Topology Math Conrad Plaut

Transcription

1 Topology Math Conrad Plaut

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3 Contents Chapter 1. Background 1 1. Set Theory 1 2. Finite and Infinite Sets 3 3. Indexed Collections of Sets 4 Chapter 2. Topology of R and Beyond 7 1. The Topology of R 7 2. Continuous Functions Other Topologies on R 14 Chapter 3. Basic Topological Concepts Bases Closed Sets and Separation Axioms Closures of Sets Continuous Functions Revisited Sequences Summary of Examples 34 Chapter 4. Products and Metrics Finite Cartesian Products Metric Spaces The General Product Topology Completeness 47 Chapter 5. Compactness Compact Spaces Compactness and Metric Spaces Finite Products of Compact Spaces 57 Chapter 6. Connectedness Connected Spaces Path Connected Spaces Local Forms of Connectedness Final Summary of Examples 71 Index 73 iii

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5 CHAPTER 1 Background 1. Set Theory The student is expected to know elementary logic and set theory. For purposes of notation we recall some basic definitions and theorems. The natural numbers are N := {1, 2, 3,...}. The integers are Z := {..., 2, 1, 0, 1, 2,...} and the rational numbers Q consist of all real numbers equal to a quotient p q, where p an q 0 are integers. The real numbers are denoted by R. We write x X to express x is an element of the set X and x / X to express the negation of this statement. We will generally use capital letters to represent sets and small letters to represent elements of sets. Given sets A and B, the union of A and B is defined to be A B := {x : x A or x B}. The intersection of A and B is defined by A B := {x : x A and x B}, and the complement of B in A (more briefly A take away B ) is defined by A\B := {x : x A and x / B}. The empty set is denoted by. If A B = then A and B are said to be disjoint. We say that A is a subset of B (written A B) if whenever x A, x B. We say that A is a proper subset of B if A B and there exists some x B such that x / A. We denote this by A B. If X is a set and A is a subset of X, we define the complement of A in X by A c := {x : x X and x / A}. Note that by definition, A c = X\A. One of the most basic theorems in set theory, reflecting the fact that the negation of an and statement is an or statement, and vice versa, is de Morgan s laws: (1.1) A\(B C) = (A\B) (A\C) and A\(B C) = (A\B) (A\C) Additional statements that we will need are are A B = B A, A (B C) = (A B) C and the identical statements with every replaced by (i.e. the commutative and associative laws for sets). The distributive laws are, for all sets A, B, C: (1.2) A (B C) = (A B) (A C) and A (B C) = (A B) (A C) The next lemma is very useful; the proof is not hard and the statement should be clear from the picture that follows it. L 1. (Disjoint Union Lemma) If A and B be are sets then (1) A B = (A\B) (A B) (B\A) where any two of the sets on the right side of the equation are disjoint. (2) A = (A B) (A\B) and if B A then this formula reduces to A = B (A\B). 1

6 2 1. BACKGROUND F 1. Disjoint Union Lemma A function f : X Y is called one-to-one (or 1-1, or an injection) provided the following holds: If f(x) = f(y) for some x, y X then x = y. The function f is called onto (or a surjection) if for every y Y there exists some x X such that y = f(x), i.e. if Y is the range of X. A function that is both 1-1 and onto is called a bijection (or one-to-one correspondence). The set X is called the domain of f. If f : X Y is a function and A X we define f(a) = {y Y : y = f(x) for some x A}, called the image of A. If B Y, we define f 1 (B) = {x X : f(x) B}, called the inverse image of B. We list now list various statements involving images and inverse images of functions: (1.3) f (A B) = f(a) f(b) and f (A B) f(a) f(b) (1.4) f 1 (A B) = f 1 (A) f 1 (B) and f 1 (A B) = f 1 (A) f 1 (B) (1.5) f 1 (A\B) = f 1 (A)\f 1 (B) and f(a)\f(b) f(a\b) The second statements in (1.3) and (1.5)are set equalities provided f is oneto-one. For f : X Y we also have that f(f 1 (B)) B with equality when f is onto, while A f 1 (f(a)) with equality when f is one-to-one. When f : X Y is a bijection, f 1 has a separate meaning, namely the inverse function of f. That is, f 1 : Y X is the unique function such that f(f 1 (y)) = y and f 1 (f(x)) = x hold for all x X and Y y. If f is not a bijection then f 1 (y) doesn t actually make sense. However, we will sometimes use this notation as shorthand for f 1 ({y}), which always does make sense and is the set {x X : f(x) = y}. For any set X, the identity function id X is the function defined by id X (x) = x for all x X. Of course id X is a bijection and is equal to its own inverse. Given functions f : X Y and g : Y Z, we define the composition of f and g to be g f : X Z, where (g f)(x) = g(f(x)) for all x X. If f : X Y is a bijection then we can translate the above two conditions for f 1 using the the identity function: f f 1 = id Y and f 1 f = id X. The composition of 1-1 functions is 1-1 and the composition of onto functions is onto. Moreover, if f and g are bijections then f g is a bijection and (f g) 1 = g 1 f 1.

7 2. FINITE AND INFINITE SETS 3 Finally, let f : X Y be a function and A X. The restriction of f to A is the function f A : A Y defined by f A (x) = f(x) for all x A. That is, the restriction of f is simply f itself with its domain reduced to the set A. Here are some refresher exercises: E 1.1. Prove the statements (1.5), including equality for the second one when f is 1-1. E 1.2. Let f : X Y and g : Y Z be functions. Prove that if g is 1-1 and g f is onto then f is onto. E 1.3. Prove that a function f : X Y is 1-1 (resp. onto) if and only if for every set A X, f 1 (f(a)) = A (resp. for every B Y, f(f 1 (B)) = B). Hint: For the if parts, let A = {a, b} where f(a) = f(b) (resp. B := {y} for some y B). E 1.4. Prove that if f : X Y is a bijection then for any y Y, f 1 ({y}) = {f 1 (y)}. E 1.5. Let f : X Y be a function. (1) Prove that if f is 1-1 and A X then f A is 1-1. (2) Give an example of a function f : X Y that is onto, with a proper subset A X such that f A is also onto. 2. Finite and Infinite Sets We will state without proof the facts that we will need. D 1. We say that sets non-empty sets A and B have the same cardinality if there exists a bijection from A to B. If A is empty we say that A is finite with cardinality c(a) = 0. If A has the same cardinality as a set {1, 2,..., n} then we say A is finite with cardinality c(a) = n. Since the identity function from {1,..., n} is a bijection, it follows that c({1,..., n}) = n. The next statement is the most important basic theorem about finite sets. It should seem reasonable (although the proof requires some tedious effort), given the Disjoint Union Lemma and the fact that when we count the elements of A and B separately we double count any elements in the intersection. T 1. Let A and B be finite sets. Then A B and A B are finite, and c(a B) = c(a) + c(b) c(a B). In particular, if A and B are disjoint then c(a B) = c(a) + c(b). C 1. If A is finite and B is a proper subset of A then B is finite and c(b) < c(a). Of course it follows from Theorem 1 that the union of two finite sets is finite. The idea of the union of two sets may be extended to that of finitely many sets, as follows: For sets A 1,..., A n, define A 1 A n := {x : x A i for some 1 i n} The intersection of finitely many sets may defined similarly, replacing for some by for all. Later we will definte unions and intersections of arbitrary collections of sets. It will be shown in an exercise that the union of finite sets is finite.

8 4 1. BACKGROUND The following statement seems almost too simple to be useful, but it is heavily used in discrete mathematics and computer science. The name comes from the following interpretation: Suppose there are n pigeons and fewer than n holes. If each pigeon is placed in a hole then at least two pigeons will share a hole. C 2. (Pigeonhole Principle) If f : X Y is a function between non-empty finite sets and c(y ) < c(x) then f is not 1-1. If a set A is not finite we say it is infinite. Note that the natural numbers N cannot be finite. In fact, if N were finite of cardinality n then the fact that N contains {1,..., n} as a proper subset would contradict Corollary 1. If A has the same cardinality as the natural numbers N, then we say A is countably infinite. If A is either finite or countably infinite then we say A is countable. Otherwise, we say that A is uncountable. A bijection from N to a set A is, in effect, counting the elements of A. After all, exactly one element is being assigned 1, a different element is assigned 2, and so on. Therefore it makes sense to refer to a set that is infinite and for which no such bijection exists as uncountable. Note that Corollary 1 has no analog for infinite sets. In fact, N is a proper subset of Z but these two sets have the same cardinality which is verified by the bijection from N to Z that takes every even natural number 2m to m and every odd natural number 2k + 1 to k. For most of human history, the idea of infinity was not considered within the realm of mathematics, and even when it was, the idea that there could be uncountable sets was shocking to mathematicians. In the 19th century Georg Cantor proved not only that any close real interval [a, b] with a < b (and hence R) is uncountable, but that there are there are infinitely many different possible cardinalities of sets. In particular, given any set S, the set of all subsets of S has a greater cardinality than S (this is not the place to make a statement like that precise). We will primarily only be interested in the distinction between countable and uncountable sets. It may seem obvious, but does require proof, that any subset of of a countable set is countable, and any set that contains an uncountable set is uncountable. It is a useful fact that a set A is countable if and only if there is a surjection σ : N A (so σ doesn t need to be injective). From this one may conclude that if B is countable and there is a surjection from B onto a set A then A is countable. E 1.6. Prove Corollary 1. You may assume that a subset of a finite set is finite. Hint: Use Theorem 1 and the Disjoint Union Lemma. E 1.7. Prove that if A = A 1 A n and each A i is finite then A is finite. Hint: Use induction. E 1.8. Prove the Pigeonhole Principle. Hint: Prove the contrapositive and use the fact that f is onto f(x) Y. 3. Indexed Collections of Sets An indexing makes precise the idea of labeling a collection of sets. This is particularly important when considering infinite collections of sets. D 2. An indexing of a collection A of sets with indexing set Λ is an onto function ι : Λ A. We normally write A λ rather than A = ι(λ) and denote the indexed collection by {A λ } λ Λ.

9 3. INDEXED COLLECTIONS OF SETS 5 Thus the indexing of A assigns to each λ a unique set A λ in the collection A. Note that it is possible that A λ = A λ when λ λ (i.e. the indexing is not 1-1). E 1. Consider the collection {( n, n)} n N of all real intervals ( n, n), where n is a natural number. Written more explicitly this is the collection {( 1, 1), ( 2, 2), ( 3, 3),...}. As another example, {[ r, r]} r (0,1) is the collection of all closed intervals having endpoints r, r, where 0 < r < 1. Given an indexed collection of sets {A λ } λ Λ, we define the intersection of the collection to be A λ := {x : x A λ for all λ Λ} λ Λ and the union of the collection to be A λ := {x : x A λ for some λ Λ}. λ Λ When Λ = N (or Λ is finite) we will often write, for example, A i (or m A i). E 2. Let A n := [ n, n]. Then n=1 A n = R. In fact, if x A n then x [ n, n] for some n, and hence x R, so n=1 A n R. For the opposite inclusion note that if x R, x [ n, n] for some n N by the unboundedness of N. But then by definition x A n. E 3. We will prove that for sets A and {A λ } λ Λ we have the following distributive law: ( ) A A λ = (A A λ ) λ Λ λ Λ We have x A ( λ Λ A λ) if and only if x A and x λ Λ A λ x A and x A λ for some λ Λ x A A λ for some λ Λ x λ Λ (A A λ ). E 1.9. Verify that if we have only two sets A 1 and A 2 (i.e. the indexing set is Λ = {1, 2}) then λ {1,2} A λ = A 1 A 2 and λ {1,2} A λ = A 1 A 2. Therefore our new more general definitions are consistent with the old ones. E Let {A λ } λ Λ be a collection of sets. Show that for each λ 0 Λ, λ Λ A λ A λ0 λ Λ A λ. E Find the intersections and unions of the following collections (no need to write down a proof, but be careful about your answers!) (1) {( n, n)} n=1 (2) {[ r, r]} r (0,1) (3) {(0, 1 i ]}

10 6 1. BACKGROUND E Prove the following generalization of the de Morgan law for arbitrary intersections A\ ( λ Λ A ) λ = λ Λ (A\A λ). Formulate the appropriate generalizations of the other de Morgan law, and the remaining distributive law not considered in the above example, but don t bother to write down proofs unless you need extra practice. If ι : Λ A is an indexing with Λ countably infinite then there is a bijection g : N Λ. The function ι g : N A is an indexing of the same collection A with (possibly more convenient) indexing set N. Reindexing does not, of course change the collection A in any way and we will therefore use N as our indexing set when proving general theorems about countable collections. Likewise if Λ is finite with cardinality n then we may assume that Λ = {1,..., n}. The next proposition requires a somewhat tedious proof, which we will omit. P 1. If {A i } i Λ is a countable collection of countable sets then A i is countable. i Λ Let Q n := { p n : p Z}. The function p p n is surjective from Z Q n and therefore Q n is countable. But Q = Q n and we obtain: C 3. The set of rational numbers Q is countably infinite. We conclude with the following observation, which illustrates why one must be so precise in mathematics. You know from earlier classes that between every two irrationals there is a rational, and it is not hard to prove that between every two rationals there is an irrational. These two subsets seem to be evenly spaced and a casual observer might take this to mean that there are the same number of irrationals as rationals. However, since we have made precise the idea of same number to mean same cardinality, which we have clearly defined, and carefully proved some theorems about this concept, we can now prove that this statement is false: We know that the rationals are countable. But since the real numbers are the union of the rationals and the irrationals, and the real numbers are uncountable, then the irrationals must be uncountable. It may well have been vague arguments like the one above, using ill-defined terms to make incorrect claims that sound reasonable and support preconceived notions, that lead many otherwise smart people to reject the proofs of Cantor near the end of the 18th century.

11 CHAPTER 2 Topology of R and Beyond In this chapter we will define the standard topology of R, and use it as a motivation to introduce, in a limited way, some basic concepts from general topology. Beginning with the next chapter we will revisit many of these basic concepts, expand on them, and add new concepts. In mathematics one encounters many sorts of functions. In calculus, the first functions that are considered are real functions functions from intervals in R back into R. Among the most important types of real functions are continuous functions. Polynomials, trigonometric and exponential functions and their inverse functions are all continuous, as are functions described by power series. Many important theorems, such as the Intermediate Value Theorem and Max-min Theorem, are true for continuous functions. The basic assumptions of topology rest on the following basic question: what additional structure is necessary for a set to have in order for some reasonable notion of continuity to be defined. The answer to this quesion is shockingly simple. A set S must have a collection of subsets, called open sets, that satisfy only the following three axioms: A 1. The empty set and the set S are open. A 2. The union of any collection of open sets is open. A 3. The intersection of a finite collection of sets is open. This collection of open sets is called the topology of S, and S, together with its topology is called a topological space. At this point it may not be clear to the reader why these three axioms have anything at all to do with continuity. After all, doesn t the definition of continuity for real functions require δ s and ε s? In fact these axioms gradually appeared as the right set of assumptions in the early part of the 20th century, as mathematicians gained experience with the incredibly wide variety of situations in which continuity is a legitimate and useful concept. In this way the important field of general topology was born. 1. The Topology of R Before we study topological spaces in the abstract, we will have a look at open sets in R and continuity of real functions. We will show that open sets in R do satisfy the above three axioms and that continuity of real functions can be equivalently expressed using only the concept of open sets. D 3. A subset U of R is called open if for every x U there is some ε > 0 such that (x ε, x + ε) U. In other words, a set is open if there is a little room in the set around each of its points. Note that the size of ε may depend on the point. In our discussions 7

12 8 2. TOPOLOGY OF R AND BEYOND about R we will often use elementary facts about intervals and inequalities. Some that we will use very frequently are contained in the next exercise. Note that by an open interval we mean the set (a, b) = {x R : a < x < b}. The definition forces a < b, which we will always assume, and we will also consider infinite open intervals in which a = and/or b =. Note that is not a real number; in this text it is simply a notation used to define intervals and limits. We use the following conventions: x > and x < for all x R. For example, (, 1) = {x : x < 1 and x > }. Since the second condition is always true we may eliminate it to get (, 1) = {x : x < 1}. E 2.1. Prove the following (be sure to consider the cases when the interval has infinite endpoints ): (1) Let x R and ε > 0. Then a (x ε, x + ε) if and only if a x < ε. (2) Suppose (a, b) is an open interval. If x < y < z and x, z (a, b) then y (a, b). (3) Prove that if a < c < d < b then (c, d) (a, b). For starters, let s observe that every open interval (a, b) is open according to our new definition. (The fact that it is called an open interval is suggestive, but certainly doesn t prove anything. As the student may well have noticed, certain words are used in different ways in different contexts in mathematics.) First suppose that a and b are finite. Let x (a, b). By definition this means that a < x < b. Let ε := min{x a, b x} > 0. Then we have ε x a x ε a and ε b x x + ε b. Now if y (x ε, x + ε), y (a, b) by a previous exercise. This proves that (x ε, x + ε) (a, b). Note that the proof is quite clear if one looks at the picture: This picture shows the case when x a < b x, and the small interval (x ε, x+ε) = (a, x + ε) clearly lies in (a, b). Pictures are extremely useful in topology as a way to decide whether or not a theorem is true and sometimes to figure out how to prove it. But of course a picture is not the same thing as a proof, and the details must always be written down. E 2.2. Sketch the corresponding picture for the cases when b x < x a and b x = x a. E 2.3. Prove that any interval of the form (, a) is open. This will follow from Axiom 2, which will be proved below, since (, a) = ( n, a), but you may not use Axiom 2 in your proof. A similar argument shows that an interval of the form (a, ) is open. On the other hand, open sets need not be open intervals. For example, the union of two intervals A = (0, 2) (4, 6) is not an open interval by Exercise But is it open? Rather than go through a rather tedious proof in this special case, n<a

13 1. THE TOPOLOGY OF R 9 we will more generally verify Axiom 2, which will prove that A is open since it is a union of open intervals. To see why Axiom 2 is true, let U := U λ, where each U λ is open. Let x U. This means that x U λ0 for some λ 0 and so there exists some ε > 0 such that (x ε, x + ε) U λ0. But adding Exercise 1.10 gives (x ε, x + ε) U λ0 U. Now we can construct lots of open sets, for example ( 1 n + 1, 1 ) n Moreover, since R = n=1 n=1 λ Λ ( n, n), we see that R is open, one of the requirements of Axiom 1. What about the other requirement of Axiom 1? There are no points in the empty set, so how can it be open? In fact, the lack of points is precisely why the empty set is open, by virtue of the vacuous hypothesis. Since there are no points in, every point x in has whatever property you want to give it, including the impossible property that there exists some interval (x ε, x + ε)! Yes, things that don t exist can satisfy impossible properties. If you still don t believe it, consider a proof by contradiction. If were not open then by definition there would have to exist some x such that (x ε, x + ε) for some ε > 0, which contradicts the fact that has no elements. We are left with proving Axiom 3. Suppose that U 1,..., U n are open sets in R. Suppose that x U 1 U n. That is, x U i for all i. So for each i there exists some ε i > 0 such that (x ε i, x + ε i ) U i. Now let ε := min{ε 1,..., ε n }, which is positive since ε = ε i for some i. Moreover, (x ε, x + ε) (x ε i, x + ε i ) U i for all i. In other words, (x ε, x + ε) n U i, which finishes the proof. D 4. The topology consisting of the open sets from Definition 3 is called the standard topology on R. Unless otherwise stated, we will always take the standard topology on R. What about infinite intersections of open sets? It is easy to see where the above proof breaks down. If we have infinitely many ε i then there might be no minimum in fact the infimum could be 0, in which case we are in trouble. Let s consider a concrete counterexample to the statement that any intersection of open sets is open. Let U i = ( 1 i, 1 i ). Then U i = {0}, which is not open. In fact, any interval (0 ε, 0 + ε) = ( ε, ε) with ε > 0, contains points (even rational numbers, by density of the rationals) different from 0 so cannot be contained in {0}. E 2.4. Prove that every open subset of R is a union of open intervals. E 2.5. Prove that the set of rational numbers is not open. For real numbers a b, we define the closed interval [a, b] := {x : a x b}. Half-open intervals such as (a, b] are defined similarly (provided a < b). We will discuss the term closed a bit more in the future. E 2.6. Prove that any closed interval [a, b] is not open. argument shows that half-open intervals are not open either. A similar

14 10 2. TOPOLOGY OF R AND BEYOND E 2.7. Prove that if A is open in R and a R, then a + A := {a + x : x A} is open. E 2.8. Prove that for any x R, {x} is not open. 2. Continuous Functions In calculus, assuming you took a course that still believes in δ s and ε s, you would have learned something like the following definition of continuity for real functions. For simplicity we will start with functions defined on all of R. For any x R, we say that f : R R is continuous at x if for every ε > 0 there exists some δ > 0 such that if x y < δ then f(x) f(y) < ε. Intuitively speaking (which might be the only way your calculus text described continuity), if points are close to x then their functional values are close to f(x). Or put another way, which we will make more precise later with a notion of limits, if y approaches x then f(y) approaches f(x). Such intuitive descriptions may help you visualize or even understand a mathematical concept, and may be all that, say, an engineer, needs to know, but as a mathematics student, complete understanding must include the ability to work with precise definitions. The next proposition shows that continuity of real functions may be equivalently defined using only the topology of R. P 2. Let f : R R. Then f is continuous at x R if and only if for every open set U containing f(x) there is an open set V containing x such that f(v ) U. P. Suppose first that f is continuous at x and let U be open containing f(x). Then there is some ε > 0 such that (f(x) ε, f(x) + ε) U. By definition of continuity, there is some δ > 0 such that if x y < δ then f(x) f(y) < ε. But x y < δ is equivalent to y (x δ, x + δ), which is an open set that we will call V. Now if y V then f(x) f(y) < ε, which is equivalent to f(y) (f(x) ε, f(x) + ε) U. That is, f(v ) U. Now suppose that for every open set U containing f(x) there is an open set V containing x such that f(v ) U. Let ε > 0. Then U := (f(x) ε, f(x) + ε) is an open set, so there exists some open V containing x such that f(v ) U. By definition of open there is some δ > 0 such that (x δ, x + δ) V. Now if x y < δ then y (x δ, x + δ) V, and (2.1) f(y) f(v ) U = (f(x) ε, f(x) + ε) Equivalently, f(y) f(x) < ε. The above proposition is a fine start, but an immediate question comes to light when thinking about problems from calculus. Often basic calculus problems, such as optimization, involve functions defined on subsets of R (often restrictions of functions to closed, bounded intervals). The problem with our earlier δ,ε-definition of continuity is best illustrated with an example. E 4. Let f be the restriction of the real function g(x) = 2x to [0, 2]. You may well have checked continuity of a function like this in calculus class. If x y < ε 2 then f(x) f(y) = 2x 2y < ε so we may just choose δ = ε 2, regardless of x. But there is a small problem. If we take, for example, x = 0 then for any ε > 0 we may choose, there are plenty of points y such that x y < ε 2 but

15 2. CONTINUOUS FUNCTIONS 11 y < 0, so f(y) is not even defined! Your calculus class may have dealt with such problems by requiring in advance that f be defined on an open interval containing the set in quesion (which in this case is of course true). But this assumption is entirely unnecessary and there is a much simpler, better way. D 5. Let A R be a set and f : A R be a function. For x A we say f is continuous at x if for every ε > 0 there exists some δ > 0 such that if x y < δ and y A then f(x) f(y) < ε. So we have simply used the definition to force f(y) to be defined! Put another way, we require (2.2) if y (x δ, x + δ) A then f(x) f(y) < ε This is a reasonable strategy if you consider intuitively what continuity is supposed to represent. As y approaches x we want f(y) to approach f(x). But even if f happens to be the restriction of some function defined outside of A, we are only interested in the behavior of f on A and so restrict our attention to y A. Note that this definition also agrees with our original definition for functions f : R R; when A = R, the extra condition y A is no longer needed. But how will Proposition 2 fare in this new setting? E 5. Returning to Example 4, suppose we try to verify the condition in Proposition 2 at x = 0. Let U := ( 1, 1). We seek an open set V [0, 2] containing 0 such that f(v ) U. But in fact there are no open sets at all that are contained in [0, 2] and contain 0, because every open interval containing 0 must contain points less than 0. In the previous example we encountered something similar to the problem with the δ,ε-definition. We solved that problem by taking an intersection in (2.2) to ensure that the points in question were in the domain of the function. We can do something similar here, replacing f(v ) U in the statement of Proposition 2 by f(v A) U, and a modification of the same proof works to generalize the proposition: T 2. Let f : A R, where A R. Then f is continuous at x A if and only if for every open set U containing f(x) there is an open set V containing x such that f(v A) U. E 2.9. Prove the above theorem. But there is yet a better way to deal with the question of functions defined on subsets, namely to give the subset its own topology, and then forget entirely that it was once a subset of something else. D 6. Let X be a topological space and A X. We define the subspace topology on A by letting a set U A be an open set in the subspace topology of A if and only if U = A V for some open set V in the topology of X. This definition has the potential to create some confusion. After all, as we will see below, for any given subset B of X, B could be open in the subspace topology of A but not open in the topology of X, vice versa, open in neither topology, or open in both. However, we would like to avoid cumbersome phrases like open in the topology of... but we cannot simply eliminate references to A and X. Some texts use phrases like U is open relative to A and V is open relative to X. But

16 12 2. TOPOLOGY OF R AND BEYOND this notation is still a bit tedious, and with a slightly higher risk of confusion we will simply use statements like U is open in A and V is open in X. We have called the above collection the subspace topology so we need to be sure that we check that it really is a topology! For the first axiom, simply note that = A and A = A X, so both and A are open in A. For Axiom 2, let {U α } α Λ be a collection of open sets in A. Then for each α, U α = A V α for some V α open in X. According to one of the distributive laws, U α = (A V α ) = A α Λ α Λ α Λ which is open in A since V α is open in X. α Λ E Prove that the subspace topology satisfies Axiom 3. E 6. Let s consider A := [0, 2] R. What sort of sets are open in A? Certainly (0, 1) is open in A, since (0, 1) is open in R and contained in A. Now V := ( 1, 1) is also open in R but not contained in A. According to the definition, V A should be open. But V A = [0, 1), which is a half-open interval. So [0, 1) is open in A but not open in R. E Let X be a topological space and U X, where U is open. Prove that a set V X is open in U if and only if V is open in X and V U. Motivated by Theorem 2 and the subspace topology, we make the following general definition: D 7. Let f : X Y be a function, where X and Y are topological spaces. For x X we say that f is continuous at x if for every open set U in Y containing f(x) there is an open set V in X containing x such that f(v ) U. If f is continuous at every point in X then we simply say that f is continuous. Theorem 2 completely reconciles our two definitions of continuity for functions f : A R, when A R, since the set A V in the statement of the theorem is precisely an open set in the subspace topology. An important property of continuity is that the restriction of a continuous function is continuous. While this theorem was never stated in calculus, it is used constantly whenever one studies the behavior of a function on a particular interval, e.g. in optimization problems. The subspace topology makes it possible to correctly state, and easily prove, this theorem. T 3. Let X and Y be topological spaces and f : X Y be a function. Suppose that A X, x A, and f is continuous at x. Then f A is continuous at x when A has the subspace topology. P. For simplicity we will let g := f A. Let U be an open set in Y containing g(x) = f(x). Since f is continuous at x there is an open set W in X such that f(w ) U. By definition, V := W A is open in A and contains x. Since f(v ) f(w ) U, the proof is finished. The notion of being continuous at a point has a relatively small roll in basic topology for various reasons. There are few theorems about functions that are not continuous everywhere (although non-continuous functions definitely arise in analysis). When dealing with a function f that is not continuous, one can at least let A be the set of all points at which f is continuous. Then according to V α

17 2. CONTINUOUS FUNCTIONS 13 the previous theorem, f A is continuous. The next theorem moves us away from the concept of continuous at a point and often provides an easier way to verify continuity. Most topology texts simply take the equivalent condition in this theorem as the definition of continuity, and we will use this condition like our definition from now on. T 4. Let f : X Y be a function, where X and Y are topological spaces. Then f is continuous if and only if for every open set U Y, f 1 (U) is open in X. Before proving this theorem, we will prove a lemma that will be useful in many situations. L 2. Let X be a topological space and U X. If for every x U there exists an open set V in X such that x V U then U is open. P. For every x U we will denote by V x an open set in X such that x V x U. It will follow that U is open if we can show that U = Now certainly x U x U then there is some V y such that y V y V x V x U, since each V x U. On the other hand, if y U V x, so U V x. P T 4. Suppose f is continuous (at every point). Let U Y be open. Let x f 1 (U). Since f is continuous, there exists some open V containing x such that f(v ) U. But then V f 1 (U) and we have verified the conditions of Lemma 2. This implies that f 1 (U) is open. Conversely, suppose that f 1 (U) is open in X for every U Y that is open. Let x X and U be any open set containing f(x). By assumption, V := f 1 (U) is open in X, contains x, and f(v ) = f(f 1 (U)) U. That is, f is continuous at x. E Let f : X Y be a constant function, where X and Y are topological spaces. That is, for some y 0 Y, f(x) = y 0 for all x X. Prove that f is continuous in two ways: show it is continuous at every point, and that it satisfies the equivalent condition of Theorem 4. Here is another important theorem about continuous functions that you will have learned in calculus (for real functions). T 5. Suppose that X, Y, Z are topological spaces and f : X Y, g : Y Z are continuous. Then g f : X Z is continuous. P. Let U be open in Z. Since g is continuous, V := g 1 (U) is open in Y, and since f is continuous, is open in X. x U x U (g f) 1 (U) = f 1 (g 1 (U)) = f 1 (V ) E Suppose that X, Y, Z are topological spaces and f : X Y, g : Y Z are functions. Prove that if f is continuous at x X and g is continuous at y = f(x) Y then g f : X Z is continuous at x. Note that this argument is a bit more complicated than the proof of Theorem 5.

18 14 2. TOPOLOGY OF R AND BEYOND D 8. Let X be a set and A X. The inclusion of A into X is the function j : A X defined by j(x) = x for all x A. Although it seems like a simple, uninteresting function, the inclusion function turns out to be quite useful in topology, especially algebraic topology. Note that if A = X then the inclusion function is just the identity function. E Let X be a topological space and A X. Prove that the inclusion function j : A X is continuous when A has the subspace topology. 3. Other Topologies on R Remember that a topological space consists of a set and a collection of subsets that comprise the topology. Given a set S, there are of course many possible ways to define a topology on S, and there are many reasons to consider multiple topologies on a given set. In addition to topologies that have proved useful in many situations, such as the standard topology on R, topologies may be defined that, while perhaps somewhat esoteric, help motivate definitions and hypotheses in theorems. A good theorem has as few hypotheses as possible to reach a given conclusion. One should always ask whether a given hypothesis is really necessary to reach the same conclusion, and one way to show that a hypothesis really is necessary is to produce an example for which the hypothesis is not satisfied, and the theorem fails to be true. In topology, producing such examples can be surprisingly diffi cult. Another good reason for the student to study strange examples is to break down any preconceived notions that might lead to errors in proofs. In topology, things that seem like they ought to be true based on one s experience with R, often are not true in general. This is not surprising given the small number of axioms we are working with, which leads to many possible examples. Later we will add many other assumptions to our topological spaces that will force them to behave a bit better. But we will start with the two most uninteresting examples. E 7. Let X be any set. We will take for our topology the collection of all subsets of X. Then the axioms are obviously all satisfied: The union or intersection of any subsets of X is again a subset of X we don t even need to restrict to finite intersections. It is also true that the sets {x} are all open sets, something that distinguishes this topology from the topology of R. Moreover, if f : X Y is any function to another topological space Y, f 1 (U) is always open (even if U is not!) since every subset of X is open. So every function with domain X to another topological space is continuous. On the other hand, it is relatively hard for functions into X to be continuous. For example, let X = R with the discrete topology. Then we can consider the identity function i : R X. This makes sense because as a set, X is equal to R; we have simply changed what we mean by an open set. Now for any x X, {x} is open in X. But i 1 ({x}) = {x} is not open in R. So i is not continuous at any point. (Its inverse function is of course continuous, as we discussed above.) The descrete topology is the largest possible topology on a set S. Is there a smallest one? Axiom 1 requires that and S be in any topology, and actually nothing more is needed: S = S and S =, so the collection {, S} already satisfies Axioms 2 and 3. This makes {, S} the smallest possible topology on S, called the trivial (or indiscrete) topology. If S has the trivial topology and

19 3. OTHER TOPOLOGIES ON R 15 f : X S is a function from another topological space to S, then f is always continuous, since f 1 ( ) = and f 1 (S) = X, both of which are open in X. Likewise, a function g : S X is unlikely to be continuous if X has plenty of open sets; see the next exercise. E Let X have the trivial topology. Show that if f : X R is continuous then f is a constant function. Hint: Suppose it is not constant, i.e. it has at least two functional values y 1 and y 2. Consider an open interval that contains y 1 but not y 2. The trivial topology is of very little interest rarely occurring in mathematics. At the opposite extreme, the discrete topology actually does pop up in significant situations. But its importance in these settings is not a result of its internal topological structure, which is uninteresting, but rather due to external factors. For example, if X is a topological space and the subspace topology of A X is discrete, this means that each a A has the property that there is some open subset V X such that {a} = V A. That is, the points in A are, in some sense, isolated in X, a fact that can sometimes be very useful. E 8. We will show that Z R, with the subspace topology, is discrete. To do this, it suffi ces to prove that any set {n} with n Z is open in Z, since (as we saw in an earlier proof) for any set A, A = {x}. But for any n, {n} = (n 1, n + 1) Z. E Prove or disprove: Q R with the subspace topology is discrete. E Prove or disprove: { 1 n } n N R with the subspace topology is discrete. E Prove or disprove: {0} { 1 n } n N R with the subspace topology is discrete We now know that every set S has at least two topologies: discrete and trivial (although if S has at most one point then these two topologies are the same). R has at least one additional topology (the standard one), and we will consider below some additional topologies on R. Given any two topologies T and T on a given set S, we can consider whether one topology is contained in the other; that is, whether one has more open sets than the other. As we will see later, this needn t always be the case. But if T T then we will say that the two topologies are comparable, T is a finer topology than T, and T is a coarser topology than T. These words will make more sense as your topological intuition builds, and we will revisit this terminology below. The use of the comparative ending er can be confusing, since the requirement here is T T, not T T. For example, if T = T then T is both finer and coarser than T. This may remind you of the terms increasing and decreasing, which both describe a real function that is constant. As in the case of increasing/decreasing functions, the modifier strictly is used: If T T then we say that T is strictly finer than T. If we happen to be considering two topologies T, T on the same set S, when necessary to avoid confusion we will sometimes use the more formal notation (S, T ) and (S, T ) to denote the different topological spaces. The discussion concerning continuous functions involving the discrete and trivial topologies motivates the x A

20 16 2. TOPOLOGY OF R AND BEYOND following intuitive observation. If T T are two topologies on a given set S and X is a topological space, with a function f : X S, then it is easier for the function f to be continuous with respect to T than with respect to T ; there are more open sets U in T than in T, so more sets f 1 (U) have to be open in X. Put another way, there are more continuous functions f : X (S, T ) than f : X (S, T ) (again, the comparative may not be strict). Likewise, there are fewer continuous functions f : (S, T ) X than f : (S, T ) X. E Suppose that S is a set with two topologies T T, and let i : S S be the identity. Which is continuous: i : (S, T ) (S, T ) or i : (S, T ) (S, T )? Given two topologies on a set S, described in different ways, it may not be simple to decide whether they are the same topology. For example, suppose we define a set U in R to be open if (2.3) for each x U there is some open interval (a, b) such that x (a, b) U. Certainly open sets in the standard topology are open in the sense of (2.3): Definition 3 requires not just an interval, but an interval of the form (x ε, x + ε) for some ε > 0. But on the other hand, we have already shown that an interval (a, b) is open in the standard topology. Now suppose that V that is open according to (2.3). Then by definition there is an open set (a, b) in the standard topology such that x (a, b) V. The conditions of Lemma 2) are met, so V must be open in the standard topology. In other words, a set is open in the standard topology if and only if it is open according to (2.3). That is, these two definitions give rise to exactly the same open sets, i.e. the same topology. The method we used to prove this is encoded in the following proposition: P 3. Let S be a set with topologies T, T. Then T T if and only if for every open set U in T and x U there is an open set V in T such that x V U. P. Suppose first that T T and let U be an open set in T. Since T T, this means that U is already an open set in T, so the conclusion follows by simply letting V = U for every x U. For the converse, let U be in T. By assuption, for each x U there is some V T such that x V U. The conditions of Lemma 2 are satisfied and this means that V is in T. The above proposition helps to explain the terminology coarser and finer. Suppose that T is coarser than T, i.e. T T. The previous proposition says that an open set U in T may be written U = V x, where x V x U and V x is in T. One imagines that U has been broken up into the smaller or finer pieces V x. It is possible that some of the sets V x cannot be similarly broken up into a union of open sets in the topology T, in which case T is strictly finer. Applying the proposition in both directions gives the following corollary: C 4. Two topologies on S are equal if and only if given an open set U in one topology and x U there is an open set V in the other topology such that x V U. x U

21 3. OTHER TOPOLOGIES ON R 17 E 9. What happens if we change the definition of open set for the standard topology on R (Definition 3), replacing the open interval (a, b) with the half-open interval [a, b)? First of all we know that [a, b) is not open in the standard topology, but clearly satisfies the new definition of open. Since we now have a different collection of sets, we need to start by checking that what we get is actually a topology. Certainly Axiom 1is satisfied (this is usually the easy axiom!). The proof of Axiom 2 is very similar to the proof for the standard topology, and is left as an exercise. For Axiom 3, we will first consider the intersection of two open sets. Suppose that U and V are open in this new sense, and let x U V. By definition there exist intervals [a, b) and [c, d) such that x [a, b) U and x [c, d) V. By checking a couple of cases, it is easy to verify that since [a, b) and [c, d) are not disjoint (each contains x) their intersection is an interval [p, q) (in fact p {a, c} and q {b, d} depending on the ordering of the points), and we have x [p, q) U V. We have shown that the intersection of any two open sets is open. We may finish the proof by induction: Suppose that the intersection of any n open sets, n 2 is open. Given open sets U 1,..., U n+1, by elementary set theory we may write n+1 U i = U n+1 n U i. Both U n+1 and n U i are open (the latter by the inductive hypothesis) and by what we just proved about the intersection of two open sets, we are finished. This topology is known as the lower limit topology (or Sorgenfrey line) and is usually denoted by R l. R 1. The method that we used at the very end of the above example clearly works in general: To verify Axiom 3 we need only check that the intersection of any two open sets is open. E Verify Axiom 2 for the lower limit topology. We have already observed that [a, b) is open in the lower limit topology but not in the standard topology of R. We will show that in fact the lower limt topology is strictly finer than the standard topology, using Proposition 3. Let U be open in the standard topology and x U. By definition of the standard topology there exists some ε > 0 such that (x ε, x + ε) U. But then x [x, x + ε) U, which is what we needed to prove. R l has many distinctive properties, and we will revisit it later. Both R and R l may be more conveniently described using the device that we will define next: D 9. Let S be a set. A collection B of subsets of S is called a basis for a topology on S if (1) For each x S there is some B B such that x B. (2) If x B 1 B 2, where B 1, B 2 B, then there is some B 3 B such that x B 3 B 1 B 2. If B is a basis, we define the topology generated by B (or more simply, the topology of B) to be the collection of all subsets U of S such that for all x U there exists some B B with x B U. In particular, the standard topology on R is defined by using the collection of open intervals as a basis, and the topology of R l is defined using the collection of intervals of the form [a, b) as a basis. In each of those cases we proved directly that the resulting open sets formed a topology. The proof that the open sets defined by

22 18 2. TOPOLOGY OF R AND BEYOND a basis form a topology is similar to the proof that the lower limit topology is a topology, and is an exercise. E Prove that the collection of sets described in Definition 9 is a topology. E Prove that if B is a basis for a topology T on X then T is precisely the collection of all sets U X such that U is a union of basis elements. In other words, to get the topology from a basis we simply take all possible unions of basis elements. (Note that the empty set is the union of any empty collection of sets!) E Prove that if X has the discrete topology then a basis for the topology is the collection B of all single-point sets {x}. Note that you not only have to prove that that B is a basis for a topology, you need to show that the topology generated by B is the discrete topology. E 10. Armed with the notion of basis, we can now proceed with the next example. Consider the set R again. Let K := { 1 n : n N} and take for our basis B all open intervals (a, b) and all sets of the form (a, b)\k. It is an exercise that B is actually a basis for a topology, called the K-topology. R with this topology is denoted by R K. Since all of the open intervals are open in the K-topology, it, like R l, is finer than the standard topology. Since ( 1, 1)\K is not open in R (it contains no interval about 0), R K is strictly finer than R. E Prove that the set B defining R K is a basis. P 4. R l and R K are not comparable. P. We need to find a set that is open in R l but not in R K, and vice versa. Consider the inteval [ 2, 1). This is a basis element, hence open in R l. We already know that there is no interval (a, b) containing 2 and contained in [ 2, 1), and similarly there cannot be a basis element (a, b)\k containing 2 and contained in [ 2, 1) since the points in K are all greater than 0. Now consider the basis element U := ( 1, 1)\K in R K. If U were also open in R l then there would have to be a basis element [a, b) such that 0 [a, b) U. But then a 0 < b and for large enough n, a 0 < 1 n < b and hence 1 n U, a contradiction. E Define a topology T on R by taking as a basis the collection of all sets of the form (a, b)\k, together with K, where K = { 1 n : n N}. Prove that this collection does form a basis and check comparability with R and R l. D 10. Let S be a set. The a set U S is said to be open in the finite-complement topology on S if S\U is finite or U =. E Use de Morgan s laws to show that the finite-complement topology satisfies the three axioms, and so is a genuine topology. Note that if S is finite, then the finite-complement topology is the same as the discrete topology; the situation is more interesting if S is infinite. E 11. Let R have the finite-complement topology, in which case we will denote it by R F. Suppose that U is open in R F. Then either U = R, U=, or R\U = {x 1,..., x n }. In the first two cases U is obviously open in the standard

23 3. OTHER TOPOLOGIES ON R 19 topology, and in the last, by re-ordering the index we may suppose that x 1 < x 2 < < x n. That is, U = (, x 1 ) (x 1, x 2 ) (x n 1, x n ) (x n, ) which is also open in the standard topology. That is, the finite-complement topology is coarser than the standard topology. Since (0, 1) has infinite complement, it is not open in the finite-complement, so the standard topology is strictly finer than the finite-complement topology. E Determine the relationship between R F and the two spaces R l and R K (i.e., not comparable, one is finer, or one is strictly finer than the other).