点集拓扑学基础（II）

我们的目的是学懂、学透微积分。但是，只懂得欧几里得平面基本知识还不够，中间缺少欧几里得平面点集拓扑的概念。

本文附件（英文）阐述了欧几里得平面点集拓扑的基本概念（II），值得一读。

袁萌陈启清 3月23日

附件：

8. Connectedness and the Intermediate Value Theorem

Definition 8.1. A topological space (X,T) is said to be connected provided that the only subsets of X which are both open and closed are the empty set and the whole space X. (By axioms (T1) and (T2) we know that the sets ∅ and X are always both open and closed.)

Many authors give the deﬁnition of connectedness of X as: if X = U ∪V where U and V are open sets which are disjoint then either U or V is the empty set. [Convince yourself of the equivalence of these deﬁnitions.] Example 8.2. The trivial topology Ttrivial on a set X is connected since in this topology the only open sets are ∅ and X. If X is a set with at least two distinct elements then the discrete topology Tdiscrete on X is not connected: because if x0 ∈ X then the singleton set {x0} is both open and closed (in the discrete topology, every subset of X is both open and closed) but {x0}6= ∅ and {x0}6= X.

Example 8.3. Consider the set A2 from example 7.7. Since [0,1] is an open and closed set which is neither ∅ nor A2, we conclude that A2 with the subspace topology is not a connected topological space. This example should give you an idea of where the term ‘connected’ comes from. (Draw a picture of A2 in R, does it look like it should be ‘connected’?) The discussion of this example A2 ⊂R leads directly to the next result.

Theorem 8.4. If A is a subset of R which is not an interval then A is not connected in the Euclidean topology. Proof. If A is not an interval then there is a real number x0 ∈ R−A such that A contains at least one real number smaller than x0 and at least one number larger than x0. Then (−∞,x0) ∩ A is both open and closed in the subspace topology on A but it is neither the empty set nor all of A.

Theorem 8.5. Every interval in the real line is connected in the Euclidean topology.

Proof. We will give the proof only in the case where the interval is a closed ﬁnite interval [a,b] where a < b. It is not diﬃcult to adapt the proof to handle the other possible intervals such as R = (−∞,+∞), (−∞,b], [a,b) and so on. Suppose that U is a nonempty subset of [a,b] which is both open and closed in the Euclidean topology. In particular, this means that [a,b]−U is open (since U is closed). By replacing U with [a,b]−U if necessary, we may assume that a ∈ U. The goal of the proof is to show that U = [a,b]; if we can show this then it follows that [a,b] is connected. Deﬁne a subset S of [a,b] by S = {s ∈ [a,b] | [a,s] ⊆ U} Observe that S ⊆ U, that a ∈ S (since [a,a] = {a}), and that if s ∈ S and a < t < s then t ∈ S. This shows that S must be an interval of the form [a,m) or [a,m] for some m ∈ [a,b]. We now consider three diﬀerent cases. Case 1: S = [a,m) and m / ∈ U. Since m / ∈ U, m is an element of [a,b]−U which is an open

subset of [a,b]. It follows that there is an > 0 such that (m−,m + )∩[a,b] ⊆ [a,b]−U. If s is an element of (m−,m + )∩[a,b] which is smaller than m then s ∈ S ⊆ U, which is impossible since m / ∈ U. Therefore Case 1 can’t happen. Case 2: S = [a,m) and m ∈ U. Then there is an > 0 such that (m−,m+)∩[a,b] ⊆ U. Let s be an element of (m−,m + )∩[a,b] which is smaller than m. Then [a,s] ⊆ U (since s ∈ S) and [s,m] ⊆ U. This implies that [a,m] = [a,s]∪[s,m] ⊆ U, and that m ∈ S, which is impossible. Therefore Case 2 can’t happen. Case 3: S = [a,m] and m < b. In this case m ∈ U since m ∈ S. So there is an > 0 with (m−,m + ) ⊆ U. Let s be an element of this open interval which is larger than m. Then s / ∈ S (since s > m) but s ∈ S (since [a,s] = [a,m]∪[m,s] ⊆ U). This contradiction shows that Case 3 can’t happen. Since none of the three cases can happen, we conclude that S = [a,m] where m = b. By the deﬁnition of the set S this means that U = [a,b], and the proof is complete. Theorem 8.6. Let X and Y be topological spaces and let f : X → Y be a function which is continuous and onto. If X is connected then Y is connected.

Proof. Let U be a subset of Y which is both open and closed. Then f−1(U) is open in X since f is continuous. Also f−1(U) is closed in X by theorem 7.2. So f−1(U) is both open and closed. As X is connected we conclude that either (1) f−1(U) = ∅ or (2) f−1(U) = X. Using the hypothesis that f is onto, in case (1) U must be empty and in case (2), U must equal Y . Thus the only subsets U of Y that are both open and closed are ∅ and Y , so Y is connected.

We can now prove the Intermediate Value Theorem: Theorem 8.7. If f : [a,b] → R is continuous and c is a real number between f(a) and f(b) then there is an x ∈ [a,b] such that f(x) = c. Proof. Let f : [a,b] → R be continuous. Then the function g : [a,b] → f([a,b]) deﬁned by g(x) = f(x) for x ∈ [a,b] is continuous by theorem 7.8, and g is onto. We know that [a,b] is connected with the Euclidean topology by theorem 8.5, and so g([a,b]) = f([a,b]) is connected by theorem 8.6. Therefore f[a,b] must be an interval by theorem 8.4. Since f(a) and f(b) are elements of the interval f[a,b], any real number c between f(a) and f(b) must be in f([a,b]). This completes the proof.

This proof of the Intermediate Value Theorem readily extends to a more general statement, which has been applied in many diﬀerent situations since Hausdorﬀ’s time. Most of these applications have to do with being able to guarantee that there will exist solutions to certain types of equations. Theorem 8.8. Let X be a connected topological space and let f : X → R be continuous with respect to the Euclidean topology on R. If a and b are elements of X and c is a real number between f(a) and f(b) then there is an x ∈ X such that f(x) = c.

9. Compactness and the Extreme Value Theorem

Definition 9.1. Let (X,T) be a topological space. An open cover of X is a collection U of open subsets of X such thatSU = X whereSU =SU∈U U. In other words, U is an open cover of X iﬀ U ⊆T and for each x ∈ X there is at least one U ∈U such that x ∈ U. If U is an open cover of X then a subcover of U is a subset V ⊆U which also covers X (that is SV = X). An open cover is ﬁnite if it contains only ﬁnitely many elements. Definition 9.2. We say that the topological space (X,T) is compact provided that every open cover of X has a ﬁnite subcover.

Example 9.3. The coﬁnite topology Tcofinite on a nonempty set X is compact. Recall that Tcofinite = {∅}∪{X −F | F is a ﬁnite subset of X} . Suppose U is an open cover of X. Then one of the elements of U has the form X −F where F = {x1,...,xk} is a ﬁnite subset of X. For each i = 1,...,k, let Ui be an element of U with xi ∈ Ui. Then {X −F,U1,...,Uk} is a ﬁnite subcover of U.

Example 9.4. The Euclidean line R is not compact. For example, the collection of open intervals U = {(n−1,n + 1) | n ∈ Z} is an open cover of the Euclidean line which has no ﬁnite subcover. (This is because each integer n is contained in the interval (n−1,n + 1) but not in any other interval in U. So any proper subset of U will not cover R.) For a diﬀerent explanation, it can also be shown that the collection of open intervals{(−n,n) | n ∈Z+} forms an open cover of R with no ﬁnite subcover.

Example 9.5. The half-open interval [0,3) is not compact with the Euclidean topology. To see this observe that U = {[0,3 − 1 n) | n ∈ Z+} is an open cover of [0,3) which has no ﬁnite subcover. (Note that [0,3 − 1 n) is open in the subspace topology on [0,3) since [0,3− 1 n) = (−1,3− 1 n)∩[0,3). The union of any ﬁnite subset of U will not include 3− 1 N for some positive integer N.)

Theorem 9.6. Every closed ﬁnite interval [a,b] in the real line is compact in the Euclidean topology.

Proof. Let U be an open cover of [a,b] with respect to the Euclidean topology on [a,b]. Let S be the subset of [a,b] deﬁned by S = {s ∈ [a,b] | there is a ﬁnite subset V ⊆U with [a,s] ⊆[V} .Then a ∈ S and if s ∈ S and a < t < s then t ∈ S. It follows that S is an inerval of the form [a,m] or [a,m) for some m ∈ [a,b] (in the second form [a,m) we assume that m > a). We consider two cases: Case 1: S = [a,m] where m < b. In this case m is an element of S so there is a ﬁnite subset V ⊆ U with [a,m] ⊂SV. Let U be an element of V containing m. Then there is 19

an > 0 such that (m−,m + )∩[a,b] ⊆ U. If t is an element of (m−,m + )∩[a,b] larger than m then [a,t] = [a,m]∪[m,t] ⊆SV. Therefore there is an element t in S which is larger than m but this can’s happen since m was assumed to be the largest element of S. This contradiction shows that Case 1 can’t happen. Case 2: S = [a,m) where m > a. Let U be an element of the open cover U that contains m. Then there is an > 0 such that (m−,m+)∩[a,b] ⊆ U. Choose t ∈ (m−,m+)∩[a,b] smaller than m. Then t ∈ S so that there is a ﬁnite subset V in U with [a,t] ⊆SV. Then [a,m] = [a,t]∪[t,m] is contained inS(V∪{U}). Since V∪{U} is a ﬁnite subset of U, this implies that m ∈ S, which is a contradiction. Therefore Case 2 can’t happen. As neither Case 1 nor Case 2 can happen, we conclude that S = [a,m] where m = b. This means that U has a ﬁnite subcover, which shows that [a,b] is compact. Theorem 9.7. If C is a subset of R which does not have a largest (or a smallest) element then C is not compact with the Euclidean topology.

Proof. Let C be a subset of R which does not have a largest element. We consider two cases according to whether or not C is ‘bounded above’. A subset of R is bounded above if it is a subset of (−∞,H) for some real number H. Case 1: Suppose that C is not bounded above. Then {(−∞,n)∩C | n ∈ Z+} is an open cover of C which can’t have any ﬁnite subcover (since C is not bounded above). Case 2: Suppose that C is bounded above. Then C has a least upper bound m. Note that m ∈ C since C does not have a largest element. In this case {(−∞,m− 1 n)∩C | n ∈Z+} is an open cover of C which has no ﬁnite subcover. In either case, we conclude that C is not compact. A similar argument using lower bounds works if C does not have a smallest element.

The next theorem categorizes exactly which subsets of R are compact with the Euclidean topology (take n = 1). A subset C of Rn is bounded if there exists a positive real number N so that C ⊂ B(0,N). Note that C ⊂ R is bounded iﬀ there are real numbers L and H such that L ≤ x ≤ H for all x ∈ C. Here L is called a lower bound for C and H is an upper bound for C. Theorem 9.8 (Heine-Borel Theorem). A subset C of Euclidean n-space Rn is compact in the Euclidean topology if and only if C is closed and bounded.

One part of the proof of the Heine-Borel Theorem follows from the next result.

Theorem 9.9. If X is a compact topological space and C is a closed subset of X then C is compact with the subspace topology.

Proof. Assume that X is a compact topological space and that C is a closed subset of X. Let U be an open cover of C. Then for each U ∈ U there is a set U0 which is open in X such that U = U0 ∩ C. Since U covers C, C is a subset ofS{U0 | U ∈ U}. Since C is a closed subset of X, X −C is open and therefore {U0 | U ∈U}∪{X −C} is an open cover of X. Since X is compact this open cover has a ﬁnite subcover which can be taken to be {U0 | U ∈ V}∪{X −C} where V is a ﬁnite subset of U. Since X −C doesn’t contain any elements of C, V must cover C. This shows that V is a ﬁnite subcover of U and that C is compact.

When n = 1 it is not diﬃcult to complete the proof of the Heine-Borel Theorem, however we will not pursue it here as we won’t need the result in our subsequent discussions. It is also not diﬃcult to prove Heine-Borel Theorem when n > 1 although this does rely on some knowledge of ‘product topological spaces’, we will describe this concept later. Theorem 9.10. Let X and Y be topological spaces and let f : X → Y be a continuous function. If X is compact then range(f) = f(X) is compact with the subspace topology coming from Y .

Proof. Let f : X → Y be a continuous function where X is compact. Let U be an open cover of f(X) using the subspace topology that f(X) inherits as a subset of Y . For each U ∈ U there is an open set U0 in Y such that U = U0∩f(X). Because f is continuous, each of the sets f−1(U0) where U ∈ U is open in X. Moreover the collection {U0 | U ∈ U} is an open cover of X since f(X) ⊂SU∈U U0. Since X is compact this open cover of X has a ﬁnite subcover V. Then the collection {U ∈U | f−1(U0) ∈V)} is a ﬁnite subcover of U. This shows that f(X) is compact.

We can now assemble a proof of the Extreme Value Theorem. Theorem 9.11. Let f : [a,b] → R be a continuous function with respect to the Euclidean topologies on [a,b] and R. Then f has a maximum value and a minimum value. Proof. Let f : [a,b] → R be a continuous function with respect to the Euclidean topologies on [a,b] and R. By theorem 9.6 the closed ﬁnite interval [a,b] is compact. By continuity of f (using theorem 9.10) this implies the range of f is compact with the Euclidean topology. By theorem 9.7 the range of f must have a largest and a smallest element, and by deﬁnition these are respectively maximum and minimum values for f on [a,b].

This proof of the Extreme Value Theorem readily extends to a more general statement Theorem 9.12. Let X be a compact topological space and let f : X → R be a continuous function with respect to the Euclidean topology on R. Then there are elements L,H ∈ X such that f(L) ≤ f(x) and f(H) ≥ f(x) for all x ∈ X.

10. Metric Spaces

Definition 10.1. Let X be a set. A metric on X is a function d : X×X → [0,+∞) which satisﬁes the following properties: (M1) d(x,y) = 0 if and only if x = y. (M2) d(x,y) = d(y,x) for all x,y ∈ X. (M3) d(x,z) ≤ d(x,y) + d(y,z) for all x,y,z ∈ X. A metric d on X is sometimes referred to as a distance function on X, and d(x,y) is called the distance from x to y. The third axiom (M3) is known as the triangle inequality. A ﬁxed set X can have many diﬀerent metrics on it. If d is metric on X then we say that (X,d) is a metric space. When it is clear which metric d is being considered then we sometimes say that X is a metric space.

Example 10.2. The Euclidean metric on the real line R is the function d : R×R → [0,∞) deﬁned by d(x,y) = |x − y| for x,y ∈ R. [Verify that this forms a metric.] More generally, the Euclidean metric on Rn is the function d given by d(x,y) =(x1 −y1)2 +···+ (xn −yn)21/2where x = (x1,...,xn) and y = (y1,...,yn).

Definition 10.3. Let (X,d) be a metric space. Let x ∈ X and let be a positive real number. The open ball of radius centered at x is the set B(x,) = {y ∈ X | d(y,x) < }. If there is any doubt as to which metric d is used to deﬁne an open ball, we write Bd(x,) in place of B(x,).

Theorem 10.4. Let (X,d) be a metric space. The collection of subsets of X given by Td = {U ⊆ X | for each x ∈ U there is > 0 such that B(x,) ⊆ U} forms a topology on X. This topology is called the metric topology on X associated with d.

Proof. We need to verify the four topology axioms (T1)-(T4). This is proof is exactly the same as the proof of theorem 3.2.

For each x ∈ X and each > 0, the open ball B(x,) is an open set in the metric topology Td. To prove this suppose that y ∈ B(x,), and put δ = −d(x,y) (which is a positive real number since d(x,y) < ). If z ∈ B(y,δ) then d(z,x) ≤ d(z,y) + d(y,x) < δ + d(x,y) = −d(x,y) + d(x,y) = . This shows that B(y,δ) ⊆ B(x,) and completes the proof that B(x,) is an open set in the metric topology. On the other hand, it is not true that every open set in the metric topology is an open ball. So the open balls are very special examples of open sets. However, the deﬁnition of the metric topology in theorem 10.4 can be seen to be equivalent to saying that every open set in Td is a union of open balls. [Can you verify this? Look back at theorem 5.3.] If B is a subset of a topology T with the property that every element of T is a union of elements of B then we say that B is a basis for the topology T. So we can say that the collection of open balls B = {B(x,) | x ∈ X and > 0} forms a basis for the metric topology on X. Here are some more examples of metric spaces. Example 10.5. Let X be any set and let d : X×X → [0,∞) be deﬁned by taking d(x,y) = 0 if x = y and d(x,y) = 1 if x 6= y. It is easy to verify that (X,d) is a metric space. Moreover notice that for x ∈ X, the open ball B(x,) is the singleton set {x} if ≤ 1. This shows that every singleton subset of X is open in the metric topology, and so Td = Tdiscrete. This metric on X is called the discrete metric.

Example 10.6. On R2, a metric ρ can be deﬁned by the equation: ρ(x,y) = max{|x1 −y1|,|x2 −y2|} where x = (x1,x2) and y = (y1,y2). [Verify the metric space axioms.] With this metric, an open ball Bρ(x,) is actually an open square with horizontal and vertical sides whose center is at x (the side length of the square is 2). Since every circle has an inscribed square and vice-versa, it can be shown that the metric topology Tρ on R2 is the same as the Euclidean topology. The moral of this example is that diﬀerent metrics on a set X may very well generate the same topology. Here’s a more careful explanation that Tρ coincides with the Euclidean topology on R2. Let d be the standard Euclidean metric on R2 given by d(x,y) = (x1 −y1)2 + (x2 −y2)21/2where x = (x1,x2) and y = (y1,y2). Then it is not hard to verify that, for each > 0 and x = (x1,x2) ∈R2, we have Bρ(x,) ⊃ Bd(x,) and Bd(x,) ⊃ Bρ(x,/ √2) . Now suppose that U is a subset of R2 which is open in the ρ–metric topologyTρ. By deﬁnition this means that for each x ∈ U there is > 0 such that Bρ(x,) ⊆ U. Then Bd(x,) is contained in U (since Bd(x,) is a subset of Bρ(x,)), and this shows that U is open in Td = Teuclid. On the other hand, suppose that U is open in the Euclidean topology on R2. This means that for each x ∈ U there is > 0 so that Bd(x,) ⊆ U. Then Bρ(x,/√2) ⊆ U and so U is open in the ρ–metric topology Tρ. Thus Tρ = Td = Teuclid, as claimed.

Example 10.7. If (X,d) is a metric space and A is a subset of X. Then the function d0 : A×A → [0,∞) deﬁned by d0(x,y) = d(x,y) for x,y ∈ A is a metric on A. Note that, for x ∈ A, Bd0(x,) = Bd(x,)∩A. From this it follows that the metric topology Td0 on A is the same as the subspace topology coming from the metric topology Td on X.

There are many advantages to being able to work with a metric space topology rather than an arbitrary topology on a space. One of these is that in metric spaces, many aspects of the topology can be described in terms of ‘convergent sequences’. For example, if X is a metric space and A is a subset of X then an element x ∈ X will be a limit point of A if and only if there is a sequence of elements in A which converges to x. Of course we haven’t deﬁned what a convergent sequence is but it can be deﬁned just like in calculus. That is, a sequence (xn) in the metric space (X,d) converges to x ∈ X iﬀ for each > 0 there is a positive integer N such that d(xn,x) < for all n ≥ N. In metric topologies, sequences can also be used to describe the notion of compactness—so in metric spaces certain basic topological deﬁnitions can be formulated in diﬀerent ways.

11. The Metrizability Problem

The deﬁnition of ‘metric space’ and its associated topology was ﬁrst introduced by Frechet about ten years before Hausdorﬀ’s more general abstract deﬁnition of ‘topological space’. The

early history of topology was dominated by a desire to understand the relationship between these two deﬁnitions. To explain more precisely, we say that a topological space (X,T) is metrizable if there is a metric d on X so that T = Td. So the basic questions became: is every topological space metrizable? if not, are there some additional axioms on a topological space that guarantee metrizability? In order to study these ‘metrizability problems’, many fundamental properties of topological spaces were discovered and interrelationships developed. For example these studies led to various ‘separation axioms’ and ‘countability axioms’ which play a role throughout the subject. In the 1920’s, a Russian mathematician Pavel Urysohn used some of these ideas to solve a special version of the metrizability problem which we will describe below (without proof). Ultimately, in the early 1950’s, a complete (although complicated) solution was found to the most general version of the metrizability problem, bringing an end to the early history of the subject. Here are two of the ‘separation axioms’ which play a role in the metrizability problem. Let (X,T) be a topological space. Then X is a Hausdorﬀ space provided that whenever x1 and x2 are distinct elements of X then there are neighborhoods U1 of x1 and U2 of x2 such that U1 ∩U2 = ∅. The topological space X is said to be normal provided that whenever C1 and C2 are disjoint closed sets in X then there are disjoint open sets U1 and U2 with C1 ⊆ U1 and C2 ⊆ U2. One can show that in a Hausdorﬀ topological space every singleton set is closed, and this weaker property is commonly referred to as the T1–axiom. Many authors take as part of the deﬁnition of a normal space that the space satisfy the T1–axiom. In the presence of this axiom (but not otherwise) it is easy to see that every normal space is Hausdorﬀ.

Theorem 11.1. Let (X,d) be a metric space. The metric topology on X is Hausdorﬀ. In fact, it is normal and T1.

Proof. We will only prove the Hausdorﬀness. Suppose that (X,d) is a metric space and consider the associated metric topology on X. Let x1 and x2 be distinct elements of X. Set to be half the distance from x1 to x2. That is = d(x1,x2)/2 which is a positive real number (by (M1) since x1 6= x2). Then x1 ∈ B(x1,) and B(x1,) is a neighborhood of x1 (since open balls are open in the metric topology). Likewise, B(x2,) is a neighborhood of x2. If z ∈ B(x1,)∩B(x2,) then d(x1,x2) ≤ d(x1,z) + d(z,x2) < + = 2= d(x1,x2), and this shows that d(x1,x2) is strictly less than d(x1,x2) which is clearly impossible. From this contradiction we conclude that B(x1,)∩B(x2,) is empty, thus showing that the metric topology is Hausdorﬀ.

One consequence of this theorem is that a topological space which is not Hausdorﬀ (or normal) can’t possibly be metrizable. For example, the trivial topology on a set with more than one elements is not metrizable. Also, the coﬁnite topology on an inﬁnite set is T1 but not Hausdorﬀ, so it is not metrizable either. Urysohn proved the following remarkable theorem:

Theorem 11.2 (Urysohn’s Lemma). Let X be a normal topological space, and let A and B be disjoint closed sets in X. Then there is a continuous function f : X → [0,1] with f(A) = 0 and f(B) = 1 where the interval [0,1] has the Euclidean topology.

And he used this result to establish:

Theorem 11.3 (Urysohn’s Metrization Theorem). Let X be a normal T1 topological space which has a basis B with countably many elements in it. Then X is metrizable.

The condition that X has a basis with countably many elements is an example of one of the ‘countability axioms’ for a topological space. This particular condition is often called second countability of X. For the record, a set B is countable (or B has ‘countably many elements’) if and only if there is a surjective function from Z+ to B. Another of the countability axioms (one that is weaker than second countability) is known as ‘separability’: a topological space is separable provided that there is a countable subset of X whose closure equals X. Urysohn’s Metrization Theorem only solves the metrizability problem for separable topological spaces— the complete solution took another thirty odd years for mathematicians to complete. [More precisely, what we are claiming here is that a separable metric space is second countable. In fact, a metric space is separable if and only if it is second countable. See if you can prove this.] A key role in the proof of Urysohn’s Metrization Theorem is played by the construction of ‘product of topological spaces’. Since this deﬁnition is very important in any graduate level course in point set topology, we include it here for completeness. Let (X,T1) and (Y,T2) be topological spaces. Then the collection of subsets of X ×Y given by {Z ⊆ X ×Y | for each (x,y) ∈ Z there are U ∈T1 and V ∈T2 with (x,y) ∈ U ×V ⊆ Z) } forms a topology Tproduct called the product topology on X×Y . Equivalently, the product topology is the topology for which the collection of sets

{U ×V | U ∈T1 and V ∈T2} forms a basis. The sets U ×V where U ∈T1 and V ∈T2 are often called open rectangles in X ×Y . Therefore, each open rectangle in X ×Y is an open set in the product topology, but not every open set need be an open rectangle. In general, a subset of X ×Y is open in the product topology iﬀ it can be expressed as a union (possibly inﬁnite) of open rectangles.

12. The Homeomorphism Problem

As the basic principles of topology evolved another fundamental problem which is philosphically very diﬀerent than the metrization problem emerged as an important central problem in the ﬁeld. The problem is frequently referred to as ‘the homeomorphism problem’. It has its roots in nineteenth century (pre-topology) mathematics and continues to have great relevance in today’s mathematics. In order to describe the problem we ﬁrst need to deﬁne what a ‘homeomorphism’ is. Recall that a function f : X → Y is a bijection if and only if it is both one-to-one and onto. Associated with each bijection f : X → Y there is an inverse function f−1 : Y → X deﬁned for each y ∈ Y by taking f−1(y) to equal x where f(x) = y. (Note that such an element x ∈ X exists since f is onto and it is unique since f is one-to-one.) The inverse function f−1

is characterized by the facts that the composition f−1 ◦f is the identity function 1X on X and that the composition f ◦f−1 is the identity function 1Y on Y . In other words, for each x ∈ X, f−1(f(x)) = x, and for each y ∈ Y , f(f−1(y)) = y. A homeomorphism between topological spaces X and Y is a bijection f : X → Y such that both f and f−1 are continuous. Another way to deﬁne what a homeomorphism between X and Y is is to say that it is a bijection f : X → Y for which a subset U is open in X if and only if f(U) is open in Y . Thus a homeomorphism from X to Y induces a bijection between the open sets in X and the open sets in Y , and this captures the idea that X and Y are essentially the same topological space with elements labelled diﬀerently. If there is a homeomorphism between X and Y then we say that X and Y are homeomorphic spaces. Being homeomorphic is easily verifed to be an equivalence relation on the collection of all topological spaces. This means that: (1) each topological space X is homeomorphic to itself (the identity function 1X is a homeomorphism); (2) if X is homeomorphic to Y then Y is homeomorphic to X (if f : X → Y is a homeomorphism then f−1 : Y → X is a homeomorphism; and (3) if X is homeomorphic to Y and Y is homeomorphic to Z then X is homeomorphic to Z (if f : X → Y and g : Y → Z are homeomorphisms then so is g◦f : X → Z). The Homeomorphism Problem can now be phrased as: Given two topological spaces determine whether or not they are homeomorphic. A propertyP of a topological space X which is expressed entirely in terms of the topology of X is called a topological property. If X and Y are homeomorphic spaces and P is a topological property then X satisﬁes P if and only if Y does. So one way to show that two topological spaces are not homeomorphic is to ﬁnd a topological property satisﬁed by one of the spaces but not the other. For example, the intervals [−1,1] and (−1,1) (with the Euclidean topology) cannot be homeomorphic because [−1,1] is compact but (−1,1) is not compact and being compact is a topological property. We have encountered a number of topological properties in our discssions. For example each of the following is a topological property: compactness, connectedness, Hausdorﬀness, metrizability.

Problem List:

Problem 1. For each positive integer n (that is, n ∈Z+) let Un = (−n,n) and let Vn = [n,n + 1]. (So, for each n, Un is an open interval in R, and Vn is a closed interval in R.) Determine each of the following:

1. U5 ∪V6 2. U5 ∩V6 3. U1 ∪U3 4. U1 ∩U3 5. V1 ∪V2 ∪V3 ∪V4 6. V1 ∩V2 ∩V3 ∩V4 7. S{Vn | n ∈Z+} 8. Tn∈Z+ Vn 9. S{Un | n ∈Z+} 10. T{Un | n ∈Z+} Problem 2. If A, B and C are the sets A = {1,2}, B = {1,3,5} and C = {4} then list all of the elements of the Cartesian product A×B×C.

Problem 3. Let A and B be sets. If A∪B is a subset of A∩B what does that imply about the relationship between the sets A and B? Clearly state your result as a Theorem, and then prove it using the deﬁnitions of intersection and union. (Suggestion: Use Venn diagrams as an aid to make your conjecture but leave them out of your ﬁnal proof.)

Problem 4. Find the domain, codomain and range of each of the following functions: a) f : R→R given by f(x) = 2 + sin(x). b) g : [0,4] →R given by g(x) = 3−2x2. c) h : R→ [−4,+∞) given by h(x) =   −3 if x < −1 −x2 if −1 ≤ x ≤ 1 x + 3 if x > 1

Problem 5. Determine whether each of the three functions in the previous problem are one-to-one. If a particular function is not one-to-one then ﬁnd two distinct elements of the domain that have the same image.

Problem 6. Determine whether each of the following subsets of R2 is an open set, or a closed set. Give some justiﬁcation for your two assertions (ie- whether it is open or not open, and whether it is closed or not closed) for each part.

a) U = R2. b) U = {(x,y) ∈R2 | y ≤ 5}. c) U = {(x,y) ∈R2 | 2 < y ≤ 5}. d) U = {(x,y) ∈R2 | x2 + y2 < 10}.

(Warning: It is not true that any set which is not open must be closed, so you have to independently check BOTH assertions in these problems.)

Problem 7. Consider the set of real numbers R. Let T be the collection of subsets of R which consists of ∅, R and every ﬁnite open interval (a,b) where a < b. In other terms, T = {∅,R}∪{(a,b) | a,b ∈R and a < b} a) Prove that T is not a topology on R. b) The collection T does satisfy three of the four axioms T1—T4. Determine which three axioms these are, and then explain why each of the three is true.

Problem 8. Consider the function f : R2 → R given by f(x1,x2) = x2 1 + x2 2. a) In the x1x2-plane draw the set f−1(B1) where B1 = {2,3}⊂R. b) In the x1x2-plane draw the set f−1(B2) where B2 ⊂R is the closed interval [2,3]. c) In the x1x2-plane draw the set f−1(B3) where B3 ⊂R is the closed interval [−3,−2]. d) Is there any real number t such that f−1({t}) is a singleton set? Explain. (Remember: A singleton set is a set with exactly one element in it. For example, {t} is a singleton set.) e) Is there any subset B ⊆R for which f−1(B) is the open disk B(x,) where x = (1,2) ∈ R2 and = 1? Explain.

Problem 9. Let f : X → Y be an onto function and let B be a subset of Y . Use the deﬁnition of image and inverse image to prove that f(f−1(B)) = B. Did you need to use the hypothesis that f is onto?

Problem 10. Consider the collection of subsets T` of the real line R given by T` = {U ⊆R| for each x ∈ U there is > 0 so that [x,x + ) ⊆ U}. a) Show that T` forms a topology on R. (Suggestion: Use the proof of Theorem 3.3 in the class notes as a model.) b) Show that every open set in the Euclidean topology Teuclid on R is also an open set in the T` topology. In other words, Teuclid ⊆T` . c) Find (and verify) an example of a set U that is open in the T` topology but not open in the Teuclid topology. The topology T` described in this exercise is called the lower limit topology on R. Parts (b) and (c) show that Teuclid (T`, so that the lower limit topology on R is strictly ﬁner than the Euclidean topology on R.

Problem 11. Let X be a set and let x0 be an element of X. Deﬁne T to be the collection of subsets of X consisting of X itself and all subsets of X which do not contain x0. Thus T = {X}∪{U ⊆ X | x0 / ∈ U} . a) Show that T forms a topology on X. b) Describe all of the closed sets in this topology which contain x0. Explain.

c) Show that if X has at least two elements then there is a subset of X which is closed in the T topology but not open. The topology on X is called the excluded point topology.

Problem 12. There’s a topology T deﬁned on the real line R by T = {∅,R}∪{(−a,a) | a ∈R and a > 1} . (You can check that this is indeed a topology but that’s not being asked as part of this problem.)

a) Describe the collection of closed sets in this topology on R. b) Determine the closure of each of the following subsets of R with respect to this topology: A1 = (2,3], A2 = (0,2), and A3 = {√2}. c) Show that the function f : (R,Teuclid) → (R,T) given by f(x) = x2 for x ∈ R is continuous. d) Show that the function g : (R,T) → (R,Teuclid) given by g(x) = x2 for x ∈ R is not continuous. (Hint: You just need to ﬁnd one Euclidean open set U for which g−1(U) is not in T.)

Problem 13. Let (X,T1) and (Y,T2) be topological spaces. (a) Show that if T1 is the discrete topology then every function f : X → Y is continuous. (b) Show that if T2 is the trivial topology then every function f : X → Y is continuous. c) Show that any constant function from X to Y is continuous.

Problem 14. Show that the discrete topology Tdiscrete on a set X is compact if and only if X is a ﬁnite set.

Problem 15. Consider the lower-limit topology T` on the real line R. (a) Show that the interval A = [1,3] (with the subspace topology from T`) is not compact by showing that {3}∪{[1,3 − 1/n) | n ∈ Z+} is an open cover of A with no ﬁnite subcover. (b) Show that the set B = {0}∪{1/n | n ∈ Z+} (with the subspace topology from T`) is compact.

Problem 16. Let X be a set. State and prove conditions on X under which each of the following topological spaces is connected: (a) the discrete topology on X. (b) the trivial topology on X. (c) the coﬁnite topology on X. (d) the lower limit topology on X = R.

点集拓扑学基础（II）

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