Let (X;d X) be a complete metric space and Y be a subset of X:Then (Y;d Y) is complete if and only if Y is a closed subset of X: Proof. {\displaystyle 1/2^{n+1}0} Metrics on spaces of functions x < . Thus, the different notions of discrete space are compatible with one another. < It suffices to show that there are at least two points x and y in X that are closer to each other than r. Since the distance between adjacent points 1/2n and 1/2n+1 is 1/2n+1, we need to find an n that satisfies this inequality: 1 1 ) x 2 < < < To see why, suppose there exists an r>0 such that d(x,y)>r whenever x≠y. d 2 , one has either {\displaystyle x,y\in E} Certainly the discrete metric space is free when the morphisms are all uniformly continuous maps or all continuous maps, but this says nothing interesting about the metric structure, only the uniform or topological structure. Then, X is a discrete space, since for each point 1/2n, we can surround it with the interval (1/2n - ɛ, 1/2n + ɛ), where ɛ = 1/2(1/2n - 1/2n+1) = 1/2n+2.

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