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Metric Spaces Definition A metric on a set X is a real-valued function d on X × X such that, for all points x, y , z ∈ X , the following properties hold: Functional Analysis Lecture 1: Metric Spaces – Metrics, Open and Closed Sets (i) d is non-negative: d(x, y ) ≥ 0; (ii) d is definite: d(x, y ) = 0 if and only if x = y ; (iii) d is symmetric: d(x, y ) = d(y , x); Bengt Ove Turesson (iv) d satisfies the triangle inequality: d(x, z) ≤ d(x, y ) + d(y , z). August 30, 2016 If d is a metric on X , then the pair (X , d) is called a metric space. Remark When the metric d on a set X is given, we will sometimes also call X a metric space. 2 / 17 1 / 17 Examples of Metric Spaces Examples of Metric Spaces Example A metric on Rd is given by d2 (x, y ) = |x − y | = X d )t j=1 |xj − yj |2 1/2 )t Example , Rd . where x = (x1 , . . . , xd and y = (y1 , . . . , yd are vectors in metric on Cd . This will be the standard metric on Rd and Cd . Other examples of metrics on Rd and Cd are dp (x, y ) = X d j=1 |xj − yj |p 1/p Let `∞ (M) denote the set of all bounded, complex-valued functions on an arbitrary set M. This is also a A metric on `∞ (M) is given by d∞ (f , g ) = sup |f (x) − g (x)|, , x∈M where f , g ∈ `∞ (M). where 1 ≤ p < ∞, and d∞ (x, y ) = max |xj − yj |. 1≤j≤d Here, the triangle inequality in the case 1 ≤ p < ∞ requires a proof, which we shall come back to later. 3 / 17 4 / 17 Examples of Metric Spaces Examples of Metric Spaces Example Example For 1 ≤ p < ∞, the set `p consists of all sequences x = (xj )∞ j=1 of P p < ∞. complex numbers such that ∞ |x | j=1 j A metric on `p is given by (a) The metric space `∞ = `∞ (N) consists of all bounded sequences x = (xj )∞ j=1 of complex numbers. By the definition in the previous example, d∞ (x, y ) = sup |xj − yj |, j≥1 x, y ∈ `∞ . dp (x, y ) = X ∞ j=1 |xj − yj |p 1/p x, y ∈ `p . , The triangle inequality for `p follows from Minkowski’s inequality: ∞ (b) The metric space c consists of all sequences x = (xj )∞ j=1 ∈ ` such that the limit limj→∞ xj exists. X ∞ (c) The metric space c0 consists of all sequences ∞ such that lim x = (xj )∞ j→∞ xj = 0. j=1 ∈ ` j=1 (d) The metric space c00 consists of all sequences ∞ such that x = 0 for all but a finite number x = (xj )∞ j j=1 ∈ ` of indices j. p |xj + yj | 1/p ≤ X ∞ j=1 p |xj | 1/p + X ∞ j=1 p |yj | 1/p , p ∞ p which holds for all sequences x = (xj )∞ j=1 ∈ ` and y = (yj )j=1 ∈ ` . The proof of this inequality shall be given later. 5 / 17 Examples of Metric Spaces 6 / 17 Balls in Metric Spaces Definition Example Let (X , d) be a metric space. If x ∈ X and r > 0, the open ball Br (x) with center x and radius r is the set (a) The space Cb (M) consists of all bounded, continuous, complex-valued functions on a set M ⊂ Rd . The metric on Cb (M) is the same as in `∞ (M), i.e., d∞ (f , g ) = sup |f (x) − g (x)|, x∈M Br (x) = {y ∈ X : d(x, y ) < r }. The corresponding closed ball is B r (x) = {y ∈ X : d(x, y ) ≤ r }. f , g ∈ Cb (M). Example (b) If K ⊂ Rd is compact (i.e., closed and bounded), then, as we know from Calculus, all continuous functions on K are bounded. We shall use the notation C (K ) for Cb (K ). If x ∈ R and r > 0, then Br (x) = (x − r , x + r ) 7 / 17 and B r (x) = [x − r , x + r ]. 8 / 17 Interior Points and the Interior Open Subsets of Metric Spaces Definition Definition Suppose that (X , d) is a metric space and A is a subset of X . Suppose that (X , d) is a metric space. A subset G of X is said to be open if all its points are interior points. A point x ∈ A is called an interior point of A if there exists a number r > 0 such that Br (x) ⊂ A. Remark Thus, G is open if G = G ◦ . A◦ The interior of A is the subset of A that consists of all interior points of A. Example Example An open ball Br (x) in a metric space (X , d) is of course open. Indeed, suppose that y ∈ Br (x) and y 6= x. Then the ball Bs (y ), where s = r − d(x, y ) > 0, is a subset of Br (x), since if z ∈ Bs (y ), then Let us show that [a, b]◦ = (a, b) if −∞ < a < b < ∞. If x ∈ (a, b), then (x − r , x + r ) ⊂ [a, b], where r = min{|x − a|, |x − a|}, showing that x ∈ [a, b]◦ . d(x, z) ≤ d(x, y ) + d(y , z) < d(x, y ) + s = r . Hence, (a, b) ⊂ [a, b]◦ . This shows that y is an interior point of Br (x). Since Br (x) ⊂ Br (x), x is also an interior point of Br (x). Since a, b ∈ / [a, b]◦ , it follows that [a, b]◦ = (a, b). 9 / 17 Open Subsets of Metric Spaces 10 / 17 Open Subsets of Metric Spaces Proposition Example Suppose that (X , d) is a metric space and A is a subset of X . Then the interior A◦ of A is an open subset of A. (a) By the previous example, any interval of the form (a, b) ⊂ R, where −∞ < a < b < ∞, is open: Take x as the midpoint of the interval and r as half its length. Proof. This is obvious if A◦ = ∅, so suppose that A◦ 6= ∅. (a) It is easy to see that any interval of the form (−∞, a) or (a, ∞), where −∞ < a < ∞, is open. Take x ∈ A◦ and choose r > 0 such that Br (x) ⊂ A. We need to show that Br (x) ⊂ A◦ . If y ∈ Br (x), then, since Br (x) is open, there exists a number s > 0 such that Bs (y ) ⊂ Br (x). Example In any metric space (X , d), the whole space X and ∅ are open. Since Br (x) ⊂ A, it follows that Bs (y ) ⊂ A and hence that y ∈ A◦ . This shows that Br (x) ⊂ A◦ . 11 / 17 12 / 17 Open Subsets of Metric Spaces The Topology on a Metric Space Proposition Remark Suppose that (X , d) is a metric space and A is a subset of X . Then the interior A◦ of A is the largest open subset of A with respect to inclusion. The following theorem shows that every metric space is what is known as a topological space. A topological space is a set X together with a collection τ of subsets of X that satisfies the axioms (i)–(iii) below. The elements of τ are called open sets. Proof. Theorem We already know that A◦ is an open subset of A and need to show that any open subset B of A is also a subset of A◦ . Suppose that (X , d) is a metric space. The collection τ of open subsets of X is a topology on X : We can obviously assume that B is non-empty. (i) X , ∅ ∈ τ ; S (ii) if Gα ∈ τ for every α ∈TA, then α∈A Gα ∈ τ ; n (iii) if G1 . . . , Gn ∈ τ , then j=1 Gj ∈ τ . If x ∈ B, then there exists a number r > 0 such that Br (x) ⊂ B. Since B ⊂ A, it follows that Br (x) ⊂ A and therefore that x ∈ A◦ . Remark We remark that the index set A in (ii) may be infinite and even uncountable. Hence, B ⊂ A◦ . 14 / 17 13 / 17 The Topology on a Metric Space Closed Subsets of Metric Spaces Proof. (i) The first S statement in the theorem is obvious. (ii) If x ∈ α∈A Gα , then x belongs to some set Gα0 , and since Gα0 is open, Br (x)S⊂ Gα0 for some r > 0. But from this it follows that Br (x) ⊂ α∈A Gα , which shows that the union is open. (iii) This is left as an exercise. Definition Remark (a) Any interval of the form [a, b] ⊂ R is closed since the complement of [a, b] is the union of the sets (−∞, a) and (b, ∞), both of which are open. Suppose that (X , d) is a metric space. A subset F of X is said to be closed if its complement F c is open. Example Notice that an infinite intersection of open sets may not be open. For instance, ∞ \ 1 1 − , = {0}, n n (b) In the same manner, one can show that all intervals of the form (−∞, a] and [a, ∞), where −∞ < a < ∞, are closed. n=1 and {0} is not an open subset of R. 15 / 17 16 / 17 Closed Subsets of Metric Spaces Theorem Suppose that (X , d) is a metric space. Then the following properties hold: (i) The sets X , ∅ are closed. T (ii) If Fα is closed for every α ∈ A, Snthen α∈A Fα is closed. (iii) If F1 . . . , Fn are closed, then 1 Fj=1 is closed. Proof. Property (i) follows from the previous theorem. We will prove that (ii) holds and leave (iii) as an exercise. By De Morgan’s law, \ c [ Fα = Fαc . α∈A α∈A Now, every set Fαc is open, so it follows that S c α∈A Fα is open. 17 / 17