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Let $X$ be a metric space such that for any two disjoint closed subsets $A,B$ , $d(A,B)>0$ , then is it true that $X$ is complete ? I was trying to prove it and have figured that if we go by contradiction say $\{x_n\}$ is a non-convergent Cauchy sequence , then if we can find two disjoint subsequences of $\{x_n\}$ , we are done because then the distance of those two subsequences would be $0$ . But I cannot find two disjoint subsequences of a non-convergent Cauchy sequence . Please help . And is there any other approach ? thanks in advance

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Let X be a metric space such that if A and B are closed and disjoint subsets of X, d(A,B)>0. Suppose that X is not complete and consider a Cauchy sequence $(x_n)$ which does not converge. $(x_n)$ has an infinite number of different terms. We can extract from $(x_n)$ two sequence $(u_n)$ and $(v_n)$ such that $u_m\neq v_n, n,m>1$, $u_n\neq u_m, n,m>0$ and $v_n\neq v_m,m,n>0$. Let $U$ be the adherence of $(u_n)_{n\in N}$ (remark that this adherence is just the sequence $(u_n)$) and $V$ the adherence of $(v_n)$, $d(U,V)=0$ contradiction.

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