is as follows. There exists a nonempty finite subset J I of indices, and for each index α J an open neighborhood V (α) N x(α) such that B x = {y α I X α : y(α) V (α) for all α J} The restriction f B x being constant means that for all y X holds (11.7) α J [y(α) V (α)] f(y) = 0 We define y X by setting (11.8) y (α) := { y(α) for α J z(α) for α I \ J We see that y B x. Hence equation (11.7) implies f(y ) = 0. On the other hand, both y S and z S. Hence lemma 29 yields f(y ) = f(z) = 1. This contradiction shows that the function f is constant. Now lemma 26 confirms that X is connected. 96

97 End of a second proof of theorem 40. If the product X is empty, we are ready. Otherwise let z X be any point. The set (11.9) D = {y X : y(α) = z(α) holds for all but finitely many α I} is dense by lemma 30 and connected by lemma 31. connected set is connected by corollary 9. A space X containing a dense Remark. We see that theorem 40 even holds without the axiom of choice. Lemma 30. Let z X be any point in the product space X = α I X α. The set D from equation (11.9) is dense in X. Proof. We check that the set D intersects every nonempty open set O X. Take any point x O. By the definition 41 of the product topology, the set O contains a basic neighborhood B x O, which is as follows. There exists a nonempty finite subset J I of indices, and for each index α J an open neighborhood V (α) N x(α) such that B x = {y α I X α : y(α) V (α) for all α J} We define x X by setting (11.10) x (α) := { x(α) for α J z(α) for α I \ J We see that x B x and x D both hold. Hence D O D B x holds for any nonempty open set. This confirms that the set D is dense. Lemma 31. Let z X be any point in the product space X = α I X α. Assume that the coordinate spaces X α are connected for all α I. The set ((11.9)) D = {y X : y(α) = z(α) holds for all but finitely many α I} is connected in X. Proof. For each finite subset J I, the set D(J) = {y X : y(α) = z(α) holds for all α I \ J} is homeomorphic to the finite product α J X α. By Corollary 10 this finite product is connected. Hence the set D(J) is connected. Since z D(J) for all J I, the spider theorem 38 yields that the union is connected. D = {D(J) : J I is a finite subset } 97

98 Definition 57 (Locally connected space). A topological space X is called locally connected at a point a X iff there exists a basis B a for the neighborhoods of point a consisting of connected sets. A topological space X is called locally connected iff the space is locally connected at all its points. Lemma 32. A topological space X is locally connected at a point a X if and only if each open neighborhood V N a contains a connected neighborhood a W V. 10 Problem Prove Lemma 32. Answer. Assume the topological space X is locally connected at point a X. Take any open neighborhood V N a. Since the connected neighborhoods are a basis for the neighborhoods at point a postulate (B3) tells that there exists a connected neighborhood W B a such that a W V. Conversely assume that each open neighborhood V N a contains a connected neighborhood a B V. Then the family B a := {B N a : B is connected} of connected neighborhoods satisfy the postulates (B1),(B2), as well as (B3). Indeed (B1) and (B2) are valid since B a N a. Moreover (B3) is valid since each open neighborhood V N a contains a connected neighborhood a B V. Proposition 41. In a locally connected space all components are open. Corollary 12. In a locally connected space all components are clopen. 10 Problem Complete the proof showing in a locally connected space all components are open. Explain why all components are both open and closed. Proof of Proposition 41. Let Q be any component of the locally connected space X. Take any point a Q. Since the entire space X is a neighborhood of a and the space is assumed to be locally connected, the point a has a connected neighborhood V N a. By the proof of Theorem 1, the component Q is the equivalence class of any of its points a Q: (11.11) Q = [a] for all a Q Moreover, the equivalence class [a] of any element a X is the union set [a] = {A X : a A and A is connected.} The connected neighborhood V N a is a possible choice for set A in this union. Hence V [a]. Together with remark (11.11) we conclude V Q. The entire last paragraph has shown that a Q V N a [V Q] In other words, we see that the set Q contains a neighborhood for each of its points. By lemma 5, the set Q is open. 98

99 Proof of Corollary 12. By Theorem 1, we know that any component Q is closed. Together with the proposition 41 proved above, we conclude that all components are indeed clopen. Lemma 33. For any a, b R, the interval [a, b] is connected in the standard topology for the reals. First variant of the proof. We assume towards a contradiction that A, B is a separation of the interval [a, b]. Without loss of separation, we may assume a A. The case that A = {a} and B = (a, b] can be excluded immediately. Indeed in that case a A B, contradicting the assumption that A, B is a separation. Let C := {x R : a x b and [a, x] A} and c := sup C The set C is nonempty since a C. Since a C [a, b], the supremum exists. We distinguish the cases: (a) a < c < b; (b) c = a; (c) c = b. and rule out each case by producing a contradiction to A B (A B) =. (a) a < c < b: The definition of the supremum implies that for all ɛ > 0, there exists x (c ɛ, c] C. Hence [a, c ɛ] [a, x] A and [a, c) A. By the definition of the closure for the standard topology of the reals, we get c A. The definition of the supremum yields [a, c + ɛ] A for all ɛ > 0. Hence there exists y [c, c + ɛ) such that y / A and hence y B. By the definition of the closure for the standard topology of the reals, this implies c B. From the last two paragraphs, we conclude c A B (A B), contradicting A B (A B) =. (b) c = a: The definition of the supremum yields [a, c + ɛ] A for all ɛ > 0. Hence there exists y [c, c + ɛ) such that y / A and hence y B. By the definition of the closure for the standard topology of the reals, this implies c B. We conclude a A B contradicting A B (A B) =. (c) c = b: The definition of the supremum implies that for all ɛ > 0, there exists x (c ɛ, c] C. Hence [a, c ɛ] [a, x] A and [a, c) A. This case implies [a, b) = A and {b} = B. A B (A B) =. Hence b A B, contradicting 99

100 Second variant of the proof. 13 We assume towards a contradiction that A, B is a separation of the interval [a, b]. If both a, b A, there exists y (a, b) B. If [a, y] would be connected, Lemma 27 would imply [a, y] A, impossible. Hence [a, y] is disconnected. We see that that A [a, y], B [a, y] is a separation of the interval [a, y]; and similarly A [y, b], B [y, b] is a separation of the interval [y, b]. Thus we are able to use the reasoning below to anyone of these separations and get a contradiction. We may assume a A and b B for the separation A, B of interval [a, b]. Let c := sup A This supremum exists. We distinguish the cases: (a) c = a; (b) a < c < b; (c) c = b. and rule out each case by producing a contradiction to A B (A B) =. (a) c = a: This case implies {a} = A and (a, b] = B. Hence a A B (A B) which is impossible. (b) a < c < b: The definition of the supremum implies that for all ɛ > 0, there exists x (c ɛ, c] A. By the definition of the closure for the standard topology of the reals c A. The definition of the supremum yields (c, b] A =. By the definition of a separation, A B = [a, b] and hence (c, b] B. The definition of the closure for the standard topology of the reals implies c B. We conclude c A B (A B), which is impossible. (c) c = b: The definition of the supremum implies that for all ɛ > 0, there exists x (b ɛ, b] A. By the definition of the closure for the standard topology of the reals b A. Since b B, we have obtain b A B (A B), which is impossible. Hence there can be no separation A, B of the interval [a, b] with a A and b B. By the remark at the beginning of the proof, there cannot be any separation A, B of the interval [a, b] at all. Hence the interval [a, b] is connected, as claimed. Lemma 34. Assume that the set S R is connected and a, b S. Then [a, b] S. Proof. Assuming there would exist a number c (a, b)\s, we put A := S [, c), B := S (c, + ] and get the separation of S. This contradicts the fact that S is connected. Hence [a, b] S. 13 I find this second proof even more tricky. It is mainly included to illustrate there is no simpler proof at all. 100

101 Theorem 2 (The general intermediate value theorem). Given a continuous map f : X R from a connected topological space X into the reals. If the values f(a) R and f(b) R are both assumed, then all intermediate values c (f(a), f(b)) are assumed as values of the function. Proof. We assume towards a contradiction that the value c lies between f(a) and f(b), but is not in the range of the function: c (f(a), f(b)) \ f(x). Define A = {x X : f(x) < c} and B = {x X : f(x) > c} Both sets are open since the intervals (, c) and (c, ) are open. Since the function f assumes the value f(a) < c as well as the value f(b) > c, both sets A and B are nonempty a f 1 (f(a)) A and b f 1 (f(b)) B Moreover A B = X. Hence {A, B} is a partition of X witnessing that X is disconnected. This is a contradiction. Hence all intermediate values c (f(a), f(b)) are assumed as values of the function: [f(a), f(b)] f(x). Definition 58 (totally disconnected set). A set in a topological space is called totally disconnected if and only if all its components are one-point sets. 10 Problem 11.9 (A mean question, in the honor of Russell). Given a continuous map f : X Q from a connected topological space X into the rational numbers. Is the following true or false: If the distinct values f(a) Q and f(b) Q are both assumed, then all intermediate values c (f(a), f(b)) are assumed as values of the function. Answer. This statement is vacuously true. Indeed such a function does not exist. The rational numbers are a totally disconnected set. Hence the only continuous functions from a connected space into the rational numbers are the constant functions. 10 Problem Show that both the set Q, as well as the set of real irrational numbers R \ Q both are totally disconnected. Show that the only continuous functions from a connected space into the rational numbers are the constant functions. Proposition 42 (The only connected subsets of the reals are the intervals). The only connected subsets of the real numbers in the standard topology are with any a, b R., R, (, a], (, a), (a, ), [a, ), [a, b], [a, b), (a, b], (a, b) 101

102 For the purpose of this proof: For any x, y R, let [x, y] = [min{x, y}, max{x, y}] denote either a one-point set or a nonempty interval. We prove the enumerated sets to be connected. Let the set S be any to the cases from the proposition. Since the empty set is connected, we may assume x S. Let a := inf S [, x] and b := sup S [x, ] and T := {[x, y] : y S} The definition of T implies S T. Because S is one of the sets from the proposition (indeed convex), the inclusion T S holds, too, and hence S = T. Since the intervals [x, y] are connected by Lemma 33, the union T = S is connected by the spider theorem, proposition 38. We prove that all connected subsets of the reals have been enumerated. Given is a connected set S R. Since the empty set is connected, we may assume x S. By the Lemma 34, we conclude that [x, y] S for all y S. Hence Let Since S = {[x, y] : y S} a := inf S [, x] and b := sup S [x, ] (a, b) R {[x, y] : y S} [a, b] R we see that the set S has to be of one of the cases from the proposition. Corollary 13 (The standard intermediate value theorem). Given a continuous map f : I R, where the domain I is anyone of the nonempty intervals enumerated in proposition 42. If the two distinct values f(a) f(b) are assumed, then all intermediate values are assumed. In other words, for all values c (f(a), f(b)), there exists z I such that f(z) = c. Corollary 14 (The range of a real-valued continuous function). For any continuous map f : X R from a connected topological space X into the reals, the range of this function is an interval. Any one of the nonempty intervals from proposition 42 may turn out to be the range f(x). Especially, a continuous function f : I R with any real interval I as domain, has such a real interval as range, too. 10 Problem Prove that any continuous mapping f : [0, 1] [0, 1] has fixed point. 102

103 Proof. In the case that either 0 or 1 is a fixed point, we are ready. Otherwise f(0) > 0 and f(1) < 1. The function g(x) := f(x) x has a positive value g(0) > 0 and a negative value g(1) < 0. Clearly, the function g : [0, 1] R is continuous. By the standard intermediate value theorem, corollary 13, there exists z (0, 1) such that g(z) = 0. Hence f(z) = z and z is indeed a fixed point. Lemma 35. The spaces [0, 1] and [0, 1] 2 are not homeomorphic. Reason. By proposition 42 the only connected subsets of the reals are the intervals. Hence in the space [0, 1], the intersection of any two connected subsets is connected. In the space [0, 1] 2 there exist two connected subsets A and B such that their intersection A B is disconnected. An example is provided by problem Hence the spaces [0, 1] and [0, 1] 2 are not homeomorphic. 103

104 12 Pathwise connected spaces Definition 59 (path). Let X be any topological space. A path in X from point a X to the point b X is a continuous map p : [0, 1] X such that p(0) = a and p(1) = b. Definition 60 (pathwise connected space). A topological space is called pathwise connected iff there exists a path between any two of its points. Lemma 36. The image of a pathwise connected topological space is pathwise connected. Proof. Let the topological space X be pathwise connected and let f : X Y be a continuous mapping to any second topological space Y. Take any two points in the image f(a) and f(b) with a X and b X. By assumption there exists a path p : [0, 1] X such that p(0) = a and p(1) = b. The composite function f p : [0, 1] f(x) is a path in f(x) from f(a) to f(b). Lemma 37. Any pathwise connected topological space is connected. First attempt. Let the topological space X be pathwise connected. We assume, towards a contradiction that the space X is disconnected. Hence there exists a partition {A, B} of space X into disjoint nonempty open sets A and B. Take any points a A and b B. Since X is pathwise connected, there exists a path p : [0, 1] X from p(0) = a to p(1) = b. By lemma 33, the interval [0, 1] is connected. By proposition 37, the continuous image of a connected set is connected. Hence each one image p[0, 1] is connected. By the lemma 27, the connected set p[0, 1] is contained in one part of the separation. There are two mutually exclusive cases: (i) either p[0, 1] A and p[0, 1] B = ; (ii) or p[0, 1] B and p[0, 1] A =. But both possibilities contradict to p(0) = a A and p(1) = b B. This contradiction shows that no separation exists. Hence the space X is connected. Make it short. Let the topological space X be pathwise connected. Take any point a X and define the set of points reachable from a: A = {b X : there exists a path p from a to b } = {p[0, 1] X : p is any path from a to b } By lemma 33, the interval [0, 1] is connected. By proposition 37, the continuous image of a connected set is connected. Hence each one image p[0, 1] is connected. Now the spider theorem 38 yields that their union set A is connected. On the other hand, the assumption that the topological space X is pathwise connected tells that X = A. Hence X is connected. 104

105 For a functors fan. We prove that the only continuous functions g : X {0, 1} are the two constants g 0 and g 1. Take any points a X and b X. Since X is pathwise connected, there exists a path p : [0, 1] X from p(0) = a to p(1) = b. The composite function g p : [0, 1] {0, 1} is continuous. By lemma 33, the interval [0, 1] is connected. Hence by lemma 26, the function g p is a constant. Hence g(b) = g(p(1)) = g(p(0)) = g(a) holds for any two points a, b X. Hence the function g is constant. By the lemma 26, this fact implies that the space X is connected. My last attempt. Let the topological space X be pathwise connected. We assume towards a contradiction that the space X is disconnected. Hence there exists a partition {A, B} of space X into disjoint nonempty open sets A and B. Take any points a A and b B. Since X is pathwise connected, there exists a path p : [0, 1] X from p(0) = a to p(1) = b. Let I 0 = {r [0, 1] : p(r) A} = p 1 (A) and I 1 = {r [0, 1] : p(r) B} = p 1 (B) Thus we get a partition {I 0, I 1 } of the unit interval [0, 1] into disjoint nonempty open sets. This is impossible since the interval [0, 1] is connected by lemma 33. This contradiction shows that no separation exists. Hence the space X is connected. Is the converse true? Is every connected set pathwise connected? Obviously, this holds in special cases, as for example subsets of the reals. But in general this statement is not true. Here is a counterexample: Lemma 38. Let X = {(x, y) R 2 : 0 < x 1 and y = sin(x 1 )} and Y = {(0, y) R 2 : 1 y 1} The union X Y is connected, but not pathwise connected. Proof. Both sets X and Y are pathwise connected, and hence connected. The set X is not closed, and X = X Y. The set Y is closed. Hence X Y = Y, and the pair X, Y is not a separation. Does there exist any other separation of X Y? Assume towards a contradiction the pair A, B to be a separation of Z = X Y. Since the set Y is connected, lemma 27 yields (i) either Y A and Y B = ; (ii) or Y B and Y A =. Without loss of generality, we may assume that Y B. Thus case (ii) occurs. Moreover Y B and A X. Since the set X is connected, too, lemma 27 yields 105

106 (i) either X A and X B = ; (ii) or X B and X A =. This time case (i) occurs. Together with the first argument we finally get A = X and B = Y. But we have seen above that the pair X, Y is not a separation. Hence the space X Y has no separation and is connected. Assume towards a contradiction that the union X Y is pathwise connected. Hence there exists a path p : [0, 1] X Y from point p(0) = (0, 0) Y to point p(1) = (1, sin 1) X. For better notation, denote the coordinate functions of the path by p(t) = (x(t), y(t)). Let t be the supremum of parameter values for which the path lies in part Y : t = sup{t [0, 1] : x(t) = 0} Since p(1) / Y, we conclude t < 1. The path is continuous at parameter t. Hence x(t ) = 0. By definition of the supremum, t > t implies x(t) > 0 and hence y(t) = sin(x(t) 1 ). We may choose any ɛ > 0, for example ɛ = 1. There exists δ > 0 such that 2 t t < δ implies x(t ) x(t) < ɛ and y(t) y(t ) < ɛ t t < t + δ implies 0 < x(t) < ɛ and sin(x(t) 1 ) y(t ) < ɛ But there exist both a value x 1 (t) (0, ɛ) for which sin(x 1 (t) 1 ) = +1, as well as a value x 2 (t) (0, ɛ) for which sin(x 2 (t) 1 ) = 1. We get the contradiction 2 = sin 1 x 1 (t) sin 1 x 2 (t) sin 1 x 1 (t) y(t ) + sin 1 x 2 (t) y(t ) < 2ɛ = 1 Hence there is no continuous path connecting point (0, 0) to point (1, sin 1) inside the set X Y. Thus the union X Y is not pathwise connected. Proposition 43. Any open, connected domain of R n is pathwise connected. Moreover, each two points of the domain can be connected by a path lying inside the domain. Proof. Let D R n be an open, connected set. We prove that for each two points a, b D, there exists a polygonal path connecting a to b, and lying inside the domain. Fix point a D, and define the set of points reachable from a: A = {x D : there exists a polygonal path p from a to x lying inside D. } The set A is open. Take any point x A. Since A D and D is open, there exists ɛ > 0 such that B(x, ɛ) D. But any point y B(x, ɛ) can be connected to point x by a straight segment since a ball is convex. Hence there exists a polygonal path p from a to x, and going on straight from x to y, and lying inside D. This tells that y A and B(x, ɛ) A. The set A is relatively closed in D, in other words A D A. Take any point x A D. Since D is open, there exists ɛ > 0 such that B(x, ɛ) D. Since x A, 106

107 there exists a point y B(x, ɛ) A. The points x and y can be connected by a straight segment, which lies inside B(x, ɛ) D. Hence there exists a polygonal path p from a to y, and then straight from y to x, and lying inside D. Hence x A and A D A. We have shown that the set A is both and relatively closed in D. Since the domain D is assumed to be connected, we conclude that A = D. Thus there exists a polygonal path from a to any point x D. 107

108 13 The Hilbert curve Figure 2: Three steps towards the Hilbert curve. Theorem 3. There exists a continuous mapping from the unit interval onto the unit square. An example of such a mapping provides the Hilbert curve. Construction of the Hilbert curve. For each natural number n, we successively construct piecewise linear mappings p n from the unit interval [0, 1] into the unit square [0, 1] 2. We agree to set p n (0) = (0, 0) and p n (1) = (1, 0) for all natural n. In the first step, the mapping p 1 is defined. We divide the unit interval into four congruent intervals [0, 1], [ 1, 1], [ 1, 3], [ 3, 1]; and the unit square into four congruent squares, each one with side length 1. These four squares are ordered in the following 2 manner: the first square Q 1 (1) has one vertex (0, 0), the fourth square Q 1 (4) has one vertex (1, 0). The squares Q 1 (2) and Q 1 (3) are chosen in a way such that two successive squares always have a common side. Obviously there are two ways to proceed, of which 108