Therefore, if n N then x n x N, and so fx ng converges to. Then there exists a positive integer N such that if m n N, then d(x m x n) on a compact metric space X is bounded. Prove that every nite subset of a metric space is complete. We denote by B(y, r) the closed ball of center y G Z and radius r > 0. ![]() If (Y,dy) is complete, so is (C(X,Y), p). In other words, we now allow unbounded continuous functions f, and we also show. Solution (a) If FXis closed and (x n) is a Cauchy sequence in F, then (x n) is Cauchy in Xand x nxfor some x2Xsince Xis. (c) Prove that a compact subset of a metric space is closed and bounded. (b) Prove that a closed subset of a compact metric space is compact. As mentioned in the text, the reason is that for unbounded, continuous functions, p(f, g) = sup defines a metric on C(X,Y). (a) Prove that a closed subset of a complete metric space is complete. ![]() We leave it to the exercise that follows to show that the given deni-tion of kx + Mk does make X/M a normed linear space. ![]() Since it is a closed subspace of the complete metric space X, it is itself a complete metric space, and this proves part 1. When X is noncompact, we have defined our metric p on the space C6(X,Y) of bounded continuous function and not on the space C(X,Y) of all continuous func- tions. M is certainly a normed linear space with respect to the restricted norm.
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