MATH 411 HOMEWORK 3 SOLUTIONS ADAM LEVINE 2.17.6. Let A, B, and Aα denote subsets of a space X. (a) Show that if A ⊂ B, then A¯ ⊂ B¯. By definition, A¯ is the intersection of all closed sets containing A. Since B¯ is a closed set that contains B and hence A, it must therefore contain A¯. (b) Show that A [ B = A¯ [ B¯. The set A [ B is a closed set that contains A [ B, so it contains both A and B, and therefore it contains both A¯ and B¯. Therefore, A [ B ⊃ A¯ [ B¯. Conversely, A¯ [ B¯ is a closed set (since it is the union of two closed sets) that contains A [ B, so it contains A [ B. (Note that we had to use the fact that finite intersections of closed sets are closed!) S S ¯ (c) Show that Aα ⊃ Aα; give an example where equality fails. S Just as in (b), the set Aα is a closed set that contains each Aα, so it contains each ¯ S ¯ Aα, and thus it contains Aα. (But we can't do the converse because an arbitrary union of closures isn't necessarily closed!) As a counterexample, let A be the closed set [ 1 ; 1] ⊂ . Then S A¯ = S A = i i R i2N i i2N i (0; 1], while S A = [0; 1]. i2N i 2.17.11. Show that the product of two Hausdorff spaces is Hausdorff. Let X and Y be Hausdorff spaces. For any two distinct points (x; y) and (x0; y0) in X × Y , we may assume that they differ in at least one coordinate. If x 6= x0, choose disjoint open sets U; U 0 ⊂ X with x 2 U and x0 2 U 0. Then U × Y and U 0 × Y are disjoint open sets in X × Y containing (x; y) and (x0; y0) respectively. Likewise, if y 6= y0, choose disjoint open sets V; V 0 ⊂ Y with y 2 V and y0 2 V 0; then X × V and X × V 0 are the needed open sets. 2.17.12. Show that a subspace of a Hausdorff space is Hausdorff. Let X be Hausdorff, and let A ⊂ X be a subspace. For any x; x0 2 A, we may find disjoint open sets U; U 0 with x 2 U and x0 2 U 0. Then U \ A and U 0 \ A are disjoint open sets in A containing x and x0 respectively. 2.17.13. Show that X is Hausdorff if and only if the diagonal ∆ = fx × x j x 2 Xg is closed in X × X. We will show an equivalent condition: that X × X − ∆ is open. If X is Hausdorff, then for any x × x0 2 X × X − ∆, we have x 6= x0. Choose disjoint open sets U; U 0 with x 2 U and x0 2 U 0. Then U × U 0 is open in X × X, and it does not contain any points of ∆. Therefore, X × X − ∆ is open. 1 2 ADAM LEVINE Conversely, if X × X r ∆ is open, then for any point x × x0 2 X × X − ∆, there is a basic open set U × U 0 containing x × x0 and disjoint from ∆. Therefore, U and U 0 are disjoint open sets containing x and x0 respectively. 2.17.17. Consider the lower limit topology and the topology given by the basisp C = f[a; b)pj a; b 2 Q; a < bg. Determine the closures of the intervals A = (0; 2) and B = ( 2; 3) in these two topologies. In the lower limit topology, we claim that the closure of any interval (a; b) is [a; b). If x < a or x ≥ b, we can easily find an open set [x; x + ) disjoint from (a; b). On the other hand, a is a limit point: any open set containingp a mustp contain an interval ¯ ¯ [a; a + ), which intersects (a; b). Thus, A = [0; 2)p and B = [ 2; 3). In the topology given by C, the onlyp difference is 2 is a limitp point of A. Indeed,p ¯ if [a; b) isp an element of C containing 2, we must have a < 2. Thus, A = [0; 2] and B¯ = [ 2; 3). 2.17.18. Determine the closures of the following subsets of the ordered square: 1 (a) A = f n × 0 j n 2 Z+g We claim that A¯ = A [ f0 × 1g. The point 0 × 1 is a limit point because any open set containing 0 × 1 must contain (0; ) × [0; 1] for some > 0, and therefore meets A. Any other point x × y 2 I × I − A can be seen to have a neighborhood (specifically, an interval in the dictionary ordering) that is disjoint from A. Namely, if x = 0 and y < 1, then we can use [0 × 0; 0 × 1). If 0 < x < 1, then choose n 2 Z+ 1 1 1 1 with n+1 ≤ x < n , and take the interval ( n+1 × 0; n × 0). If x = 1 and y > 0, then take (1 × 0; 1 × 1]. 1 1 (b) B = f(1 − n ) × 2 j n 2 Z+g We claim that A¯ = B [ f1 × 0g. The argument is similar to the previous example. (c) C = fx × 0 j 0 < x < 1g Any point x × y with 0 < y < 1 has a neighborhood distinct from C, as do the points 0 × 0 and 1 × 1. On the other hand, any neighborhood of a point x × 0 (with 0 < x ≤ 1) or x × 1 (with 0 ≤ x < 1) must include an entire vertical strip (as in (a)), and thus intersect C. Thus, C¯ = (0; 1] × f0g [ [0; 1) × f1g 1 (d) D = fx × 2 j 0 < x < 1g 1 1 Any point x × y with 0 < y < 2 or 2 < y < 1, or with x = 0 and 0 ≤ y < 1, or with x = 1 and 0 < y ≤ 1, has a neighborhood disjoint from D. On the other hand, any neighborhood of x × 0 (with 0 < x ≤ 1) or of x × 1 ( with 0 ≤ x < 1) must include entire vertical segments, and thus intersects D. Thus, D¯ = D [ fx × 0 j 0 < x ≤ 1g [ fx × 1 j 0 ≤ x < 1g: 1 (e) E = f 2 × y j 0 < y < 1g 1 1 Any point x × y with 0 ≤ x < 2 or 2 < x <≤ 1 has a neighborhood disjoint from 1 1 ¯ 1 E. The points 2 × 0 and 2 × 1 are limit points. Therefore, E = f 2 × y j 0 ≤ y ≤ 1g. MATH 411 HOMEWORK 3 SOLUTIONS 3 2.18.2. Suppose that f : X ! Y is continuous. If x is a limit point of A ⊂ X, is it necessarily true that f(x) is a limit point of f(A)? Suppose f is a constant function: f(x) = y0 for all x 2 X. Then for any limit point x of A, we have f(x) = y0. Since f(A) has only one point, that point cannot be a limit point. 2.18.7(a). Suppose that f : R ! R is \continuous from the right," i.e., limx!a+ f(x) = f(a) for each a 2 R. Show that f is continuous when considered as a function from R` to R. As a review from calculus, we say that limx!a+ f(x) = b if for all > 0, there is a δ > 0 such that for all x with a < x < x + δ, we have jf(x) − bj < . Now, assume that limx!a+ f(x) = f(a) for each a 2 R. Let U ⊂ R be an open set. For any a 2 f −1(U), choose some > 0 such that (f(a) − , f(a) + ) ⊂ U. By assumption, there is a δ > 0 such that f([a; a + δ)) ⊂ (f(a) − , f(a) + ) ⊂ U, and −1 −1 therefore [a; a + δ) ⊂ f (U). Thus, f (U) is open in R`. Note: We will see in Chapter 3 that the only continuous maps from R to R` are constant maps. 2.18.8. Let Y be an ordered set in the order topology. Let f; g : X ! Y be continuous. (a) Show that the set fx j f(x) ≤ g(x)g is closed in X. In class we proved the following: Given an ordered set Y , for any two distinct elements a; b 2 Y with a < b, there are open sets U; V such that a 2 U, b 2 V , and for all c 2 U and d 2 V , c < d. Now, let us show that the set A = fx j f(x) > g(x)g (which is the complement of the one above) is open in X. Suppose x 2 A. We may find open sets U and V as above with g(x) 2 U and f(x) 2 V . Then g−1(U) and f −1(V ) are open in X, and g−1(U) \ f −1(V ) contains x and is contained in A. Thus, A is open as required. (b) Let h: X ! Y be the function h(x) = minff(x); g(x)g. Show that h is contin- uous. By the preceding discussion, the sets C1 = fx j f(x) ≤ g(x)g and C2 = fx j f(x) ≥ g(x)g are closed, and their union equals X. The function h equals f on C1 and g on C2, and these agree on the overlap. Hence, by the pasting lemma, h is continuous..