Homework Set #2 Math 440 Topology Topology by J. Munkres

Transcription

1 Homework Set #2 Math 440 Topology Topology by J. Munkres Clayton J. Lungstrum October 26, 2012

2 Exercise 1. Prove that a topological space X is Hausdorff if and only if the diagonal = {(x, x) : x X} is closed in X X with the product topology. Let X be a Hausdorff topological space. Consider ; we know is closed if and only if its complement is open. To that end, let (x, y) (X X) \. Since (x, y) is not on the diagonal, the points are not the same, i.e., x y. Thus, in X, there exist open sets U and V such that x U and y V such that U V = (since X is Hausdorff). Since U and V don t share any elements in common, (U V ) =, but U V is open in X X as it is the product of open sets. Since the point (x, y) is arbitrary, this implies C is open, hence is closed. Conversely, let be closed. Then, for any (x, y) C, there exists a basic open set U V containing (x, y) such that (U V ) =. Then U and V don t share any elements in common, hence U V = where x U and y V. Thus, X is Hausdorff. 1

3 Exercise 2. Write the 1-point compactification of the natural numbers with the discrete topology as a subspace of [0, 1]. The 1-point compactification of the natural numbers with the discrete topology as a subspace of [0, 1] is { } 1 {0}. n n=1 It is clear this is the space we want, for consider the function h : N [0, 1] (where N is the 1-point compactification of N) defined by h(n) = { 1 n : n N 0 : n =. h is clearly injective, it is surjective onto its image, and it and its inverse are clearly continuous as their domains are discrete topological spaces (where every function is continuous). Thus, h is a homeomorphism. 2

4 Exercise 3. Let X be a connected, locally connected, locally compact Hausdorff space. Prove that for every x, y X, there is a compact connected subset containing them. Let U be a cover of X, then a chain in U is a finite indexed set U 1, U 2,..., U N U such that, for all i = 1,..., N 1, we have that U i U i+1. It is called a chain from x to y in U when we have x U 1 and y U N. Now, a space X is connected if and only if for every open cover U of X, we have a chain between any pair of points of X. Fix x 0 X and define O to be the set of all y X such that there is a chain from x 0 to y. O is nonempty, as any x X is covered by some U U, then U 1 is a chain from x to x, so x O. Now I will show O is open and closed, thus it must be the whole space (otherwise we could separate X by it and its complement). Open: Let y O and let x U 1,..., U N be a chain (from U) for it. Then for every z in U N, that same chain will be a chain from x to z, so that z O as well. Thus, U N O, and every point of O is an interior point. Note that we do not even need the cover to be open, just that the interiors of the sets cover X. Closed: Suppose that y / O, and let U be an element from U that covers y. Suppose that some z U is in O, and again, let x U 1,..., U N be a chain from x to z so we have z U N. But then the chain x U 1,..., U N, U is a chain from U as well, because all intersections are non-empty in the beginning by assumption, and U N U N+1 is non-empty, as both contain z, and this would be a chain from U from x to y, so that y O, a contradiction to what we assumed. Thus, U misses O entirely, so O is closed. But now the connectedness of X forces O = X (there is only one nonempty closed and open set), and we have what we wanted in the chain condition, as x was arbitrary. Having this, we can quickly see that between any two points, we can find such a chain, and using the properties of locally compact and locally connected, we can find compact and connected neighborhoods inside such a chain, and then we take the union of these neighborhoods. Being a finite union, the set is still compact, and since each subsequent set has nonempty intersection with the previous one, the union is connected, thus we re done. 3

5 Exercise 4. Prove that a countable closed subset of a locally compact Hausdorff space has an isolated point. Let A be a closed subset of a locally compact Hausdorff space with no isolated points. Let x A. Then, since X is locally compact, x has a compact neighborhood, call this set V. As a subset of a Hausdorff space, clearly V is closed, so V A is closed. We claim V A is compact. This is immediate as we take any open cover of V A, say {B i } i I. Then (V A) C i I B i is an open covering of V, a compact set, hence there is a finite subcover. Therefore, we have a finite subcover of the open cover of {B i } i I, hence V A is compact. As a subspace of X, V A is also Hausdorff. Thus, V A is a compact Hausdorff set with no isolated points, hence V A must be uncountable. Since V A A, A is uncountable. Taking the contrapositive of this, we see that if we have a countable closed subset of a locally compact Hausdorff space, it must have an isolated point. 4

6 Exercise 5. Let X be a compact Hausdorff space, C U, where C is a component of X and U is open. Prove that there is a set V which is both open and closed such that C V U. First, suppose that a component C can be characterized by the intersection of all closed and open sets containing C. Thus, suppose C = i I F i where each F i is a closed and open set containing C. Then U C is covered by {Fi C } i I. Since U is open, its complement is closed, and a closed subset of a compact space is compact, hence U C is compact. Thus, there must be a finite collection, relabeling if necessary, such that Thus, U C C N i=1 F C i. N F i, i=1 which is a finite intersection of closed and open sets, hence is closed and open. To show that a component can be characterized by the intersection of all closed and open sets containing it in a compact, Hausdorff topological space is clear, though, as the following argument shows. Let x C, a component, and let E be the intersection of all closed and open sets containing C, that is, the F i above. Clearly E is closed as it is the intersection of closed sets. Suppose E = A B with A and B closed and disjoint in E (thus they are closed in X). Without loss of generality, assume x A. Recall that compact and Hausdorff implies normal, so there exists disjoint open subsets U and V such that A U and B V ; hence, we have E U V. Now, the union of the Fi C s form an open cover (as each Fi C is also closed and open) of (U V ) C. Since (U V ) C is compact, there is a finite subcover. Thus, we have E N i=1 F i U V by construction. For simplicity, let F = N i=1. Now we claim that U F is closed and open. It is clearly open as a finite intersection of open sets. Also, any point in its closure would be in F, thus in U V, but it cannot be in V as U and V are disjoint. Hence, U F is closed. Therefore, x U F since x A U. We also have x E F, thus U F is just one of the closed and open sets containing x, therefore E U F. But the B must be empty as U and V are disjoint, and in particular, U F and V are disjoint. Thus, there is no separation of E, hence E is connected. Conversely, a connected set must be contained in some component, as a component is a maximally connected subset. 5