Institut für Analysis WS2019/20 Prof. Dr. Dorothee Frey 31.10.2019 M.Sc. Bas Nieraeth Functional Analysis Solutions to exercise sheet 3 Exercise 1: Totally boundedness (a) Suppose that (X; d) is a totally bounded metric space. Show that X is separable. (b) Let (xn)n2N be a Cauchy sequence in a metric space (X; d). Show that the set fxn : n 2 Ng is totally bounded. n (c) Show that a subset of R is totally bounded if and only if it is bounded. (d) Equip R with the metric d(x; y) := minf1; jx − yjg. Show that (R; d) is bounded, but not totally bounded. Solution: (a) For each n 2 N we can find a finite collection of points Fn ⊆ X such that [ 1 X ⊆ B(x; ): (1) n x2Fn D := S F Now set n2N n. As a countable union of finite sets, this set is again countable. To prove that X is separable, it remains to show that D is dense in X. Let y 2 X and " > 0. To conclude that D = X, we need to show that B(y; ") \ D 6= ;. 1 Pick n 2 N large enough such that n < ". Then by (1), we can find an x 2 Fn such that 1 1 y 2 B(x; n ). Then d(x; y) < n < " so that x 2 B(y; "). Since also x 2 D, we conclude that B(y; ") \ D 6= ;, as desired. The assertion follows. (b) Let " > 0 and let N 2 N such that d(xn; xm) < " whenever n; m ≥ N. In particular, this implies that xn 2 B(xN ;") for all n ≥ N. Thus, since xn 2 B(xn;") for n 2 f1;:::;Ng, we have N [ fxn : n 2 Ng ⊆ B(xn;"): n=1 Thus, fxn : n 2 Ng is totally bounded, as asserted. (c) The implication that a totally bounded set is bounded is true in general. The difficult n direction to prove is that bounded sets in R are totally bounded. One might be tempted n to use the fact that the closure of a bounded set in R is compact and thus, since compact sets are in particular totally bounded, the result follows. This argument is however circular, n since the proof of the fact that closed and bounded sets in R are compact makes use of n the fact that bounded sets in R are totally bounded. Thus, one should prove the directly. — Turn the page! — We first prove this for R. Suppose A ⊆ R is a bounded. Then there is some r > 0 such that A ⊆ (−r; r). It now suffices to show that (−r; r) is totally bounded to conclude that A is, since subsets of totally bounded sets are totally bounded. Let " > 0. Then consider the set of points F := fn" : n 2 Zg \ (−r; r) and note that F S is finite (check this). Moreover, we have (−r; r) ⊆ t2F B(t; "). Indeed, if x 2 (−r; r), then there is a unique n 2 Z such that n" ≤ x < (n + 1)". In particular, this implies that jx − n"j < " and that jx − (n + 1)"j < ". Since either n" (when x ≥ 0) or (n + 1)" (when S x < 0) lies in (−r; r), we conclude that x 2 t2F B(t; "). The assertion follows. Now that we have established the result for R, the general result follows from the following result: Lemma 1. Let (X; d) and (Y; ρ) be metric spaces and let A ⊆ X and B ⊆ Y be totally bounded. Then A × B is also totally bounded with respect to any of the equivalent metrics 1 0 0 0 p 0 p p dp((x; y); (x ; y )) = d(x; x ) + ρ(y; y ) for p 2 [1; 1) 0 0 0 0 d1((x; y); (x ; y )) = maxfd(x; x ); ρ(y; y )g on X × Y . Indeed, using this lemma with the p = 2 product metric allows us to use induction on the n n dimension n to conclude that any set of the form (−r; r) ⊆ R for r > 0 is totally bounded. n Since any bounded set A ⊆ R lies in such a set, any bounded set is totally bounded. Proof of Lemma 1. We will prove the result for the product metric d1. The other cases follow from this case and the fact that for any p 2 [1; 1) we have 0 0 1 0 0 dp((x; y); (x ; y )) ≤ 2 p d1((x; y); (x ; y )): 1 1 − p − p Indeed, from this estimate it follows that if the balls Bd1 ((x1; y1); 2 ");:::;Bd1 ((xN ; yN ); 2 ") covers A × B, then so do the balls Bdp ((x1; y1);");:::;Bd1 ((xN ; yN );"), as desired. The metric d1 works nicely with respect to products, since in this case Bd1 ((x; y); r) = Bd(x; r) × Bρ(y; r): 0 0 0 0 Indeed, we have (x ; y ) 2 Bd1 ((x; y); r) if and only if maxfd(x; x ); ρ(y; y )g < r if and 0 0 0 0 only if d(x; x ) < r and ρ(y; y ) < r if and only if x 2 Bd(x; r), y 2 Bρ(y; r) if and only if 0 0 (x ; y ) 2 Bd(x; r) × Bρ(y; r), proving the equality. Now let " > 0. Since both A and B are totally bounded, we can find finite sets F ⊆ A and S S G ⊆ B such that A ⊆ x2F Bd(x; ") and B ⊆ y2G Bρ(y; "). But then [ [ A × B ⊆ Bd(x; ") × Bρ(y; ") = Bd1 ((x; y);"): x2F; y2G (x;y)2F ×G Since F × G is finite, this proves that A × B is totally bounded, proving the lemma. Exercise 2: Compactness (a) Let (X; d) be a compact metric space and let A ⊆ X be closed. Show that A is compact. (b) Let (X; d) be a metric space and let B ⊆ X be relatively compact. Show that any subset A ⊆ B is also relatively compact. (c) Let (X; d), (Y; ρ) be metric spaces and suppose that X is compact. Show that if f : X ! Y is a continuous bijective map, then its inverse f −1 : Y ! X is also continuous. (d) Let (X; d), (Y; ρ) be compact metric spaces. Show that X × Y is also compact. Solution: For all of these result it is possible to both use a proof using open sets and a proof using sequences. (a) The proof using sequences here is completely straightforward. We present the proof using open covers here. Suppose (Ui)i2I is an open cover of A. Since A is closed, the set XnA is open in X. This means that the collection (Ui)i2I [ fXnAg is an open cover of X. Since X is compact, we can find a finite F ⊆ I such that (Ui)i2F [ fXnAg covers X. But then (Ui)i2F is a finite subcover of A, proving that A is compact. (b) This follows from part (a). Indeed, if B is relatively compact this means per definition that B is compact. Since A ⊆ B, the set A is a closed subset of the compact set B. By part (a), this implies that A is compact. Hence, A is a relatively compact, as asserted. (c) To show that f −1 is continuous, we will show that for each closed set F ⊆ X we have that (f −1)−1(F ) = f(F ) is closed in Y . Let F ⊆ X be closed. Then, since X is compact, it follows from part (a) that F is compact. Since the continuous image of a compact set is again compact, the set f(F ) is compact in Y . Finally, since compact sets are closed, we conclude that f(F ) is closed, as desired. This proves the continuity of f −1. Alternatively, one can prove this with sequences. Indeed, suppose (yn)n2N is a sequence in −1 −1 Y with limit y 2 Y . We conclude that f is continuous, we need to show that f (yn) ! −1 −1 −1 f (y) in X. Writing xn := f (yn), x := f (y), this means it remains to show that xn ! x in X. For this we will use Exercise 1(b) of Exercise sheet 0. Thus, we need to show that any subsequence (xnj )j2N has a further subsequence (xnj )j2N that converges to x to conclude that the sequence (xn)n2N itself converges to x. Let (xnj )j2N be a subsequence of (xn)n2N. Since X is compact, the sequence (xnj )j2N has (x ) x~ 2 X x~ = x a subsequence njk k2N that converges to some , and it remains to show that . f f(x ) ! f(~x) k ! 1 f(x ) = y ! By continuity of , we have njk as . But since also njk njk y = f(x), the fact that limits are unique implies that f(x) = f(~x). Since f is injective, we conclude that x =x ~. This proves the desired result. (d) First we prove this result with sequences. Suppose ((xn; yn))n2N is a sequence in X × Y . Since X is compact, the sequence (xn)n2N has a convergent subsequence (xnj )j2N with limit x 2 X. Next, using the fact that Y is compact, the sequence (ynj )j2N has a convergent (y ) y 2 Y x ! x k ! 1 subsequence njk k2N with limit . Then, since also njk as , we have that (x ; y ) ! (x; y) k ! 1 njk njk as in X×Y . Thus we have found that the sequence ((xn; yn))n2N has a convergent subsequence.