Mainstream Mathematics

Transcription

1 Mainstream Mathematics Jeroen Goudsmit th December In this document we develop some of the basics of mainstream mathematics. First we delve into basic analysis, and using some ideas from topology we proceed to prove some elementary theorems. All the proofs can be given formal underpinnings, and the reasoning is rational in that the steps can be justified by (classical) logic. It is important to see that most proofs in a way spell out the details as given to us through geometric intuition. Oen it is quite useful to draw a small sketch, even though a sketch is not a precise representation of the situation at hand. A good way to acquire mathematical understanding is working out proofs by yourself. is may take time, and you will probably not always succeed, certainly not right away. Many lemma s below are simply stated, and a proof is completely le out. It is up to you to find these proofs. Do write them down. Most of the material here is also covered in the Utrecht University courses Inleiding Analyse and Inleiding Topologie, see the respective lecture notes by van den Ban () and Crainic (). Munkres () is a nice reference on topological maers..! is document is still under development. It quite possibly contains some errors, and some other parts might be unclear. When you find a part which is unclear or which you believe to be wrong, please report this to me so that I may fix it. e first three sections are near their desired state, but the final two will probably change some more. If you want to print this document, please do not yet print these final sections.

2 Contents Proximity Continuity Topology. Hausdorff Spaces Invariance Constructions Closures Compactness. Uniform Continuity Completeness & Compactness Index References

3 Section I Proximity One has an idea of distance, of how far an object is away from another object. Our first task is to codify this in mathematics. We will use the real numbers as a given, and assume many properties, all of which will feel quite natural. Later on we make these things more explicit. We write R n for the n-dimensional real space, as is customary. We will oen refer to this as Euclidian n-space. Note that the distance of two points p and q in real n-space is the same as the distance between 0 and p q. is laer expression is just the size of the vector p q, with which we are already familiar. Let us define this precisely. Definition. (Euclidian Norm). Given a point p R n, we define its Euclidian norm as p = p 2 + p p2 n. is Euclidian norm has many properties, all of which easily derivable from the above definition. As an exercise, try to prove all of the below. In the future, whenever you see a red square in the margin, there is something le to prove. Prove it. Lemma. (Elementary Properties of Euclidian Norm). e Euclidian norm:. (i) is non-negative, and zero only if its input is zero; (ii) respects scalars in that r p = r p for all p R n and r R; (iii) satisfies the triangle inequality, i.e. p + q p + q for all p, q R n. Using the observation made above, that is, the distance between two points is the distance between the origin and their difference, we can now define distance. Definition. (Euclidian Distance). e Euclidian distance between two points p and q in Euclidian n-space is defined as p q. We denote this distance by d (p, q). As with the norm function, the distance function has several properties. Note that we never need to expand the definition of the norm to prove these properties, it suffices to use Lemma.. Lemma. (Elementary Properties of Euclidian Distance). e Euclidian distance measure: (i) is non-negative, and zero if and only if its inputs are equal; (ii) is symmetric, in that d (p, q) = d (q, p) for all p, q R n ; (iii) satisfies the triangle inequality, i.e. d (x, z) d (x, y) + d (y, z) for all x, y, z R n. Proof. All of these properties follow immediately from the properties on the Euclidian norm. Indeed, we know that d (p, q) 0 from Lemma..(i). Also, if 0 = d (p, q) = p q then p q = 0 by the same property, which entails p = q. ese two observations prove (i). Property (ii) follows immediately from Lemma..(ii), as demonstrated by the following calculation. Finally, we can compute d (p, q) = p q = q p = q p = d (q, p) d (x, z) = x z = (x y) + (y z) x y + y z = d (x, y) + d (y, z ), whence (iii) follows from Lemma..(iii). Let us look at several very concrete examples of the Euclidian distance measure. When we consider the real line, that is R = R, it is not very difficult to see what the Euclidian distance map amounts to.

4 Above we see a small segment of the real line. Consider 4 and on this line, and note that d ( 4, ) = ( 4 ) 2 = 5 = 4. It is good to note here that the Euclidian distance between two points is merely the absolute value of their difference. Now consider the plane, R 2 = R R, as partially depicted in Figure. All points, and exactly those points, that fall within the circle below satisfy d (x, 0) 4. Of course, d (x, 0) = x holds for all x. Definition. (Open Ball). For each point p R and each positive number r > 0 R we define the open ball of radius r centered at p as B (p; r) := { x R n d (p, x) < r }. Definition. (Open Set). A subset U R is said to be open whenever for each point p U there is an ϵ > 0 such that B (p; ϵ) U. Try to think of some open sets, for instance on the real line. When you first encounter a definition, try to think of examples and non-examples. q p Figure : Balls in the real plane Lemma. (Open Balls are Open). For ea point p R n and r > 0 the open ball B (p; r) is open. e proof idea is simple enough, as illustrated in Figure. Given some point q in an open ball around p, there always is a ball around q that fits in this other ball. e radius of the laer ball is the difference between the radius of the former and the distance between p and q. is construction is demonstrated for two points in the aforementioned picture. Proof. Let q B (p; r) be a point within the open ball. We need to show that there exists some ϵ > 0 such that B (q ; ϵ) B (p; r). Simply take ϵ to be some positive value less than equal to r d (p, q). Of course, such a value exists, as d (p, q) < r. To show that B (q ; ϵ) B (p; r), pick some x B (q ; ϵ). All is well when we can show that d (p, x) < r. But this follows easily, as illustrated below. d (p, x) d (p, q) + d (q, x) < d (p, q) + ϵ d (p, q) + (r d (p, q)) = r

5 Consider the following lemma. Try to figure out why it would not work when n = 0. Lemma. (Singletons are Not Open). Let p R n be some point, and assume n. Now {p} is not open. Proof. Write e for any unit vector in R n, say, 0,..., 0. Suppose that {p} were open. en there would be some ϵ > 0 such that B (p; ϵ) {p}. Take such an ϵ, and note that p+ 2 ϵ e B (p; ϵ). Yet it is obvious that p+ 2ϵ e {p}, a clear contradiction. Consequently, {p} is not open. Lemma. (Characterization of Opens). A set U is open in R n if and only if U is the union of open balls.. Corollary.. Both R n and are open.. Corollary.. e union of any set of open sets is again open. at is to say, if U is a set of open sets then. U = U U U is open. Lemma.. Any interval of the form (a, b) for a < b real numbers on the Euclidian line is open.. Lemma.. Let U and V be open sets. Now U V is open as well.. Corollary.. Let A be a finite subset of R n for n. Now A is open if and only if A is empty... Figure : Increasingly smaller intervals around the origin. Although the union of any set of open sets is again open, it is not at all the case that the intersection of any set of open sets is again open. Below we consider the situation as illustrated in Figure. From the picture in becomes quite clear that no interval can be contained within the intersection of all open intervals centered around a point. Example.. Consider the open intervals I r := ( r, r), and note that U := {I r r > 0} is a set of open sets. It is clear that p r>0 I r holds if and only if p = 0. Consequently, the intersection of all I r for r > 0 is a singleton set. By Lemma. we know that this intersection is not open. Definition. (Neighbourhood). A set A R n is said to be a neighbourhood of a point p R n when there is some open U A such that p U. In particular, each open ball centered around p is an open neighbourhood of p. Moreover, each point p contained within an open set U has U as a neighbourhood. e name is suggestive, and this is meant to be so. When every neighbourhood of a point p coincides with a given set, the point p is bound to be close to this set. Consider for instance the set I = [0, ), which consists of all real numbers x such that 0 x <. e point intuitively lies very close to this interval I. Note that every open set which contains must coincide at least partially with I. Definition. (Limit Point). A point p R n is said to be a limit point of a subset A R n when for each ϵ > 0 we have B (p; ϵ) A. Lemma.. Let A R n be a set and let be p R n a point. Now p is a limit point of A if and only if the intersection. of ea open neighbourhood of p with A is nonempty. Corollary.. A set A R n is open if and only if each point has an open neighbourhood.. Lemma.. Let A R n be a set. Define U to be the set of all non-limit points of A. at is to say, Now U is open. U := {p R n p is not a limit point of A}.

6 Proof. Consider some point p U. We need to prove that some ϵ > 0 exists such that B (p; ϵ) U. Suppose no such ϵ exists. is means that for each ϵ > 0 we know that B (p; ϵ) U. Now as U A = R n, it follows that B (p; ϵ) A. But this entails that p is a limit point! A clear contradiction, so some open ball centered around p contained within U exists, proving U to be open. Some sets contain all points nearby. at is to say, some sets contain all their limit points. Let us formally define these sets. Definition. (Closed Set). A set A R n is said to be closed whenever its complement is open. At first glance, the above definition does seem to contain the claimed. But we can prove these two notions to be equivalent. Before we do this, let us look at an example of a closed set. Lemma. (Closed Intervals are Closed). Let a b R be arbitrary. Now [a, b] is closed. Proof. Consider U := R [a, b], and let p U be arbitrary. We seek some ϵ > 0 such that B (p; ϵ) U. Let ϵ be half of the minimum of d (a, p) and d (b, p). is proves that B (p; ϵ) [a, b] =, so B (p; ϵ) U as desired. Lemma.. A set is closed if and only if it contains all of its limit points. Proof. Let A R n be a subset of Euclidian n-space. To prove the one direction, suppose that A contains all of its limit points. By Lemma. we know that the set of non-limit points of A is open. Now note that if a point is not not a limit point, we know it to be a limit point of A, and hence included in A. Now to prove the converse, suppose that A is closed. is means that U := R n A is open. Let p be a limit point. If p A then there is nothing le to prove. So instead, suppose p A. is immediately entails that p U. Now this gives some ϵ > 0 such that B (p; ϵ) U, for U was assumed to be open. As a consequence we know B (p; ϵ) to be disjoint from A, contradicting the assumption that p is a limit point. is proves that all limit points of A lie within A, as desired Figure : Graph of x x on the domain U = [0, ) (, 2]. When one considers a function f : R n A R m, this function need not always sensibly extend to all of R n. ink for instance of the function f : [0, ) (, 2] R, x x x as depicted in Figure. is function is clearly not defined at, but we do know what value f approaches when geing arbitrarily close to. is idea can be made formal. Definition. (Limit). Let f : R n A R m be a function and let p A and q R m be points. e limit of f at p is said to equal q when for all ϵ > 0 there exists a δ > 0 such that for all x A with d (p, x) < δ it holds that d (q, f(x)) < ϵ. We denote this as lim x p f(x) = q. When we write f : R n A R m we mean that f is a function with domain A and codomain R m, where A is a subset of R n.

7 Lemma.. Let f : A R m R n be a function and let p R n and q R m be points. Now lim x p f(x) = q. precisely if for all ϵ > 0 there is a δ > 0 su that: f ( B (p; δ) A ) B (q ; ϵ) () Let us again look at the example of Figure, and prove that our previous statement can actually be made formal. Lemma.. Define the function f : [0, ) (, 2] R, x x x It now holds that lim x f(x) = 2. Proof. Let ϵ > 0 be arbitrary. We now seek a δ > 0 such that () holds. To this end, take some x R. First observe that x = ( x)( + x), and that + x >. As a consequence, assuming that x < δ and x [0, ) (, 2] ensures that x < δ. We can now compute ( d f(x), ) = 2 f(x) 2 = x x 2 = 2 ( x) ( x) 2 ( x) = 2 x + x 2 ( x) = ( x) 2 2 ( x)( + x) < δ 2 < δ. From the above it is clear that choosing δ equal to ϵ ensures that () holds. It is important to note that we really do need that x 0, which is ensured by assuming x to lie in f s domain. Example.. Consider the function f : R {a} R, x x2 a 2 x a for some fixed a R. We know that (x a)(x + a) = x 2 a 2, so it might seem that f is actually the function x x + a. Yet this is not the case, because on x = a we get a nonsensical expression in the definition of f. However, we do have lim x p f(x) = p + a for all p R. Indeed, let p R and ϵ > 0 be arbitrary. Now take some z a with d (z, p) < ϵ and compute ( z 2 a 2 ) d z a, p + a = is proves that f (B (p; ϵ)) B (p; ϵ). z 2 a 2 z a (p + a) = (z + a) (p a) = z p < ϵ. e notation of a limit is quite suggestive, in that lim x p f(x) = q may lead one to believe that there only one such q exists. In general, this is not the case. Consider for instance some function [0, ] R, say f defined as mapping x to x 2. We can legitimately write lim x 5 f(x) = 34. Indeed, for any ϵ > 0 we see that B (5; ) [0, ] =, so its direct image under f is equal to the emptyset. is is surely contained within B (34; ϵ). In this case, 5 clearly is not a limit point of [0, ]. For suppose it were, then by Lemma. it would be contained in [0, ] is this interval were closed. But the interval is closed, as demonstrated in Lemma.. Lemma.. Let f : R n A R m be a function and let p R n and v, w R m be points. Suppose that. lim f(x) = v, lim x p f(x) = w. x p If p is a limit point of A then v = w.

8 Section II Continuity Some functions behave especially well with respect to taking limits. Definition. (Continuous). A function f : X R n is said to be continuous at p X when for every ϵ > 0 there is a δ > 0 such that for all x X with d (x, p) < δ one has d (f(x), f(p)) < ϵ. e function f itself is continuous when it is continuous at all points in its domain. B (f(p); ϵ). B (p; δ) Figure : A continuous function. In Figure we give an example of a function that is everywhere continuous (i.e. continuous) and Figure depicts a function which is not continuous at a specific point. e right-hand side function sends x to if x and to if x >. As a consequence, no maer how small an interval you choose around, f will always map half of it to and half of it to. So for small ϵ, no δ exists. We can of course re-formulate the above using open balls. e following lemma gives this reformulation, which in a way hides away the universally quantified variable x in the above definition. Lemma.. A function f : X R n is continuous at p X if and only if for every ϵ > 0 there is a δ > 0 su that. f(b (p; δ) X) B (f(p); ϵ). Equivalently, f (B (f(p); ϵ)) B (p; δ) X. is definition is reminiscent of the definition of a limit. And justifiably so, as illustrated by the following lemma. Lemma.. A function f : X R n is continuous at p X if and only if lim x p f(x) = f(p).. Many functions are continuous. Let us consider the following example. Lemma.. e map + : R 2 R is continuous. Proof. Let p = (x, y) R 2 and ϵ > 0 be given. Let q = (x, y ) R 2 be arbitrary such that d (p, q) < 2ϵ. is entails that x x, y y < δ. We observe that proving the desired. d (x + y, x + y ) = (x + y) (x + y ) = x x + y y < 2 ϵ + 2 ϵ = ϵ,

9 . B (f(p); ϵ) Figure : A function not continuous at. ere is yet another way to formulate continuity. Given a function f we have a map f which sends subsets of the codomain to subsets of the domain. In the following lemma we show that a function is continuous if and only if the map f restricts nicely to opens. e proof idea is illustrated in Figure. Lemma.. Let A R n be a subset of Euclidian n-space and let f : A R m be a function. Now f is is continuous if and only if the f (U) is the intersection of an open with A for ea open U R n. Proof. First observe that if X Y then f (X) f (Y ). Moreover, X f (f(x)) and X = f(f (X)) hold for all X. Now suppose that f is continuous and let U R m be an open set. We need to show that f (U) is an intersection of an open set and A. To this end, let p f (U) be arbitrary. When we can show that there is an open ball centered at p such that the intersection of this ball and A is contained within f (U), then we are done. As U is open and f(p) U, we know of an ϵ such that B (f(p); ϵ) U. By continuity this yields, via Lemma., a δ > 0 such that f(b (p; δ) A) B (f(p); ϵ). Applying both our observations we derive the following, from whence the desired follows. p B (p; δ) A f ( f(b (p; δ) A) ) f ( B (f(p); ϵ) ) f ( U ) f (U) f U p. f(p) B (p; δ) B (f(p); ϵ) Figure : Continuity entails that pre-images of opens are open.

10 To prove the converse, let p A and ϵ > 0 be arbitrary. Realize that B (f(p); ϵ) is open, hence there is some open U R n such that U A = f (B (f(p); ϵ)). Now as p is contained in the right-hand side of this equation, we know that p U. is gives us some δ > 0 such that e proof is now finished due to Lemma.. B (p; δ) A U A = f ( B (f(p); ϵ) ). Below some properties that ought not to be too hard to prove, le as an exercise for your entertainment. Lemma.. For all p R n we have lim x p x = p.. Lemma.. For all p R n and q R m we have lim x p q = q.. Lemma.. Let f and g be functions A R n R m and let p R n and a, b R m be points. Suppose that. lim f(x) = a, lim x p f(x) = b. x p It now holds that: (i) lim x p ( f(x) + g(x) ) = a + b; (ii) lim x p ( f(x) g(x) ) = a b. Corollary.. Let f : A R n R( m be a function ) and let λ R be a constant. For any p R n and q R m we. have that if lim f(x) = q then lim λ f(x) = λ q. x p x p Corollary.. Every polynomial is continuous..

11 Section III Topology Above we saw that the Euclidian distance measure gives us a notion of proximity. We defined what it means to be an open set, and used this definition to characterize some behavior. We now generalize away everything but four elementary properties of open sets. Surprisingly enough, the resulting theory is fairly rich. We will not delve into it deeply, but use it as a comfortable position from whence we can prove some desirable results. Definition. (Topological Space). A topological space is a pair X, τ where X is a set and τ is a set of subsets of X such that the following hold: (i) the empty set is included in τ; (ii) the set X is included in τ; (iii) if A, B τ then A B τ; (iv) if U τ then U τ. We will oen denote the topological space simply by its underlying set X, whenever the topology is clear from the context. Elements of τ are called open, and a set is called closed if its complement is open. e set τ is said to be a topology on X. Our first task is to retrieve what we had before. To this end, we consider the Euclidian topology on Euclidian n-space. Definition. (Euclidian Topology). Let X R n be a subset of Euclidian n-space. We define the Euclidian topology on X to be given as X, τ where U τ if and only if there is some V R n open in the sense of Definition. such that U = V X. We need to check that this Euclidian topology is in fact a bonafide topology. First, the empty set and X are open as a direct consequence of Corollary.. If U and U 2 are open in X (in the new sense), then there are V, V 2 open in the old sense such that U i = V i X. Now see that U U 2 = (V X) (V 2 X) = (V V 2 ) X, so the le-hand side is open (in the new sense) because V V 2 is open (in the old sense) due to Lemma.. Similarly, the final property follows from Definition.. From now on, whenever we speak of opens within R n, we assume that these sets are open with respect to the justdefined Euclidian topology on Euclidian n-space. It is an easy exercise to see that these are exactly the same as opens in the sense of Definition.. Example. (Prime Spectrum). Let us consider a fairly wild example of a topological space. ere is no need to understand this space yet at this point. Do revisit it later. Let X contain exactly the prime numbers and 0. For a natural number n N we can define U(n) := {p X p n}. A set U X is defined to be open in X whenever U = U(n) for some n N. We can easily see that U() = X and U(0) =. Indeed, nothing but is a divisor of and everything is a divisor of 0. e intersection of two opens U(n) and U(m) is given by U(lcm(n, m)). To see this, note that p n or p m holds if and only if p is a divisor of the least common multiple of n and m. e union of arbitrarily many opens can be found analogously, by taking the greatest common divisor of a set. Try to work this out in greater detail.. is example would be fairly trivial when all subsets of X were open. To show that this is not the case we prove that no singleton is open. So let p X be arbitrary, and suppose that {p} = U(n) for some n N. is means that x n if and only if x = p holds for all x X. We can exclude the case where n = 0, because then U(0) = {p} would follow, a contradiction. Take p and p 2 to be two distinct primes larger than n. On the positive natural numbers we

12 know that a b entails a b, so it is easy to see that p n and p 2 n holds. As a consequence, p, p 2 U(n) = {p}, which means that p = p = p 2. Yet p and p 2 were assumed to be distinct, a contradiction. e space X is quite similar to what an algebraic geometer would call the prime spectrum of Z. ese kinds of spaces play an important role in modern geometry and number theory, see for instance Hartshorne (). Definition. (Continuous). Let X, τ and Y, σ be topological spaces. A function f : X Y is said to be continuous whenever f (U) σ for all U τ. We will oen simply say that f is a map from X to Y to mean precisely this. Lemma.. Let A R n and B R m be subsets. Now f : A B is continuous if and only if the corresponding function A R m is continuous in the sense of Definition.. Proof. e desired follows from Lemma., another characterization of the old notion of continuity.. Lemma.. e identity function on any space is continuous. Proof. Let X be some topological space and consider the identity map id X : X X which maps x to x. Now note that id X (U) = U, so the former is open exactly if the laer is, for they are equal. Lemma.. If f : X Y and g : Y Z are both maps, then so is their composition g f = gf : X Z.. Definition. (Subspace Topology). Let X be a topological space and Y X a subset of X. e subspace topology. on Y has as opens exactly the intersection of opens of X with Y. We now know of two ways to define a topology on a subset of Euclidian n-space. Either we immediately use Definition., or we use the above definition to inherit a subspace topology from the Euclidian topology on the whole n-space, as given to us via Definition.. Both of these topologies coincide.. Lemma.. Suppose that Z Y X are sets and X is also a topological space. Endow Y with the subspace topology. inherited from X. Now the subspace topology on Z as inherited from Y equals that as inherited from X. When generalizing away from the real line to Euclidian n-space, we replaced intervals with balls. In doing this, we lost some of the ordering properties we had. Let us generalize away from the real line in another way, by first capturing the desirable order properties, and then defining a suitable generalization for intervals. Definition. (Total Order). A total order is a pair X, where X is set and is a binary relation on X satisfying the following laws: transitivity if x y and y z then x z; anti-symmetry if x y and y x then x = y; totality for all x, y X we have x y or y x. Given such a total order, we can define the interval between x and y as the set of all p X such that x p y and denote it [x, y]. In a similar way we dan define all other types of intervals. Examples of total orders are, amongst others, the real numbers, the rational numbers and the natural numbers, all with the usual ordering. We can equip a total order with a topology. Definition. (Order Topology). e order topology on a total order X, is the least topology for which the all open intervals (a, b) with a, b X are open. Note that such a topology always exists. Although in some cases it might be fairly uninteresting, for instance when we take the order to be N.

13 Lemma.. e order topology on N is discrete, that is to say, all subsets of N are open. Proof. is immediately follows when each singleton is open, so let us show this. Take n N arbitrary, and note that ( n, n + ) only contains n, but it is open. us {n} is open. When we take R with the usual order, we may wonder whether the order topology on R coincides with the usual Euclidian topology. It does. Lemma.. e Euclidian topology on R is the least topology su that all open intervals are actually open. Proof. Let R, τ be some topology on R such that the open intervals are open. We wish to prove that all opens of R with the Euclidian topology are included in τ. is is not very difficult, as U is open in R (with the Euclidian topology) whenever U is the union of open balls, and open balls are simply open intervals. From this the desired is immediate.. Hausdorff Spaces Euclidian space is not just any topological space; it has a lot more interesting properties. One of those properties is the following. is property expresses that any pair of separate points can be separated by disjoint opens. Definition.. Let X be a topological space. We say that the space X is Hausdorff (or, X is a Hausdorff space) when for all p q X there exist opens p U p and q U q such that U p and U q do not overlap. ϵ. p d (p, q) q e proof below is illustrated by Figure. Figure : Distinct points can be separated by non-overlapping opens. Lemma.. Ea Euclidian space is Hausdorff. Proof. Consider Euclidian n-space R n. Now let p and q be arbitrary non-equal points. is means that there is a positive distance r := d (p, q) > 0 between them, as follows from Lemma..(i). Choose some ϵ < 2r, it is now clear that B (p; ϵ) and B (q ; ϵ) have an empty intersection. ese are the desired opens. Lemma.. Any subspace of a Hausdorff space is again Hausdorff.. ere are also spaces which are not Hausdorff. Many of them show up quite naturally in mathematics. e following are examples of non-hausdorff spaces.

14 Example. (Indiscrete Topology). Let X be any set. e indiscrete topology on X is defined as the topology whose. only open sets are X and. is is a valid topology, but it is not Hausdorff. Example. (Half-Infinite Topology). Consider the following topology on the real line R. Take as opens the sets. {x R x t} for each t R. Again, this is a valid topology, but is is not Hausdorff. Example. (Prime Spectrum is not Hausdor). Recall the space X of Example.. We will prove that this space is not Hausdorff. Consider the distinct points 2 and 3. Suppose we have opens 2 U(n) and 3 U(m), such that U(m) U(n) is empty. Convince yourself that U(lcm(n, m)) is the previously mentioned intersection. Any prime larger than n m does not divide this least common multiple, hence the intersection can not be empty. In Lemma. we proved uniqueness of limits, assuming we take a limit of f approaching a limit point of its domain. e proof of that lemma is partially reminiscent of the above proof of Lemma., and not by coincidence. Indeed, the referenced proof can be split up in two steps: Every Euclidian space is Hausdorff and e limit of any function approaching a limit point of its domain is uniquely determined whenever the codomain is Hausdor. Definition. (Limit). Let A X be an inclusion of topological spaces and let f : A Y be a function between topological spaces. Now q Y is the limit of f as it approaes p X, wrien lim x p f(x) = q, precisely if for all neighbourhoods A of q one has a neighbourhood B of p such that A f (B ). We now know two definitions of what it means to be a limit of f approaching some point p. Indeed, we know of Definition. and we can specialize Definition. to the same case. Whenever one has two definitions of the same notion in two different seings with an overlapping domain of quantification, as is the case here, it is oen a good idea to check whether they coincide. Lemma.. e new and old notion of a limit coincide. Proof. We prove that for each function f : R n X R m and each p X we have that lim f(x) = a x p () in the old sense if and only if it is the case in the new sense. Let us get right down to business, and assume that () holds in the old sense. Let B be a neighbourhood of a. is means that there is some ϵ > 0 such that B (a; ϵ) B. As a consequence of our assumption we know of the existence of some δ > 0 such that for all x X with d (x, p) < δ we have d (f(x), a) < ϵ. Rewriting this in terms of open balls we see that f (B (a; ϵ)) contains B (p; δ). e laer clearly is a neighbourhood of p, proving the desired. e converse is easier, we leave it as an exercise.. We can now easily formulate the following, and the proof is not all that hard. Lemma.. Let f : X Y be a function between topological spaces. If Y is Hausdorff and lim x p f(x) = q and lim x p = q 2 then q = q 2. Proof. Suppose that q and q 2 are unequal. As Y is Hausdorff this gives us V around q and V 2 around q 2 without overlap. By definition of the limit, this yields A and A 2 such that p A i f (V i ) for i =, 2. We can compute p A A 2 f (V ) f (V 2 ) f (V V 2 ), from whence it is immediate that V overlaps V 2. is yields a contradiction, so q = q 2 must follow. We are not done yet, as the above lemma does not instantiate to Lemma.. Indeed, there exists some X R m such that not all limit points of X are included in X. An easy example of this would be the (open) unit interval within the real line. So to fully generalize our old lemma, we first need the notion of a limit point in an arbitrary. topological space.

15 .. Figure : Closed subsets in R 2. Definition. (Limit Point). Let X be a topological space and let A X be a subset. Now p X is a limit point of A whenever all neighbourhoods of p overlap with A. Of course, we again need to show that the above coincides with the old notion of a limit point. As one would suspect, it does. Lemma.. Let A R n be a subset of Euclidian n-space. Now p R n is a limit point of A in the sense of Definition.. if and only if it is a limit point in the sense of Definition.. From here on the generalization of Lemma. is quite clearly the following. Lemma.. Let A X be a subspace of a topological space, Y a topological space, and let f : A Y be a function. between topological spaces. Finally, let a A be a limit point of A. If lim x a f(x) = p and lim x a f(x) = q and Y is Hausdorff, then p = q. It is oen interesting to look at a lemma, and wonder what the minimal amount of required structure is. ink for instance of the following statement: Let f : R n R be a continuous function. en f ({c}) is closed for all c R. Proof. e set R {c} is open. Indeed, if p R {c} then ϵ = d (p, c) > 0, so B (p; ϵ) R {c}. is means that the pre-image under f of this set is open as well, due to the continuity of f. By the following equation we now know the desired. f ({c}) = f (R) f (R {c}). Note that this statement entails that the sets depicted in Figure are closed. is follows for they are the zero-sets of polynomials. To be precise, the le-hand figure is a circle (the zero-set of x 2 y 2 ) and the right-hand figure is Bernoulli s Lemniscate (the zero-set of (x 2 y 2 ) 2 2(x 2 y 2 )). We could generalize R n to a topological space X, but then the second sentence of the above proof would be uer nonsense. Of course, if we could prove that {c} were closed, then the desired would follow immediately from the remaining lines. In fact, even the following holds. Lemma. (Pre-image of Closed is Closed). Let f : X Y be a map and let B Y be closed. Now f (B) is. closed. But when can we prove that {c} is closed? is property is related to the space being Hausdorff. A space is Hausdorff when each pair of distinct points has distinct open neighbourhoods. We can make the laer statement slightly more. lax to obtain the desired seing. Lemma.. Let X be a topological space. e following are equivalent: (i) Ea singleton in X is closed; (ii) For ea pair of unequal points p and q in X there is an open U X su that p U and q U.

16 Proof. Suppose that (ii) holds. Now let {p} X be a singleton. Consider its complement X {p}, we wish to show that it is open. To this end, pick some point q X {p}. When we can find an open q U X {p} we are done. By (ii) we know of an open U such that q P but p U, so we have shown (i) to hold. To prove the converse, suppose that (i) holds. Let p, q X be unequal and define U := X {q}. By (i) we know U to be open, and it is clear that p U and q U, proving (ii). Lemma.. If X is Hausdorff, then ea singleton in X is closed. Proof. Let p q be distinct points in X. Now there are disjoints opens U and V containing p and q respectively. Suppose that q U, then q U V =, a contradiction. is proves (ii) of the above lemma, from whence the desired is immediate.. Invariance Two spaces may be defined differently, yet still be virtually the same. Recall that when one considers sets, N does not differ much from {2n n N}. e reason for this is that there is a bijection between these sets. In the topological seing, identifying spaces that are bijective would be a gross oversimplification. It is quite possible to have distinct topologies on the same underlying set, such that some properties are not shared. Try to think of some examples of this.. e proper type of map to consider is a homeomorphism, which we define below. Not only is this map a continuous bijection, but it also maps opens to opens. Due to these properties we know this kind of map to preserve topological properties, such as being Hausdorff. Definition. (Homeomorphism). A map f : X Y is said to be a homeomorphism whenever there exists a map g : Y X such that fg = id Y and gf = id X. Such an g is said to be inverse to f, and the spaces X and Y are said to be homeomorphic. Note that if f is a homeomorphism and g and g 2 are inverse to f, then g = g 2 must follow. Indeed, g (y) = g fg 2 (y) = g 2 (y) holds for all y Y, so g and g 2 are the same map. We can thus safely speak of the inverse of f, and denote this map by f. is should not be confused with the pre-image of some set under f, which we denote in the same way. e context ought to uniquely determine which of the two notions is meant. e identity map is a homeomorphism, and the composition of two homeomorphisms is again a homeomorphism. Lemma.. e relation of being homeomorphic is an equivalence relation.. Lemma.. Every homeomorphism is both injective and surjective.. Let us now look at some examples of homeomorphisms and thus of homeomorphic spaces. Example.. For any b R one has that the function R R which maps x to x + b is a homeomorphism.. Example.. Let a R be a number. e function R R which maps x to ax is a homeomorphism exactly if a. is not equal to 0. Example.. Consider some topological space X which has but finitely many points. Now X is not homeomorphic. to any (non-empty) interval, as an immediate corollary of Lemma.. In particular, no point is homeomorphic to an interval. Example.. e interval (a, b) is homeomorphic with any other interval (c, d) for any choice of a, b, c, d R. where a b and c d. Example.. e space R and (0, ) are homeomorphic. More elaborately, there is a homeomorphism between. the Euclidian real line, and the Euclidian unit interval. is requires several steps. First we define a function f : (0, ) R which maps x to x + x = 2x x(x ). One needs to show that this function is in fact continuous. We also need an inverse, and this inverse needs to be continuous.

17 All interesting properties are preserved under homeomorphism, as we illustrate below. In a sense, this means that these properties could be seen as defined on topological spaces up to homeomorphism. Lemma.. Let f : X Y be a homeomorphism. Now f(u) is open (in Y ) if and only if U is open (in X). Proof. For convenience, write g for the inverse of f. Now note that the pre-image of U under f is simply the image of U under g, and an analogous statement holds where the roles of f and g are switched. As a consequence, if U is open, then g (U) = f(u) must be open. Moreover, if f(u) is open, then so is f (f(u)) = g(f(u)) = id X (U) = U. is proves the desired. Lemma.. Let f : X Y be a homeomorphism, and let g : X Z be any function. Now g is continuous if and. only if gf is. Lemma.. If X and Y are homeomorphic, then X is Hausdorff if and only if Y is... Constructions is section describes the sensible topology one would put on the cartesian product of two topological spaces. Although interesting, this section can be skipped at first without great harm. Definition. (Product). Given two topological spaces X and Y, we can form the product-space X Y whose underlying set is X Y and whose opens are of the form U V where U and V are sets of opens in X and Y respectively. U U, V V Lemma.. Let n and m be natural numbers. Now R n R m is isomorphic to R n+m. Proof. e isomorphism is the obvious map on points. We need to check that the isomorphism is continuous. To do this, it suffices to check this for open balls. So consider B (p; ϵ) for some p R n+m and ϵ > 0. Note that p = p,..., p n, p n+,..., p n+m, and write p := p,..., p n and p := p n+,..., p n+m. We claim that B ( p ; 2 ϵ) B ( p ; 2 ϵ) is contained within B (p; ϵ). To see this, let x, y be an arbitrary element of the former. We now compute d ( x, y, p) = (x p ) (x n p n ) 2 + (x n+ p n+ ) (x n+m p n+m ) 2 (x p ) (x n p n ) 2 + (x n+ p n+ ) (x n+m p n+m ) 2 = d (x, p ) + d (y, p ) < ϵ, which that x, y B (p; ϵ). It is similarly easy to prove that the inverse map is continuous as well.. Below a characterization of the product of two topological spaces. It is not necessary to grasp the following at this point. Lemma.. For ea pair of topological spaces X and Y there exists a space T and maps π X : T X, π Y : T Y su that for all maps f : Z X and g : Z Y there is a unique map h : Z T su that π X h = f and π Y h = g.

18 Z. f T h g π X π Y X X Proof. Let X and Y be given. e desired space T is X Y, and the maps π X : X Y X and π Y : X Y Y are the obvious projection maps. One needs to check that these maps are actually continuous.. Now let maps f : Z X and g : Z Y be given as stated above. We define the map h as below. h = f, g :=: Z X Y, z f(x), g(x) It is not hard to see that is a bonafide map. To this end, it suffices to let U X and V Y be opens, and show that h (U V ) is open as well. So suppose that z Z is such that h(z) U V. is simply means that f(z) U and g(z) V. Rewriting this yields z f (U) and z g (V ), so z f (U) g (V ). Conversely, any z in the intersection of these two pre-images gets mapped to U V. Finally, we need to show that this map is unique with respect to satisfying π X h = f and π Y h = g. () So suppose that h : Z X Y is some map that satisfies (). is entails that h(z) = f(z), g(z) = f, g (z), hence h = f, g as desired. Corollary.. Suppose that one has maps f : X Y and g : X 2 Y 2. en there exists a unique map f g = h : X X 2 Y Y 2 such that π Y hπ X = f and π Y2 hπ X2 = g. Corollary.. Suppose that f : R n R and g : R m R are maps. en so is the map f + g : R n+m R defined as mapping x, y to f(x) + g(y). Proof. Consider the composite below. f g. R n R m R R R R n+m e map f g comes from the above corollary. e first map is the homeomorphism of Lemma., and the one-but last map is another instance of this homeomorphism. e one missing map comes from Lemma., which we know to be continuous by Lemma.. is entire composition can easily be seen to equal the composition as given in the lemma-statement. e desired is now but an immediate consequence of Lemma.. Lemma.. Suppose that the spaces X and Y respectively are homeomorphic to the spaces X 2 and Y 2. Now X Y. is homeomorphic to X 2 Y 2. R 2. Closures Every set is contained within a closed set. In the Euclidian case we saw that given a set A we can consider the set of all of its limit points, and this set in turn is closed. is set is the least closed set containing A. Let us generalize this to topological spaces. We first consider the dual notion, namely that of the largest open set contained within A.

19 Definition. (Interior). e interior of a subset A X within a topological space X is the greatest open contained within A. When it exists, it is denoted by int A. We first of all show that the interior always exists. Lemma.. Let A X be a subset of a topological space X. e interior int A is given as the union of all opens contained in A. Proof. We need to show that the union of all open subsets contained within A is the greatest open subset of A. To this end, write U for the set of opens contained within A. First note that this collection is nonempty, as it contains the empty set. We see that U is an open set, as guaranteed by the definition of a topological space. All we have to do is show that there is no greater open contained in A. So take some V A that we assume to be open. en V U, and consequently we know V U. is proves the desired. Lemma.. e interior of an open set the open set itself.. Corollary.. e operation of taking the interior is idempotent. at is, if you take the interior twice, then you just get the original interior. Lemma.. If A B X then int A int B for any topological space X.. Closed sets behave dually to open sets. at is to say, the union of two closed sets is open, and the intersection of arbitrarily many of them is closed. Moreover, the empty set and the entire space are both closed. We already saw that arbitrary intersections of opens do not yield opens, let us here illustrate how the arbitrary union of closed sets does not yield a closed set. Example.. Write I r for the closed interval [ r, r] on the real line. We can take the union { } I r = x R x [ r, r] for some 0 r < = (, ), 0 r< which is not closed. Indeed, is a limit point of this interval, yet (, ). But Lemma. tells us that the limit points of a closed interval are contained within it, so (, ) could not possibly be closed. Our example ended with quite an interesting observation: an open interval is not closed. One might mistake this to mean that a set is open if and only if it is closed, this of course is nonsense in general. We can prove the following. Lemma.. Consider A R n. Now it is not the case that A is both open and closed.. Definition. (Closure). e closure of a subset A X within a topological space X is the least closed set containing A. When it exists, it is denoted clos A. e closure is dual to the interior, in the following very precise sense. Lemma.. Let A X be a subset of a topological space. Now clos A = X int X A. Proof. First, note that int X A must be open, so X int X A is closed. Secondly we remark that A clos A. Indeed, if a A then a X A, and as int X A X A it now follows that a X int X A. Finally, suppose that A B is a closed set. en X B is open, and of course X B X A. As a consequence, we know that X B int X A, so B X int X A as desired. Corollary.. Let A X be a subset of a topological space. e closure of A is the intersection of all closed sets. of X containing A. Recall the notion of a limit point from Definition.. In the Euclidian seing we know that a closed subset contains all its limit points. In the generic seing this holds as well, and is a direct corollary of the following.

20 Lemma.. Let A X be a subset of a topological space. Now a clos A if and only if a is a limit point of A. Proof. Take an arbitrary point a X and suppose that it is a limit point of A. We wish to show that a clos A. We proceed by contradiction, that is, we assume that a clos A. is means that there is some closed set A B such that a B. As a consequence, we have know that the open X B is an open containing a which does not overlap with A. is is a clear contradiction with the assumption that a is a limit point of A. It thus follow that a clos A. Consider some topological space X, and fix a point p X. Let us spend the remainder of this section inspecting the closure of the singleton {p}, and sets which are similar to it in a way. You can skip this without serious repercussions at first reading. In the case that X is Hausdorff we know by Lemma. that {p} itself is closed, whence taking its closure does not change anything. Yet in general, in a non-hausdorff space, things may happen. We do at least know that the following property holds of closures of singletons. Definition. (Irreducible Closed). A closed subset = T X of a topological space X is said to be irreducible when for any pair of closed sets A and B the fact A B = T entails T = A or T = B. Lemma.. Let X be a topological space and let p X be a point. e closure of {p} is irreducible. Proof. Let A and B be disjoint closed sets whose union together makes T := clos{p}. We know that p A or p B. As T is the least closed set containing p, and A and B are both closed, it follows that either T = A or T = B. is proves the desired. Lemma.. Let X and Y be homeomorphic spaces. Now X is irreducible if and only if Y is.. Given two points, one can wonder whether the closures of these points coincide. In some types of topological spaces the laer entails that the two points are equal. is is of course the case in a Hausdorff space, as follows directly from the extensionality of sets. Let us generalize a bit. We say that a space X is a Kolmogorov space when each for each pair of points p, q X at least one of them has an open neighbourhood not containing the other. One can prove that this property is entailed by either of the equivalent properties in Lemma.. Recall that these properties are satisfied by Hausdorff spaces, so a Hausdorff space is always Kolmogorov. Lemma.. Let X be a Kolmogorov space. For any pair of points p, q X we know clos{p} = clos{q} if and only. if p = q. Example. (Prime Spectrum is Kolmogorov). Recall X from Example.. Let p q X be distinct points. First suppose that neither p nor q are zero. In this case it is easy to see that U(p) does not include p, but does include q. Now if p = 0 then pick an arbitrary prime distinct from q, and call this prime n. e open U(n) does include q, but it does not include 0. is proves that X is Kolmogorov. From the above it is not hard to see that X does not satisfy the properties of Lemma.. Indeed, there is no nonempty open excluding 0. For suppose that 0 U(n), then this means that 0 n. But 0 n is only true when n = 0, and U(0) =. Convince yourself that this means that the closure of {0} is all of X.. e following illustrates that we could not have generalized any further. Lemma.. Let X be a topological space, and suppose that for any pair of points we know p = q if and only if clos{p} = clos{q}. en X is Kolmogorov. Proof. Let p and q be a pair of distinct points. We need to find an open neigbhourhood around one of these points not containing the other. More formally, we wish to prove: there is an open U such that p U and q U or p U and q U. ()

21 Suppose () is not the case. en for every open U we have that either p U and q U or p U and q U. is means that p U holds if and only if q U holds. Clearly, this also entails that a closed set contains p if and only if it contains q. As a consequence we know that clos{p} = clos{q}, so p = q must follow. is is in contradiction with the assumption that p q, which means that () does hold, proving the desired. Lemma.. Let X and Y be homeomorphic spaces. Now the space X is Kolmogorov if and only if the space Y is.. e above Lemma. entails that if some irreducible set is the closure of some point, that point must be unique. But is every irreducible set the closure of a point? In general, this is not the case. Below we give a non-example, as arising from algebraic topology. Note that the example in particular entails that the space is not Hausdorff. Example. (Zariski Topology on the Real Plane). Let Z be the Euclidian plane, where opens are exactly the non-zero sets of polynomials. In symbols, a set U is open whenever U = U(I) := { x, y R n for all f I we have f(x, y) 0 } for some set I of polynomials in two variables. Of course, a closed set is now exactly the zero-set of a bunch of polynomials. Convince yourself that this is a bonafide topology. As a polynomial yields a continuous function, and. as {0} is closed in R with the Euclidian topology, we know that the usual Euclidian topology on R 2 has at least as many closed sets as the space Z. We claim that there exists an irreducible closed set in Z which is not the closure of a point. is set will be the zero-set of the polynomial x, i.e. the y-axis. is set is indeed irreducible, for suppose that it is the union of the non-zero sets of polynomials f and g. en it would also be the non-zero set of the polynomial f g. But this can only happen if f = x or g = x. Any point p on the y-axis is closed, for it is the zero-set of the polynomials x and y p. So the closure of any point on the y-axis is that point itself, and never will be the entire y-axis. A space in which the irreducible sets are the closures of unique points are said to be sober. Any Hausdorff space in particular is sober. Many spaces one encounters in analysis are Hausdorff, so why are we interested in properties strictly weaker than being Hausdor? Well, many spaces from algebraic geometry are not Hausdorff, but they are in fact sober. Moreover, sober spaces are especially well-suited for point-free topology, see for instance Johnstone () and Johnstone (). Lemma.. Let X be a topological space. If X is Hausdorff, then X is sober. Proof. By Lemma. and Lemma. we need but to prove that each irreducible set is the closure of a singleton. is of course comes down to proving that each irreducible set is a singleton. So suppose that T is irreducible and distinct p and q exist in T. en there are opens U and V around p and q respectively such that U V =. Take A := (X U) T and B := (X V ) T and note that both A and B are closed. We can compute A B = ( (X U) T ) ( (X V ) T ) = ( (X U) (X V ) ) T = ( X (U V ) ) T = (X ) T = T, so it must follow that T = A or T = B. But this means that p T = X U p or q T = X V q, a clear contradiction. Hence no such distinct points p and q exist in T, which means that T must be a singleton. Example. (Prime Spectrum is Sober). Recall the space X from Example., and also remember that this space was shown to not be Hausdorff in Example.. We use X as an example of a non-hausdorff space which is sober. Any closed set is of the form X U(n), that is to say, it is the set of divisors of n. It is clear that U(n) = U(m) if and only if n and m have the same divisors in X. We claim that T = X U(n) is irreducible if and only if n X.