It follows that a y is a subset of all closed subsets

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It follows that ¯ A Y is a subset of all closed subsets of Y containing A , thus is contained in ¯ A Y . The result follows. 3. Let X 1 , . . . , X n be metric spaces with distance functions d 1 , . . . , d n respec- tively. Let X = X 1 × · · · × X n = { p = ( p 1 , . . . , p n ) : p i X i , i = 1 , . . . , n } . Consider the following three functions from X × X to R . In the definition, p = ( p 1 , . . . , p n ), q = ( q 1 , . . . , q n ) X . δ 1 ( p, q ) = n X i =1 d i ( p i , q i ) . δ 2 ( p, q ) = ˆ n X i =1 d i ( p i , q i ) 2 ! 1 / 2 . δ ( p, q ) = max { d i ( p i , q i ) : i = 1 , . . . , n } . (a) Prove all three metrics for X are equivalent and a sequence { p m } m =1 , where p m = ( p m 1 , . . . , p mn ) converges to p = ( p 1 , . . . , p n ) if and only if each sequence { p mi } m =1 converges in X i for i = 1 , . . . , n . (b) Prove that all three metrics for X have the same Cauchy sequences. ( Equivalent metrics do NOT necessarily have the same Cauchy sequences so this is not an immediate consequence of the previous point. It is however essentially the same proof as one uses for the previous point.) (c) Prove that X is complete in any one of the three metrics if and only if it is complete in the others, and this happens if and only if each one of the spaces X 1 , . . . , X n is complete. (d) Explain how all this applies to R n . In particular, show that R n is complete. Solution. It will be convenient to prove first the following lemma: 3
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Lemma 1 Let X be a set and let d 1 , d 2 be distance functions on X . As- sume there exist real positive constants a, b such that a d 1 ( x, y ) d 2 ( x, y ) b d 1 ( x, y ) for all x, y X. Then d 1 d 2 in the sense of the definition of Problem 1. Moreover both distance functions have the same Cauchy sequences. Proof. The easiest thing to show is that both metrics have the same con- vergent sequences and the same Cauchy sequences. That’s almost obvious, I think (therefore I am). Assume { p n } is a sequence in X . Assume first it converges with respect to d 2 , say to p . Let ² > 0 be given. Then ²/a > 0 and there is N such that d 2 ( p n , p ) < ²/a whenever n N . If n N , then d 1 ( p n , p ) < a d 2 ( p n , p ) < a ² a = ². The sequence converges with respect to p 1 . Assume next the sequence is Cauchy with respect to d 2 . We essentially repeat the same argument. Let ² > 0 be given. Then ²/a > 0 and there is N such that d 2 ( p n , p m ) < ²/a whenever n, m N . If n, m N , then d 1 ( p n , p m ) < a d 2 ( p n , p m ) < a ² a = ² and the sequence is Cauchy with respect to d 1 . To prove this we only used that p 1 (1 /a ) p 2 . Since p 2 b p 1 , switching the roles of p 1 , p 2 , replac- ing 1 /a by b we see that if a sequence converges with respect to p 1 , or is Cauchy with respect to p 1 , then it converges or is Cauchy, respectively, with respect to d 2 . The rest is easy. But there is a minor detail that perhaps has to be covered. Let us define a stronger equivalence relation for two metrics d 1 , d 2 in a set X by: d 1 = d 2 iff there exist real positive constants a, b such that a d 1 ( x, y ) d 2 ( x, y ) b d 1 ( x, y ) for all x, y X.
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