Topology Homework Assignment 1 Solutions

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1 Topology Homework Assignment 1 Solutions 1. Prove that R n with the usual topology satisfies the axioms for a topological space. Let U denote the usual topology on R n. 1(a) R n U because if x R n, then B 1 (x) R n. 1(b) U because if x, then B 1 (x) R n. 2. Suppose that A is an index set (possibly infinite), and Then if V = {V α : α A} U. x V = α A V α, there is some α 0 A, such that x V α0. Since V α0 is open, there is a real number r > 0, such that B r (x) V α0. But then and hence V is open. 3. Suppose that and B r (x) α A V α = V, {V 1, V 2,... V k } U, x k i=1v i. Since each V i is open, for each i = 1,... k, there is a real number r i such that B ri (x) V i. 1

2 Furthermore, since B ri (x) B rj (x) when r i < r j, if r = min{r i : i = 1,..., k}, for i = 1,... k. Hence and k i=1v i, is open. B r (x) B ri (x) V i, B r (x) k i=1v i, 2. Prove that the ɛ δ definition of continuity is equivalent to the open set definition of continuity (i.e., f : R n R m is continuous iff f 1 (V ) is open for each open subset V R m ). We first show that f is continuous at x 0 R n in the ɛ δ definition iff for each open set V containing f(x 0 ), there is an open set U containing x 0 such that f(u) V. So suppose first that f is continuous at x 0 in the ɛ δ definition, and let V be an open set containing f(x 0 ). Since R m has the usual topology, there exists a real number ɛ > 0 such that B ɛ [f(x 0 )] V. Since we re assuming f is continuous at x 0 in the ɛ δ definition, there is some real number δ > 0 such that But and x x 0 < δ f(x) f(x 0 ) < ɛ. {x R n : x x 0 < δ} = B δ (x 0 ), {y R n : y f(x 0 ) < ɛ} = B ɛ [f(x 0 )]. So if x B δ (x 0 ), then f(x) B ɛ [f(x 0 )]. Since B ɛ [f(x 0 )] V, we have that x B δ (x 0 ) f(x) V. That is, f[b δ (x 0 )] V. Since B δ (x 0 ) is open, we can let U = B δ (x 0 ). 2

3 Conversely, suppose that for each open set V containing f(x 0 ), there is an open set U containing x 0 such that f(u) V. Let ɛ > 0. Then B ɛ [f(x 0 )] is an open set containing f(x 0 ). Hence there is an open set U containing x 0 such that f(u) V. Since R n has the usual topology, there is a δ > 0, such that B δ (x 0 ) U. But then f[b δ (x 0 )] f(u) B ɛ [f(x 0 )]. Since and we have that B δ (x 0 ) = {x R n : x x 0 < δ}, B ɛ [f(x 0 )] = {y R n : y f(x 0 ) < ɛ}, x x 0 < δ f(x) f(x 0 ) < ɛ. Now suppose that f is continuous at each point x of R n in the ɛ δ definition. We want to show that if V is an open subset of R m, then f 1 (V ) is an open subset of R n. Let V be an open subset of R m. Then for each f(x) V, by our preceding result, there is a δ x > 0, such that But then f[b δx (x)] V. f 1 (V ) = x f 1 (V )B δx (x) because every x f 1 (V ) belongs to the right-hand side, and every B δx is contained in the left-hand side. Since f 1 (V ) is a union of open sets, f 1 (V ) is open. Conversely suppose that for each open subset V R m, f 1 (V ) is open in R n, and let x 0 R n and ɛ > 0. Then B ɛ [f(x 0 )] is an open subset of R m. So f 1 {B ɛ [f(x 0 )]} is an open subset of R n. Thus, since x f 1 {B ɛ [f(x 0 )]}, there must exist an open ball B δ (x 0 ) f 1 {B ɛ [f(x 0 )]}. Thus if x x 0 < δ, (i.e., if x B δ (x 0 ), then f(x) f(x 0 ) < ɛ since f(x) B ɛ [f(x 0 ). 3

4 3. Suppose (X, U) is a topological space and Y X. Define a family of subsets V of Y by V V iff there exists a set U U such that V = U Y. Prove that V is a topology on Y. V is called the relative topology on Y, and unless we state otherwise, you should assume that a subset of a topological space has the relative topology. 1(a) Since Y X, Y = Y X, and since X is an element of U, Y is open in the relative topology. 1(b) Since = Y, and since is an element of U, is open in the relative topology. 2. Let A be an index set and let V = {V α : α A} be a family of elements of the relative topology. Then for each element V α V, there is an open set U α U, such that Thus V α = U α Y. α A V α = α A (U α Y ) = ( α A U α ) Y. But U α is open in X. So V α = ( U α ) Y is open in the relative topology. 3. Let {V 1, V 2,... V k } be a finite subset of the relative topology. Then for i = 1,... k, there is an element U i U, such that V i = U i Y. Thus k i=1v i = k i=1(u i Y ) = ( k i=1u i ) Y. Since U i is open in X, V i = ( U i ) Y is open in the relative topology. 4. Using subsets of euclidean spaces, find topological spaces X and Y and a function f : X Y, such that (a) f is continuous but not open. (b) f is open but not continuous. 4

5 Can you find a function f satisfying 4a or 4b that is both one-to-one and onto? Let f : [0, 2π) S 1 by f(t) = (cos(t), sin(t)). Then f is both one-toone and onto, for if (x, y) S 1, we can rewrite (x, y) in exponential form e it, where t is a unique value in [0, 2π), and f(t) = (x, y). Thus, f has a unique inverse g : S 1 [0, 2π). Since the component functions, cos(t) and sin(t), are continuous, f is continuous. (Technically, we need to know about product spaces in order to prove this result.) However, f is not open. For open subsets of S 1 (in the relative topology on S 1 as a subset of R 2 ) are unions of curved open intervals. To see this observe that any open subset of R 2 is a union of open balls. Since an open ball intersects S 1 in a curved open interval (or all of S 1 ), any open subset of S 1 is a union of open intervals. But open sets in [0, 2π) are unions of intersections of open intervals with [0, 2π). Thus, [0, π) is open in [0, 2π), and f([0, π)) = {(x, y) S 1 : y > 0} {(1, 0)}, which isn t a curved open interval, and hence isn t open. The function g, on the other hand is open, but it s not continuous. Since f is continuous, if U is open in S 1, f 1 (U) = g(u) is open in [0, 2π). However, g 1 ([0, π)) = f([0, π)) isn t open. So g is not continuous. 5. Suppose X and Y are topological spaces, and f : X Y is continuous. If A is a compact subset of X, show that f(a) is a compact subset of Y. (Use the open cover definition of compactness.) So compactness is a topological property: if Y is homeomorphic to X (denoted Y X), and X is compact, then so is Y. We proved this in class. 6. Suppose X is a topological space. Show that a subset C of X is closed iff C contains all of the points that are arbitrarily close to C. (Recall that the point x is arbitrarily close to C if every open neighborhood of x contains a point of C.) We proved this in class. 5

6 7. Suppose that X and Y are topological spaces and f : X Y. Show that f is continuous iff for each closed set C Y, f 1 (C) is closed a closed subset of X. Suppose first that f is continuous i.e., for each open set V in Y, f 1 (V ) is open in X. Let C be a closed subset of Y. Then Y C is open in Y. Furthermore, f 1 (Y C) = X f 1 (C), by Theorem 2.6 on page 21 of the set theory handout. But, by assumption f is continuous. So f 1 (Y C) is open in X. Thus, is closed in X. X [X f 1 (C)] = f 1 (C) Conversely, suppose that for each closed set C in Y, f 1 (C) is closed in in X. Also suppose that V is an open subset of Y. Then, by assumption, is closed, and hence is open. That is, f is continuous. f 1 (Y V ) = X f 1 (V ) X [X f 1 (V )] = f 1 (V ) 6