Metric Space Topology (Spring 2016) Selected Homework Solutions HW1 Q1.2. Suppose that d is a metric on a set X. Prove that the inequality d(x, y) d(z, w) d(x, z) + d(y, w) holds for all w, x, y, z X. Proof. Let w, x, y, z X be arbitrary. By the triangle inequality applied twice in succession, d(x, y) d(x, z) + d(z, y) d(x, z) + d(z, w) + d(w, y). Subtracting d(z, w) from both sides yields d(x, y) d(z, w) d(x, z) + d(y, w). Now, the equality above holds for all points w, x, y, z. So swapping x with z, and y with w, and using the symmetry of the metric d, we obtain d(z, w) d(x, y) d(x, z) + d(y, w). Flipping signs on this inequality yields d(x, z) d(y, w) d(x, y) d(z, w). So we have shown d(x, z) d(y, w) d(x, y) d(z, w) d(x, z) + d(y, w), which is equivalent to the inequality we hoped to show. HW1 Q1.8. Let F(S) be the set of all finite subsets of a set S. Define a function d on F(S) F(S) by the rule d(a, B) = #( (A, B)) for all A, B F(S), where denotes symmetric difference and # denotes cardinality. Is d a metric? We claim d is a metric, and prove our assertion below. Proof. Clearly d(a, B) 0 for all A, B F(S), with d(a, A) = #( ) = 0. If d(a, B) = #( (A, B)) = 0, then (A, B) =, from which we conclude A = B. Since (A, B) = (B, A), we have d(a, B) = d(b, A). So to finish the proof, we need only check the triangle inequality for d. So let A, B, C F(S) be arbitrary. We know (A, B) = (A\B) (B\A); let us first consider A\B. We note that A\B = [(A\B) C] [(A\B)\C] (C\B) (A\C). Similarly, B\A (C\A) (B\C). So we have 1
2 (A, B) = (A\B) (B\A) (C\B) (A\C) (C\A) (B\C) = (A\C) (C\A) (C\B) (B\C) = (A, C) (C, B). It follows that d(a, B) = #( (A, B)) #( (A, C) (C, B)) #( (A, C)) + #( (C, B)) = d(a, C) + d(c, B). HW2 #1. Suppose (X, d) is a metric space and e(x, y) = Prove that e is a metric on X. d(x, y) 1 + d(x, y) for every x, y x. Proof. Since d is a metric to begin with, the positivity and symmetry conditions for e obviously hold. So we need to check the triangle inequality for e. To that end, suppose x, y, z 0 are three nonnegative real numbers, and further suppose that x y + z. Since adding nonnegative terms can only make the right side bigger, it follows that x y + z + 2yz + xyz. Now, adding (xy + xz + xyz) to both sides yields the inequality x + xy + xz + xyz y + xy + yz + xyz + z + xz + yz + xyz, in other words, x(1 + y)(1 + z) y(1 + x)(1 + z) + z(1 + x)(1 + y). Dividing through above by the positive number (1 + x)(1 + y)(1 + z), we obtain x 1 + x y 1 + y + z 1 + z. To finish the proof, we simply observe that if a, b, c are any three points in X, then we can set x = d(a, b), y = d(a, c), z = d(c, b). Since d is a metric, x y + z, and now the above line implies that the triangle inequality holds for e. HW2 #2. For (X, e) in the previous problem, prove that X has finite diameter.
d(x, y) Proof. Let x, y X be arbitrary. Then d(x, y) < 1+d(x, y), and hence e(x, y) = 1 + d(x, y) < 1. So 1 is a lower bound for the set {d(x, y) : x, y x}. Therefore diam(x) = inf{d(x, y) : x, y x} 1. HW2 #3. For the metric space (Ñ, d) in Example 1.1.12, show that Ñ has exactly one accumulation point. Proof. We will show is the sole accumulation point. To see this, compute the distance from to Ñ\{ } = N: 3 dist(, N) = inf {d(, n) : n N} { } 1 = inf n : n N. Now the infimum in the last line above is at least 0, since 0 < 1 for every n N. On n the other hand, notice that for every ɛ > 0, ɛ fails to be a lower bound: because there are always integers n N for which n > 1, whence 1 < ɛ. So the infimum above is exactly 0. ɛ n This shows is an accumulation point. On the other hand, if n N, we claim n is an isolated point of Ñ (and hence is the only accumulation point). To see this, we must show dist(n, Ñ\{n}) > 0. So, for our fixed n, let us consider the following quantity: B = 1 n 1 n+1 = 1 n(n+1). Now consider an arbitrary m Ñ\{ }. Either m =, or m N and m < n, or m N and m > n. If m =, then d(n, m) = d(n, ) = 1 > 1 1 = B. n n n+1 If m N and m < n, then m n 1. Therefore 1 1, whence d(n, m) = 1 1 m n 1 m n 1 1 = 1 > 1 = B. n 1 n (n 1)n n(n+1) 1 If m N and m > n, then n + 1 m. Therefore 1, whence d(n, m) = 1 1 n+1 m n m 1 1 = B. n n+1 By checking all cases, we have shown that B is a lower bound for the set {d(n, m) : m Ñ\{n}}. It follows that dist(n, Ñ\{n}) B > 0, and n is an isolated point as claimed. This completes the proof. HW2 #4. For the metric space (R 2, d) in Example 1.1.15, show that there exists exactly one isolated point of R 2. Proof. First, we claim (0, 0) is an isolated point. To see this, note that by definition of the metric, d((0, 0), x) 1 for every x R 2, x (0, 0). It follows that dist((0, 0), R 2 \{(0, 0)}) 1 > 0, i.e. (0, 0) is isolated. On the other hand, we claim every other point x R 2, x (0, 0) is an accumulation point. To see this, let e denote the usual metric on R 2, and observe that by the definition of d, we have (R 2 \{(0, 0)}, d) = (R 2 \{(0, 0)}, e). That is to say, d and e agree everywhere
4 except at the origin. Now let ɛ > 0 be arbitrary. The set of points y for which e(x, y) < ɛ is clearly infinite; so find such a y satisfying y (0, 0). Then d(x, y) = e(x, y) < ɛ. It follows that dist(x, R 2 \{x}) = inf{d(x, y) : y R 2, y (0, 0)} cannot be equal to any positive ɛ (since no such ɛ is a lower bound for {d(x, y) : y R 2, y (0, 0)}). So dist(x, R 2 \{x}) = 0, i.e. x is an accumulation point. HW3 #1. Let (X, d) be a metric space. Let x X and let r be a positive real number. Define the set Show that B is open and diam(b) 2r. B = {y X : d(y, x) < r}. Proof. Let b B; we will show b / B. Since b B, d(b, x) < r. Set R = d(b, x), and set t = r R, so t > 0. We claim that for every z B c, d(b, z) t. Indeed, if this were not the case, we would have d(b, z) < t, and therefore by the triangle inequality we would have d(x, z) d(x, b)+d(b, z) < R + t = r, contradicting the fact that z B c. So d(b, z) t for all z B c. It follows that dist(b, B c ) = inf{d(b, z) : z B c } t > 0, and hence b / B. Since B was arbitrary, we have shown B B = and hence B is open. HW3 #2. Let (X, d) be a metric space. For any x X, show that x iso(x) if and only if {x} is open. Proof. For any x X, x iso(x) dist(x, X\{x}) > 0 dist(x, {x} c ) > 0 x / {x} {x} {x} = {x} is open. HW3 #3. Let X be any set and let d be the discrete metric on X (see Example 1.1.7). Show that every subset of X is open. Conclude that every subset of X is closed. Proof. First note that for each x X, x iso(x) since dist(x, X\{x}) = inf{d(x, y) : y X, y x} inf{1} = 1. Therefore {x} is open by the result in Problem 2 above. Now let S X be an arbitrary subset. Then S = {{x} : x S} is the union of its collection of singleton subsets. Hence, S is a union of open sets and therefore open. Since S was arbitrary, every subset of X is open.
Likewise, if S X is an arbitrary subset, then the argument above shows that S c is open. Therefore S is closed. Therefore every subset of X is closed. HW3 #4. Consider the metric space (Ñ, d) as defined in Example 1.1.12. Let A N (so / A). Show that A if and only if A is an infinite set. Proof. First suppose A is an infinite subset of N. It means that A is not bounded above. So if ɛ > 0 is arbitrary, we may find n A for which n > 1 and hence 1 < ɛ. Therefore ɛ n dist(, A) = inf{d(, n) : n A} = inf{ 1 : n A} = 0. So A A. n Conversely, suppose A is a finite subset of N. In this case, dist(, A) = inf{d(, n) : n A} = min{d(, n) : n A} > 0, since the infimum of a finite set is just the minimum, and all distances d(, n) for n A are strictly positive. So / A. Since / A by hypothesis, we conclude / A. HW3 #5. Again consider (Ñ, d) as in Example 1.1.12. Let A Ñ and suppose A. Show that A is open if and only if A is cofinite (that is, Ñ\A is a finite set). Proof. First suppose A is cofinite. Then A c is finite, and therefore closed. So A is open. Conversely, suppose A is not cofinite. Then A c is an infinite set, and / A c by hypothesis. Then by the result of Problem 4 above, A c. So A c A c. It follows that A c is not closed, and therefore A is not open. HW3 #6. Consider Q as a metric space with the usual metric. Let a and b be irrational numbers such that a < b. Define the set Is E open? Is E closed? Prove your answers. E = {q Q : a < q < b} We claim E is both open and closed, and prove it below. Proof. To show E is clopen, it suffices to prove that E = (whence E = E o = E). So let x Q; we will show x / E. There are three cases: either x < a, or x > b, or a < x < b. First suppose x < a. If q E is arbitrary, then a < q, and hence d(x, q) = q x > a x. It follows that So x / E. dist(x, E) = inf{d(x, q) : q E} a x > 0. Next suppose x > b. For every q E, we have q < b and therefore d(x, q) = x q > x b. So So again x / E. dist(x, E) = inf{d(x, q) : q E} x b > 0. Lastly, suppose a < x < b. Set R = min{x a, b x}. For every p E c, we have either p < a or p > b. If p < a, then d(x, p) = x p > x a R. If p > b, then d(x, p) = p x > b x R. So we have shown that 5
6 dist(x, E c ) = inf{d(x, p) : p E c } R > 0. So x / E. Since we checked all cases, E is empty and E is clopen. HW4 #1. Show that every subset of Ñ is open or closed, but not every subset is clopen. Proof. Let A Ñ. Note that for each n N, n is an isolated point of Ñ by HW2 #3. So each singleton {n} is open by HW3 #2, for n N. Thus, if A N, then A = n A{n} is a union of open singletons and is therefore an open set. On the other hand, if A N, then A. It then follows that A c N, whence A c is open by our previous remarks. So A is closed. Since A was arbitrary, we have shown every subset of Ñ is either open or closed. On the other hand, Ñ has non-clopen subsets. One example is the subset N. N is open in Ñ, but the closure of N contains by HW3 #4. So N = Ñ N, whence N is not closed. It follows from these arguments that the complementary set { } is a closed set which is not open. HW4 #2. Show that the open subsets of the subspace R {0} of R 2 with the usual metric are precisely those subsets of the form U {0} where U is open in R. Show also that none of those sets, except the empty set, is open in R 2. Let us start by proving the following lemma. Lemma. Suppose (X, d) and (Y, e) are metric spaces and f : X Y is an isometry of X onto Y. Then U X is open in X if and only if f(u) Y is open in Y. Proof. Let U X be an arbitrary subset. Since f is a bijection, for any point x U there corresponds some point y f(u), such that y = f(x). Now since f is an isometry, i.e. a distance-preserving bijection, we compute the following equality for all x U and y = f(x) f(u): dist(y, Y \f(u)) = inf{e(y, w) : w Y \f(u)} = inf{e(f(x), f(z)) : f(z) Y \f(u)} = inf{e(f(x), f(z)) : z X\U} = inf{d(x, z) : z X\U} = dist(x, X\U). Given the equality above, it is easy to finish proving the lemma: U is open in X x U dist(x, X\U) > 0 y f(u) dist(y, Y \f(u)) > 0 f(u) is open in Y.
Proof of HW4 #2. Note that R is isometric to R {0} via the isometry f defined by f(x) = (x, 0), for all x R. (The fact that f is an isometry is extremely easy to check.) So by the lemma we just proved, a subset W of R {0} is open if and only if U = f 1 (W ) is open in R, if and only if W = U {0} for some open U R. On the other hand, no such set is open in R 2, unless U =. If U is nonempty, then there exists (x, 0) U {0}. Every open ball about (x, 0) will intersect the complement of R {0}, so (x, 0) (U {0}). This shows U {0} is not open. HW4 #3. Suppose X is a metric space and S X. Show that S is dense in X if and only if S has nonempty intersection with every open ball of X. Proof. First suppose S is dense and let [x; r) be some open ball in X (for some x X, r R + ). Since S is dense, x S. It follows then that S [x; r). Conversely, suppose S has nonempty intersection with every open ball of X. Let x X be arbitrary. Then for every r R +, S [x; r) by hypothesis. Therefore x S. Since x was arbitrary, we have shown X S, which implies S is dense. HW4 #4. Find all dense subsets of Ñ. Prove you have found them all. We claim N and Ñ are the only two dense subsets of Ñ. Proof. It is obvious that Ñ is dense in Ñ. To see that N is dense, simply note that N is an infinite set and hence N by HW3 #4; so Ñ N. Now if A is any other subset of Ñ other than N or Ñ, then A must be missing some integer, i.e. there exists n N with n / A. But we have shown in previous problems that {n} is open; hence {n} is an open ball in Ñ that has empty intersection with A. By the previous problem, A is not dense. HW5 #1. Prove the Bounded Monotone Convergence Theorem: If (x n ) is a bounded monotone sequence in R (with the usual metric), then (x n ) converges to some point in R. Proof. Either (x n ) is nondecreasing or nonincreasing; first suppose nondecreasing. Set z = sup{x n : n N}. Note that since (x n ) is bounded above by some number M by hypothesis, we have z M < and hence z R. We claim that x n z. To see this, let r R + be arbitrary. By our definition of z, and our bonus homework characterizing supremums, we know that there exists m N for which x m > z r. But then since (x n ) is a nondecreasing sequence, we have z r < x m x n z for all n m. In other words, we see that tail m (x n ) (z r, z] [z; r). Since r was arbitrary, we have shown x n z. The nonincreasing case is very similar and we leave it to the reader. HW5 #2. Prove the Squeeze Theorem: Suppose (a n ), (b n ), and (c n ) are sequences in R (with the usual metric), and suppose there exists an M N such that for all n M, a n b n c n. If a n L and c n L for some L R, then b n L. 7
8 Proof. Let r R + be arbitrary. Since a n L, it means there exists m 1 N such that tail m1 (x n ) [L; r) = (L r, L + r). Likewise since c n L, there exists m 2 N for which tail m2 (x n ) [L; r) = (L r, L + r). Now set N = max(m, m 1, m 2 ). Then for every n N, the term b n satisfies a n b n c n by hypothesis, and also b n lies in both tails mentioned above. So we have L r < a n b n c n < L + r, so b n (L r, L + r). Therefore tail N (x n ) [z; r). Since r was arbitrary, b n L and the theorem is proved. HW6 #1. Let X and Y be sets, let A X and B Y, and let f : X Y be a function. Prove that f 1 (f(a)) A and f(f 1 (B)) B, but it is not necessarily the case that f 1 (f(a)) = A, nor that f(f 1 (B)) = B. Proof. Let a A. Then f(a) f(a), and therefore a f 1 (f(a)). Since a A was taken arbitrarily, we have shown A f 1 (f(a)). However, it is not necessarily the case that f 1 (f(a)) = A: for instance one could take the function f : R R defined by f(x) = x 2 for all x R, and observe that f 1 (f([0, 2])) = f 1 ([0, 4]) = [ 2, 2] [0, 2]. Next let y f(f 1 (B)). It means there exists an x f 1 (B) so that f(x) = y. Since x f 1 (B), f(x) B. Therefore y B. So we have shown f(f 1 (B)) B. But equality need not hold: for instance take f as in the previous paragraph, and observe that f(f 1 ([ 1, 0])) = f({0}) = {0} [ 1, 0]. HW6 #2. Let X, Y, and f be as in the previous problem. Prove that f 1 (f(a)) = A for all A X if and only if f is an injection. Proof. ( ) First assume f 1 (f(a)) = A for all A X. Let x, y X be arbitrary and assume f(x) = f(y). Then f({x}) = {f(x)} = {f(y)} = f({y}), and therefore by hypothesis {x} = f 1 (f({x})) = f 1 (f({y})) = {y}. So x = y. This shows f is injective. ( ) On the other hand assume f is injective, and let A X be arbitrary. By the previous problem, f 1 (f(a)) A, so we need only show that f 1 (f(a)) A. So let x f 1 (f(a)) be arbitrary. It means f(x) f(a). I.e. there exists an a A so that f(a) = f(x). But since f is injective, this implies a = x. Therefore x A in the first place. So f 1 (f(a)) A as claimed. HW6 #4. Prove that every affine map f is a continuous bijection. Deduce as an immediate corollary that the inverse mapping f 1 is also continuous. Proof. Let f : R R be an affine map, so f(x) = mx + b for all x R, for some m, b R with m 0. (f is injective.) Let x, y R be arbitrary and assume f(x) = f(y). Clearly mx+b = my+b implies x = y.
(f is surjective.) Let z R be arbitrary. Setting x = z b m m 0), we see that f(x) = z. 9 (which makes sense because (f is continuous.) Fix any x R and any ɛ R +. Set δ = ɛ. Assume z R satisfies m x z < δ. Then f(x) f(z) = mx + b mz b = m x z < m δ = ɛ. So f is continuous at x, for each x R. (f 1 is continuous.) The inverse mapping is given by f 1 (y) = 1 y b, which is also an m m affine map, hence continuous by our earlier remarks. HW6 #6. Prove that the map f : R R defined by f(x) = x 3 for all x R is continuous. { } Proof. Fix any x R and any ɛ R + ɛ. Let δ = min 1,, and let z R be 3x 2 + 3 x + 1 such that x z < δ. Then we have f(x) f(z) = x 3 z 3 = x z x 2 + xz + z 2. Since x z 1, we have z x + 1 by the triangle inequality, and hence the above implies: f(x) f(z) x z [x 2 + x z + z 2 ] x z [x 2 + x ( x + 1) + ( x + 1) 2 ] = x z [3x 2 + 3 x + 1] < δ[3x 2 + 3 x + 1] ɛ. Thus f is continuous at x, for every x R. HW6 #7. Prove that the function f : R R defined by f(x) = 0 iff x Q, and f(x) = 1 iff x R\Q, is not continuous at any point in R. Proof. Let x R; we will show f is not continuous at x. Set ɛ = 1, and let δ R + be arbitrary. Now either x Q or x R\Q. Both Q and R\Q are dense in the real line. Hence, we can find z (x δ, x + δ) so that x and z are not both rational. Then by the definition of the map f, we get f(x) f(z) = 1 ɛ. So f fails continuity at x as claimed. HW6 #8. Prove that the function f : R R defined by f(x) = 0 iff x Q, and f(x) = x iff x R\Q, is continuous at 0. Proof. Let ɛ R +. Set δ = ɛ. Let z R be such that z = z 0 < δ = ɛ. Either z Q or z / Q. If z Q, then f(0) f(z) = 0 0 = 0 < ɛ. If z / Q, then f(0) f(z) = 0 z = z < ɛ. So f is continuous at 0.