Chapter 1. Metric spaces Proofs covered in class P. Karageorgis [email protected] 1 / 37 Theorem 1.1 – Technical inequalities Suppose that x,y 0 and let a, b, c be arbitrary vectors in Rk. ≥ 1 1 1 Young’s inequality: If p,q > 1 are such that p + q = 1, then xp yq xy + . ≤ p q Proof. Let y 0 be fixed and consider the function ≥ xp yq f(x)= + xy, x 0. p q − ≥ Since f ′(x) = xp−1 y, this function is decreasing when xp−1 < y − and increasing when xp−1 > y, so it attains its minimum value at 1/(p−1) the point x∗ = y . One may now easily check that f(x∗) = 0 and this implies that f(x) 0 for all x 0, as needed. ≥ ≥ 2 / 37 Theorem 1.1 – Technical inequalities Suppose that x,y 0 and let a, b, c be arbitrary vectors in Rk. ≥ 2 1 1 H¨older’s inequality: If p,q > 1 are such that p + q = 1, then k k 1/p k 1/q p q ai bi ai bi . | | · | |≤ " | | # " | | # Xi=1 Xi=1 Xi=1 Proof. The result is clear when either a or b is zero. When a, b are both nonzero, we need to show that k x y 1, where i=1 i i ≤ − − k 1/p P k 1/q p q xi = ai ai and yi = bi bi . | | · " | | # | | · " | | # Xi=1 Xi=1 Young’s inequality gives x y 1 xp + 1 yq for each i. Once we now i i ≤ p i q i add these inequalities, we find that k x y 1 + 1 = 1. i=1 i i ≤ p q P 3 / 37 Theorem 1.1 – Technical inequalities Suppose that x,y 0 and let a, b, c be arbitrary vectors in Rk. ≥ 3 Minkowski’s inequality: If p> 1, then d (a, b) d (a, c)+ d (c, b). p ≤ p p Proof. First, we use the triangle inequality in R to find that k k d (a, b)p a c a b p−1 + c b a b p−1. p ≤ | i − i| | i − i| | i − i| | i − i| Xi=1 Xi=1 p 1 1 Letting q = p−1 , we also have p + q = 1, so H¨older’s inequality gives d (a, b)p d (a, c) d (a, b)p−1 + d (c, b) d (a, b)p−1. p ≤ p p p p This already implies Minkowski’s inequality whenever d (a, b) = 0 p 6 and the inequality holds trivially whenever dp(a, b) = 0. 4 / 37 Theorem 1.2 – Main facts about open sets 1 If X is a metric space, then both ∅ and X are open in X. 2 Arbitrary unions of open sets are open. Proof. First, we prove 1 . The definition of an open set is satisfied by every point in the empty set simply because there is no point in the empty set. This means that ∅ is open in X. To show that X is open in X, let x X and consider the open ball B(x, 1). This is a ∈ subset of X by definition, so X is open in X. Next, we prove 2 . Suppose that the sets Ui are open in X and let x be a point in their union. Then x U for some i. Since U is ∈ i i open in X, there exists ε> 0 such that B(x, ε) U . This implies ⊂ i that B(x, ε) U , so the union is open in X as well. ⊂ i i S 5 / 37 Theorem 1.2 – Main facts about open sets 3 Finite intersections of open sets are open. Proof. Suppose that the sets Ui are open in X and let x be a point in their intersection. Then x U for all i. Since U is open in X ∈ i i for each i, there exists ε > 0 such that B(x, ε ) U . Let ε be the i i ⊂ i smallest of the finitely many numbers εi. Then ε> 0 and we have B(x, ε) B(x, ε ) U ⊂ i ⊂ i for all i. This shows that the intersection is open in X as well. 6 / 37 Theorem 1.2 – Main facts about open sets 4 Every open ball is an open set. Proof. Consider the ball B(x, ε) and let y B(x, ε) be arbitrary. ∈ Then d(x,y) < ε and so the number r = ε d(x,y) is positive. To − finish the proof, it suffices to show that B(y, r) B(x, ε). ⊂ Suppose then that z B(y, r). Since d(y, z) < r, we have ∈ d(x, z) d(x,y)+ d(y, z) < d(x,y)+ r = ε ≤ and so z B(x, ε). This shows that B(y, r) B(x, ε). ∈ ⊂ 7 / 37 Theorem 1.2 – Main facts about open sets 5 A set is open if and only if it is a union of open balls. Proof. Suppose first that U is a union of open balls. Then U is a union of open sets by part 4 , so it is open itself by part 2 . Conversely, suppose that U is an open set. Given any x U, we ∈ can then find some ε > 0 such that B(x, ε ) U. This gives x x ⊂ x B(x, ε ) U { }⊂ x ⊂ and we can take the union over all possible x U to find that ∈ U B(x, ε ) U. ⊂ x ⊂ ∈ x[U Thus, U is a union of open balls and the proof is complete. 8 / 37 Theorem 1.3 – Limits are unique The limit of a sequence in a metric space is unique. In other words, no sequence may converge to two different limits. Proof. Suppose x is a convergent sequence which converges to { n} two different limits x = y. Then ε = 1 d(x,y) is positive, so there 6 2 exist integers N1,N2 such that d(x ,x) < ε for all n N , n ≥ 1 d(x ,y) < ε for all n N . n ≥ 2 Setting N = max N ,N for convenience, we conclude that { 1 2} 2ε = d(x,y) d(x,x )+ d(x ,y) < 2ε ≤ n n for all n N. This is a contradiction, so the limit is unique. ≥ 9 / 37 Theorem 1.4 – Main facts about closed sets 1 If a subset A X is closed in X, then every sequence of points ⊂ of A that converges must converge to a point of A. Proof. Let x be a sequence of points of A that converges and { n} let x be its limit. Suppose x X A. Since X A is open, there ∈ − − exists some ε > 0 such that B(x, ε) X A. Since x converges ⊂ − n to x, there also exists an integer N such that x B(x, ε) for all n N. n ∈ ≥ This implies x X A for all n N, which is contrary to the fact n ∈ − ≥ that x A for all n N. In particular, we must have x A. n ∈ ∈ ∈ 10 / 37 Theorem 1.4 – Main facts about closed sets 2 Both ∅ and X are closed in X. 3 Finite unions of closed sets are closed. 4 Arbitrary intersections of closed sets are closed. Proof. First, we prove 2 . Since the sets ∅, X are both open in X, their complements X, ∅ are both closed in X. To prove 3 and 4 , one needs to use De Morgan’s laws n n X U = (X U ), X U = (X U ). − i − i − i − i i[=1 i\=1 \i [i For instance, consider finitely many sets Ui which are closed in X. Their complements X U are then open in X, so the same is true − i for their intersection. Using the first De Morgan’s law, we conclude that the union of the sets Ui is closed in X. This proves 3 and one may similarly prove 4 using the second De Morgan’s law. 11 / 37 Theorem 1.5 – Composition of continuous functions Suppose f : X Y and g : Y Z are continuous functions between → → metric spaces. Then the composition g f : X Z is continuous. ◦ → Proof. We show that g f is continuous at any x X. Let ε> 0 ◦ ∈ be given. Since g is continuous at f(x), there exists δ > 0 with d (f(x),y) < δ = d (g(f(x)), g(y)) < ε. Y ⇒ Z Since f is continuous at x, there also exists δ′ > 0 with d (x,x′) < δ′ = d (f(x),f(x′)) < δ. X ⇒ Y Once we now combine the last two equations, we find that d (x,x′) < δ′ = d (g(f(x)), g(f(x′))) < ε. X ⇒ Z This shows that the composition g f is continuous at x. ◦ 12 / 37 Theorem 1.6 – Continuity and sequences Suppose f : X Y is a continuous function between metric spaces → and let x be a sequence of points of X which converges to x X. { n} ∈ Then the sequence f(x ) must converge to f(x). { n } Proof. Let ε> 0 be given. Then there exists some δ > 0 such that d (x,y) < δ = d (f(x),f(y)) < ε. X ⇒ Y Since xn converges to x, there also exists an integer N such that d (x ,x) < δ for all n N. X n ≥ Once we now combine the last two equations, we find that d (f(x ),f(x)) < ε for all n N.