CHAPTER 1. Metric Spaces. 1. Definition and examples

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1 CHAPTER Metric Spaces. Definition and examples Metric spaces generalize and clarify the notion of distance in the real line. The definitions will provide us with a useful tool for more general applications of the notion of distance: Definition.. A metric space is given by a set X and a distance function d : X X R such that i) (Positivity) For all x, y X 0 d(x, y). ii) (Non-degenerated) For all x, y X 0 = d(x, y) x = y. iii) (Symmetry) For all x, y X d(x, y) = d(y, x) iv) (Triangle inequality) For all x, y, z X d(x, y) d(x, z) + d(z, y). Examples: i) X = R, d(x, y) = x y. ii) X = R 2 = R R, x = (x, x 2 ), y = (y, y 2 ) d (x, y) = x y + x 2 y 2. iii) X = R 2, x = (x, x 2 ), y = (y, y 2 ) d 2 (x, y) = ( x y 2 + x 2 y 2 2) 2. iv) Let X = {p, p 2, p 3 } and d(p, p 2 ) = d(p 2, p ) =, d(p, p 3 ) = d(p 3, p ) = 2,

2 2. METRIC SPACES d(p 2, p 3 ) = d(p 3, p 2 ) = 3. Can you find a triangle (p, p 2, p 3 ) in the plane with these distances? v) Let X = {p, p 2, p 3 } and d(p, p 2 ) = d(p 2, p ) =, d(p, p 3 ) = d(p 3, p ) = 2, d(p 2, p 3 ) = d(p 3, p 2 ) = 4. Can you find a triangle (p, p 2, p 3 ) in the plane with these distances? vi) The French railway metric (Chicago suburb metric) on X = R 2 is defined as follows: Let x 0 = (0, 0) be the origin, then d 2 (x, y) if there exists a t R such that x = ty and x 2 = ty 2 d SNCF (x, y) =. d 2 (x, x 0 ) + d 2 (x 0, y) else Exercise: Show that the railroad metric satisfies the triangle inequality. It is by no means trivial to show that d 2 satisfies the triangle inequality. In the following we write 0 = (0,..., 0) for the origin in R n. CS Lemma.2. Let x, y R n, then ( ) ( ) 2 2 x i y i x i 2 y i 2 Lemma.3. On R n the metric satisfies the triangle inequality. d 2 (x, y) = ( x i y i 2 ) 2 Proof. Let x, y, z R n. Then we deduce from Lemma.2 CS d(x, y) 2 = x i y i 2 = (x i z i ) (y i z i ) 2 = (x i z i ) 2 2 (x i z i )(y i z i ) + d(x, z) 2 + 2d(x, y)d(y, z) + d(y, z) y i z i 2

3 . DEFINITION AND EXAMPLES 3 = (d(x, z) + d(y, z)) 2. Hence, d(x, y) d(x, z) + d(y, z) and the assertion is proved. More examples: () Let n be a prime number. On Z we define dd n (x, y) = n max{m N : m divides x-y}. The n-adic metric satisfies a stronger triangle inequality (2) Let p <. Then defines a metric n R n. (3) For p = dd n (x, y) max{dd n (x, z), dd n (z, y)}. d p (x, y) = d (x, y) = also defines a metric on R n. ( x i y i p ) p max x i y i,..,n Project : Let < p, q < such that /p+/q =. Show Minkowski s inequality. Mink (.) xy xp p + yq q holds for all x, y > 0. Hint: the function f(x) = ln x is convex on (0, ). Proof of the triangle inequality for d p. The triangle inequality for p = is obvious. We will fist show ( ) ( ) p q mink2 (.2) x i y i x i p y i q whenever p + q =. Let t > 0. We first observe that x i y i = = tp p tx i t y i x i p + t q q p tx i p + q t y i q y i q.

4 4. METRIC SPACES What is best choice of t? Make x i p = t q y i q i.e. This yields t p x i y i t p t p+q = y i q. x i p x i p = ( ) ( q = y i q ( n ) p y i q p+q ( n ) p x i p p+q ) q x i p x i p Now, we proof the triangle inequality. Let x = (x i ), (y i ) and z = (z i ) in R d. Then we apply (.2) mink2 d d d p (x, y) p = x i y i p x i y i p ( x i z i + z i y i ) d x i y i p x i z i + d x i y i p z i y i ( d ) ( q d ( x i y i p ) q ) ( p d z i x i p + ) p z i y i p. However, = /p + /q implies p = p/q and thus q(p ) = p. Hence we get d p (x, y) p d p (x, y) p (d p (x, z) + d p (z, y)). If x y we may divide and deduce the assertion.

5 2. CLOSED AND COMPACT SETS 5 2. Closed and Compact Sets Let (X, d) be a metric space. We will say that a subset A X is closed if X \ A is open. closed Proposition 2.. Let (X, d) be a complete metric space and C X a subset. C is closed iff every Cauchy sequence in C converges to an element in C. Proof: Let us assume C is closed and that (x n ) is a Cauchy sequence with elements in C. Let x = lim n x n be te limit and assume x / C. Since X \ C is open B(x, ε) X \ C for some ε > 0. Then there exists an n 0 such that d(x n, x) < ε for n > n 0. In particular, x n0 + B(x, ε) and thus x n0 + / C, a contradiction. Now, we assume that every Cauchy sequence with values in C converges to an element in C. If X \ C is not open, then there exists an x / C and no ε > 0 such that B(x, ε) X \ C. I.e. for every n N, we can find x n C such that d(x, x n ) < n. Hence, lim x n = x C but x / C, contradiction. The most important notion in this class is the notion of compact sets. We will say that a subset C X is compact if For every collection (O i ) of open sets such that C i O i = {x X i I x O i } There exists n N and i,..., i n such that C O i O in. In other words Every open cover of C has a finite subcover.

6 6. METRIC SPACES Definition 2.2. Let X O i be an open cover. Then we say that (V j ) is an open subcover if X j V j all the V j are open and for every j there exists an i such that V j O i. It is impossible to explain the importance of compactness right away. But we can say that there would be no discipline Analysis without compactness. The most clarifying idea is contained in the following example. Proposition 2.3. The set [0, ] R is compact. Proof: Let [0, ] i O i. For every x [0, ] there exists an i I such that x O i. Since O i is open, we can find ε > 0 such that x B(x, ε) O i. Using the axiom of choice, we fine a function ε x and i x such that x B(x, ε x ) O ix. Let us define the relation x y if x < y and The crucial point here is to define y x e x + e y. S = {x [0, ] x,..., x n : 2 x x n x}. We claim a) sup S S and b) sup S =. Ad a): Let y = sup S [0, ]. Then there exists an x S with y ε y < x y. Then obviously x y. Since x S, we can find Thus y S. 2 x x n x y. Ad b): Assume y = sup S <. Let 0 < δ = min(e y, y). Then y + δ y = δ ε y + ε y+δ.

7 2. CLOSED AND COMPACT SETS 7 By a), we find 2 x x n y y + δ and thus y + δ S. Contradiction to the definition of the supremum. Assertion a) and b) are proved. Therefore we conclude S and thus find x,..., x n such that 2 x x n. Let x 0 = 2 and x n+ =, then by definition of, we have [x j, x j+ ] B(x j, ε xj ) B(x j+, ε xj+ ) O ixj O ixj+ for j = 0,.., n. Thus, we deduce [ n 2, ] [x j, x j+ ] j=0 Doing the same trick with [0, ], we find 2 n+ j=0 O ixj. [0, ] m+ j=0 O ix j n+ j=0 O ixj and we have found our finite subcover. subcom Proposition 2.4. Let B X be closed set and C X be a compact set, then B C is compact Proof: Let B C O i be an open cover. then C (X \ B) i O i is an open cover for C, hence we can find a finite subcover C (X \ B) O i O in. Thus B C O i O in is a finite subcover.

8 8. METRIC SPACES ccc Lemma 2.5. Let (X, d) be a metric space and D X be a countable dense set in X, then for every subset C X and every open cover C i O i we can find a countable subcover of balls. Proof: Let us enumerate D as D = {d n n N}. Let x C and find i I and ε > 0 such that x B(x, ε) O i. Let k > 2. By density, we can find an n N such that ε Then Let us define d(x, d n ) < 2k. x B(d n, 2k ) B(x, k ) B(x, ε) O i. M = {(n, k) i I B(d n, 2k ) O i}. Then M N 2 is countable and hence there exists a map φ : N M which is surjective (=onto). Hence for V m = B(d φ (m), 2φ 2 (m) ), φ, φ 2 the 2 components of φ we have C m V m and (V m ) is a countable subcover of balls of the original cover (O i ). main Theorem 2.6. Let (X, d) be a metric space and D subset X be countable dense subset. Let C X be a subset. Then the following are equivalent i) a) Every Cauchy sequence of elements in C converges to a limit in C. b) For every ε > 0 there exists points x,..., x n X such that C B(x, ε) B(x, ε). ii) Every sequence in C has a convergent subsequence. iii) C is compact. Proof: i) ii). Let (x n ) be a sequence. Inductively, we will construct infinite subset A A 2 A 3 and y, y 2, y 3,... in X such that l Aj : d(x l, y j ) < 2 j.

9 2. CLOSED AND COMPACT SETS 9 Put A 0 = N. Let us assume A A 2 A n and y,..., y n have been constructed. We put ε = 2 n 2 and apply condition i)b) to find z,..., z m such that C B(z, ε) B(z m, ε). We claim that there must be a k m such that A n (k) = {l A n x n B(z k, ε)} has infinitely many elements. Indeed, we have A n () A n (m) = A n. If they were all finite, then a finite union of finite sets would have finitely many elements. However A n is infinite. Contradiction! Thus, we can find a k with A n (k) infinite and put A n+ = A n (k) and y n+ = z k. So the inductive procedure is finished. Now, we can find an increasing sequence (n j ) such that n j A j and deduce d(x nj, x nj+ ) d(x nj, y j ) + d(y j, x nj+ ) < 2 2 j j = 2 j because n j A j and n j+ A j+ A j. Thus (x nj ) is Cauchy. Indeed, be induction, we deduce for j < m that d(x nj, x nm ) d(x nj, x nj+ ) + d(x nj+, x nj+2 ) d(x nm, x nm ) m 2 j k=0 2 k = 2 j. This easily implies the Cauchy sequence condition. x C. We got our convergent subsequence. ii) iii): By Lemma ccc 2.5, we can assume that By a) it converges to some C k O k and O k s open. If we can find an n such that C O O n the assertion is proved. Assume that is not the case and choose for every n N an x n C \ O O n. According to the assumption, we have a convergent subsequence, i.e. lim k x nk = x C. Then x O n0 for some n 0 and there exists a ε > 0 such that B(x, ε) O n0.

10 0. METRIC SPACES By convergence, we find a k 0 such that d(x, x nk ) < ε for all k > k 0. In particular, we find a k > k 0 such that n k > n 0. Thus x nk B(x, ε) O n0 O O nk. Contradicting the choice of the (x n ) s. We are done. iii) i)b) Let ε > 0 and then C B(x, ε). x C thus a finite subcover yields b). iii) i)a) Let (x n ) be a Cauchy sequence. Assume it is not converging to some element x C. This means cccc (2.) x C ε(x) > 0 n 0 n > n 0 d(x n, x) > ε. Then C x C B(x, ε(x) 2 ). Let C B(y, ε(y ) 2 ) B(y, ε(y ) 2 ) be a finite subcover (compactness). Then there exists at least one k m such that A k = {n N d(x n, y k ) < ε(y k) 2 } is infinite. Fix that k and apply the Cauchy criterion to find n 0 such that d(x n, x n ) < ε(y k) 2 for all n, n > n 0. By ( 2.), cccc we can find an n > n 0 such that d(x n, y k ) > ε(y k ). Since A k is infinite, we can find an n > n 0 in A k thus ε(y k ) < d(x n, y k ) d(x n, x n ) + d(x n, y k ) < ε(y k) 2 + ε(y k) 2 = ε(y k ). A contradiction. Thus the Cauchy sequence has to converge to some point in C. Corollary 2.7. Every intervall [a, b] R with a < b R is compact

11 2. CLOSED AND COMPACT SETS Proof: It is easy to see that X \ [a, b] is open. Hence, by Proposition closed 2. [a, b] is complete, i.e. i)a) is satisfied. Given ε > 0, we can find k >. For m > k(b a) we ε derive Thus the Theorem main 2.6 applies. [a, b] m B(a + j k, ε). j=0 cube Lemma 2.8. Let r > 0 and n N, the set C r = [ r, r] n is compact. Proof: Let x / C r, then there exists an index j {,.., n} such that x j > r. Let ε = x j r and y R n such that then max x i y i < ε,,..,n y j = y j x j + x j x j y j x j > x j ε = r. thus y / C r. Hence, C r is closed and according to Proposition rncomp??, we deduce that C r is complete. For n = and ε > 0, we have seen above that for k > ε m [ r, r] B( r + j k, ε). j=0 and m > 2r k Therefore [ r, r] n B (( r + j k,.., r + j n ), ε). k j,...j n=0,...m Thus i)a) and i)b) are satisfies and the Theorem 2.6 main implies the assertion (The separable dense subset is Q n.) Theorem 2.9. Let C R n be a subset. The following are equivalent ) C is compact. 2) C is closed and there exists an r such that (That is C is bounded.) Proof: 2) ) Let C B(0, R). C B(0, R) [ R, R] n be a closed set. Since [ R, R] n is compact, we deduce from Proposition subcom 2.4 that C is compact as well.

12 2. METRIC SPACES ) 2) Let C subset R n be a compact set. According to Theorem main 2.6 i)b), we find C B(x, ) B(x m, ) thus for r = max,..,m (d(x i, 0) + ) we have C B(0, r). Moreover, by Theorem main 2.6 i)a) and Proposition closed 2., we deduce that C is closed.

13 2. CLOSED AND COMPACT SETS 3 Proof: ii) : Let x X and 0 < ε <, then there exists a ε f > 0 such that ε d(x, y) < ε f d(f(x), f(y)) < 3( + g(x) ) <, and ε g > 0 such that d(x, y) < ε g d(g(x), g(y)) < ε 2( + g(x) ) <. Let y X such that d(y, x) < min{ε f, ε g, } = δ. Then we deduce form ε < fg(x) fg(y) f(x) g(x) g(y) + f(x)g(y) f(y)g(y) f(x) g(x) g(y) + f(x) f(y) g(x) + f(x) f(y) g(x) g(y) < ε 3 + ε 3 + ε2 9 < ε.