豪斯多夫及其点集拓扑学
记得，60多年前，袁萌在南京大学图书馆阅览室整天“死啃”豪斯多夫点集拓扑学（1921年英文版。但是，原版书不外借）。
至今，时间过去了100多年，豪斯多夫当年首创的基本概念至今没有丝毫改变。令人惊叹不已也。
大家知道，现代微积分学的理论基础就是点集拓扑，豪斯多夫的贡献由此可见一那。
请见本文附件，
袁萌 陈启清 3月4日
附件：
Introduction to Point-Set Topology
By KC Border v. 2018.10.03::13.35
Abstract（摘要）
These notes are gathered from several of my other handouts, and are a terse introduction to the topological concepts used in economic theory. For further study I recommend Willard [4] and Wilanksy [3]. You may also be interested in my on-line notes on metric spaces [2].
Contents 1
Topological spaces 1
2 Relative topologies 2
3 Neighborhoods, interiors, closed sets, closures 2
4 Bases 3
5 Product topology 3
6 Continuous functions 3
7 Homeomorphisms 4
8 Compactness 4
9 Topological vector spaces 6
1 Topological spaces You should know that the collection of open subsets of Rm is closed under finite intersections and arbitrary unions. Use that as the motivation for the following definition. 1 Definition A topology τ on a nonempty set X is a family of subsets of X, called open sets satisfying 1. ∅∈ τ and X ∈ τ.
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2. The family τ is closed under finite intersections. That is, if U1,...,Um belong to τ, then ∩m i=1 Ui belongs to τ. 3. The family τ is closed under arbitrary unions. That is, if Uα, α ∈ A, belong to τ, then ∪α∈A Uα belongs to τ. The pair (X,τ) is a topological space. The topology τ is a Hausdorff topology if for every two distinct points x,y in X there are disjoint open sets U, V with x ∈ U and y ∈ V. The collection of open sets in Rm is a Hausdorff topology. A property of X that can be expressed in terms of its topology is called a topological property.
2 Relative topologies 2 Definition (Relative topology) If (X,τ) is a topological space and A ⊂ X, then (A,τA) is a topological space with its relative topology, where τA ={G∩A : G ∈ τ}. Not that if τ is a Hausdorff topology, then τA is also a Hausdorff topology.
3 Neighborhoods, interiors, closed sets, closures 3 Definition Theset A isaneighborhoodof x ifthereisanopenset U satisfying x ∈ U ⊂ A. We also say that x is an interior point of A. The interior of A, denoted intA, is the set of interior points of A. 4 Lemma A set is open if and only it is a neighborhood of each of it points. Proof: Clearly an open set is a neighborhood of each of its points. So assume the set G is a neighborhood of each of it points. That is, for each x ∈ G there is an open set Ux satisfying x ∈ Ux ⊂ G. Then G =∪x∈G Ux is open, being a union open sets. 5 Exercise The interior of any set A is open (possibly empty), and is indeed the largest open set included in A. □ 6 Definition A set is closed if its complement is open. The closure of a set A, denoted A or clA, is the intersection of all the closed sets that include A. 7 Exercise Theunionoffinitelymanyclosedsetsisclosedandtheintersectionofanarbitrary family of closed sets is closed. □ 8 Exercise The closure of A is the smallest closed set that includes A. □ 9 Lemma ApointxisnotinA,thatis, x ∈(A)c,ifandonlyifthereisanopenneighborhoodU of x disjoint from A.
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Proof: (⇐=) If x ∈ U, where U is open and U ∩A = ∅, then the complement Uc is a closed set including Ac, so by definition Ac ⊂ Uc. Thus x / ∈ Ac.( = ⇒ ) Since A is closed, if x / ∈ A, then (A)c is an open neighborhood of x disjoint from A, so a fortiori disjoint from A. 10 Definition The boundary of a set A, denoted ∂A, is A∩Ac. 11 Corollary ∂A = A\intA. Proof: By Lemma 9, intA =(Ac)c. Thus A\intA = A∩Ac = ∂A. 4 Bases 12 Definition A family G of open sets is a base (or basis) for the topology τ if every open set in τ is a union of sets from G. A neighborhood base at x is a collection N of neighborhoods of x such that for every neighborhood G of x there is a neighborhood U of x belong to N satisfying x ∈ U ⊂ G. In a metric space, the collection of open balls {Bε(x): ε > 0, x ∈ X} is base for the metric topology, and {B1/n(x): n > 0} is a neighborhood base at x. Given a nonempty family A of subsets of X there is a smallest topology τA on X that includes A, called the topology generated by A. It consists of arbitrary unions of finite intersections of members of A. If A is closed under finite intersections, then A is a base for the topology τA. 5 Product topology 13 Definition If X and Y are topological spaces, the collection sets of the form U×V, where U is an open set in X and V is an open set in Y, is closed under finite intersections, so it is a base for the topology it generates on X ×Y, called the product topology. 6 Continuous functions 14 Definition Let X and Y be topological spaces and let f: X → Y. Then f is continuous if the inverse image of open sets are open. That is, if U is an open subset of Y, then f−1(U) is an open subset of X. This corresponds to the usual ε-δ definition of continuity that you are familiar with. 15 Lemma Afunction f: X → Y iscontinuousifandonlyiftheinverseimageofeveryclosed set is closed. 16 Lemma If f: X → Y is continuous, then for every A ⊂ X, we have f(A)⊂ f(A). Proof: Since f is continuous and f(A) is closed, f−1(f(A))is a closed set that clearly includes A,andsoincludesitsclosure A. Thatis, A ⊂ f−1(f(A)),so f(A)⊂ f(f−1(f(A)))= f(A). v. 2018.10.03::13.35
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7 Homeomorphisms 17 Definition Let X and Y be topological spaces. A function f: X → Y is a homeomorphism if it is a bijection (one-to-one and onto), is continuous, and its inverse is continuous. If f is homeomorphism U ↔ f(U) is a one-to-one correspondence between the topologies of X and Y. Thus X and Y have the same topological properties. They can in effect be viewed as the same topological space, where f simply renames the points.
8 Compactness Let K be a subset of a topological space. A family A of sets is a cover of K if K ⊂ ∪ A∈A A. If each set in the cover A is open, then A is an open cover of K. A family B of sets is a subcover of A if B ⊂ A and K ⊂∪A∈BA. For example, let K be a subset of R, and for each x ∈ K, let εx > 0. Then the family A ={(x−εx,x+εx): x ∈ K} of open intervals is a open cover of K. 18 Definition A set K in a topological space X is compact if for every family G of open sets satisfying K ⊂ ∪G (an open cover of K), there is a finite subfamily {G1,...,Gk} ⊂ G with K ⊂∪k i=1 Gi (a finite subcover of K). 19 Lemma If (X,τ) is a topological space and K ⊂ A ⊂ X, then K is a compact subset of (A,τA) if and only if it is a compact subset of (X,τ). Proof: Assume K is a compact subset of (X,τ). Let G be a τA-open cover of K in A. For each G ∈ G there is some UG ∈ τ with G = UG∩A. Then{UG : G ∈ G}is a τ-open cover of K in X, so it has a finite subcover UG1,...,UGk. But then G1,...,Gk is a finite subcover of K in A. The converse is similar. There is an equivalent characterization of compact sets that is sometimes more convenient. A family A of sets has the finite intersection property if every finite subset{A1,...,An}of A has a nonempty intersection,∩n i=1 Ai ̸= ∅. 20 Theorem A set K is compact if and only if every family of closed subsets of K having the finite intersection property has a nonempty intersection. Proof: Start with this observation: Let A be an arbitrary family of subsets of K, and define A ={K\A : A ∈ A}. By de Morgan’s Laws∩A∈A A = ∅ if and only if K =∪B∈A B. That is, A has an empty intersection if and only if A covers K. ( =⇒ ) Assume K is compact and let F be a family of closed subsets of K. Then F is a family of relatively open sets of K. If F has the finite intersection property, by the above observation, no finite subset of F can cover K. Since K is compact, this implies that F itself cannot cover K. But then by the observation F has nonempty intersection.
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( ⇐= ) Assume that every family of closed subsets of K having the finite intersection property has a nonempty intersection, and let G be an open cover of K. Then G is a family of closed having an empty intersection. Thus G cannot have the fintiie intersection property, so there is a finite subfamily G0 of G with empty intersection. But then G0 is a finite subfamily of G that covers K. Thus K is compact. 21 Lemma A closed subset of a compact set is compact. Proof: Let K be compact and F ⊂ K be closed. Let G be an open cover of F. Then G∪{Fc} is an open cover of K. Let {G1,...,Gk,Fc} be a finite subcover of K. Then {G1,...,Gk} is a finite subcover of F. 22 Lemma A compact subset of a Hausdorff space is closed. Proof: Let K becompact,andlet x / ∈ K. ThenbytheHausdorffproperty,foreach y ∈ K thereare disjoint open sets Uy and Vy with y ∈ Uy and x ∈ Vy. By compactness there are y1,...,ykwith K ⊂∪k i=1 Uyi = U. Then V =∩k i=1 Vyi is an open set satisfying x ∈ V ⊂ Uc ⊂ Kc. Thatis, Kc is a neighborhood of x. Since x is an arbitrary member of Kc, we see that Kc is open (Lemma 4), so K is closed. 23 Lemma Let f: X → Y be continuous. If K is a compact subset of X, then f(K) is a compact subset of Y. Proof: Let G be an open cover of f(K). Then {f−1(G) : G ∈ G} is an open cover of K. Let {f−1(G1),...,f−1(Gk)} be a finite subcover of K. Then {G1,...,Gk} is a finite subcover of f(K). 24 Lemma Let f: X → Y be one-to-one and continuous, where Y is a Hausdorff space and X is compact. The f: X → f(X) is a homeomorphism, where f(X) has its relative topology as a subset of Y. Proof: We need to show that the function f−1: f(X) → X is continuous. So let G be any open subset of X. We must show that (f−1)−1(G)= f(G) is open in f(X). Now Gc is a closed subset of X, and thus compact. Therefore f(Gc) is compact, and since Y is Hausdorff, so is f(X),so f(Gc)isaclosedsubsetof Y. Now f(X)∩f(Gc)c = f(G),so f(G)isopenin f(X). 25 Weierstrass’s Theorem If K is compact and f: K → R is continuous, then there exists a point x∗ in K that maximizes f. That is, (∀x ∈ K) [f(x∗)⩾ f(x)]. Proof: Since f iscontinuous, Fα ={x ∈ K : f(x)⩾ α}= f−1([α,∞))isclosedforeach α ∈ R, as inverse images of closed sets are closed for a continuous function. Let A = {f(x) : x ∈ K} denote the range of f. Then F = {fα : α ∈ A} is family of nonempty closed subsets of K having the finite intersection property.1 Since K is compact, M = ∩ α∈A Fα is nonempty. Let x∗ belong to M and let x be any point in K. Set α = f(x). Then x∗ ∈ Fα, which means that f(x∗)⩾ α = f(x). That is, x∗ maximizes f over K. 1Let α = max{α1,...,αn}. Then Fα = Fα1 ∩···∩Fαn, and Fα ̸= ∅ since α belongs to A, the range of f.
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Notethatthisproofworksifweonlyassumethateach Fα isclosed, thatis, that f isupper semicontinuous. The following result is well known. 26 Heine–Borel–Lebesgue Theorem A subset of Rm is compact if and only if it is both closed and bounded in the Euclidean metric. This result is special. In general, a subset of a metric space may be closed and bounded without being compact. (Consider the coordinate vectors in ℓ∞.) 9 Topological vector spaces For a detailed discussion of topological vector spaces, see chapter five of the Hitchhiker’s Guide [1]. But here are some of the results we will need. 27 Definition A (real) topological vector space is a vector space X together with a topology τ where τ has the property that the mappings scalar multiplication and vector addition are continuous functions. That is, the mappings (α,x)7→ αx from R×X to X and (x,y)7→ x+y from X×X to X are continuous. (Where, of course, R has its usual topology, and R×X and X ×X have their product topologies.) 28 Lemma If V is open, then V +y is open. Proof: Since f: x 7→ x−y is continuous, V +y = f−1(V) is open. 29 Lemma If V is open, and α ̸=0, then αV is open. Proof: Since f: x 7→(1/α)x is continuous, αV = f−1(V) is open. 30 Definition A set C in a vector space is circled or radial if αC ⊂ C whenever |α|⩽1. 31 Lemma Let V be a neighborhood of zero. Then there is an open circled neighborhood U of zero included in V. Proof: The mapping f: (α,x)7→ αx is continuous, and f(0,0) = 0, the inverse image f−1(V) is a neighborhood of 0. Thus there is an δ > 0 and an open neighborhood W of 0 such that (−δ,δ)×W ⊂ f−1(V). This implies that for any α with |α| < δ and x ∈ W, we have αx ∈ V. In other words αW ⊂ V. Set U = ∪ α:0<|α|<δ αW Then U ⊂ V, U is circled, and U is open, being the union of the open sets αW. v. 2018.10.03::13.35
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32 Lemma Let T : X → Y bealineartransformationbetweentopologicalvectorspaces. Then T is continuous on X if it is continuous at 0. Proof: It suffices to prove that T is continuous at each point x. So let V be an open neighborhood of T(x). Then V −T(x) is an open neighborhood of 0. Since T is continuous at 0, the inverse image T−1(V −T(x)), is a neighborhood of 0, so T−1(V −T(x))+x is a neighborhood of x. But by linearity, T−1(V −T(x))+x = T−1(V), and we are done. References [1] C.D.AliprantisandK.C.Border.2006. Infinitedimensionalanalysis: Ahitchhiker’sguide, 3d. ed. Berlin: Springer–Verlag.
[2] K. C. Border. 2013. What to remember about metric spaces. http://www.its.caltech.edu/~kcborder/Notes/MetricSpaces.pdf
[3] A. Wilansky. 1998. Topology for analysis. Mineola, NY: Dover. Unabridged republication of the work originally published by Ginn and Company, Waltham, MA in 1970. [4] S. Willard. 1970. General topology. Reading, Massachusetts: Add