If every (proper) closed subset of a metric space is complete, then is the whole space necessarily complete as well?
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@copper.hat Just made the same mistake: $(-1..0]$ is closed in $(-1..2)$. – k.stm Mar 10 '14 at 17:51
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@k.stm: Good catch. – copper.hat Mar 10 '14 at 17:52
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Yes. Let $X$ be a non-complete space. Consider a non-convergent Cauchy sequence $(a_n)$ in $X$. Let $x\in X$ be any point. The point $x$ is not an accumulation point of $(a_n)$, because otherwise $(a_n)$ (being Cauchy) would converge to $x$. Therefore there is an open neighbourhood $U$ of $x$ and an $N\in\mathbb N$ such that $a_k\notin U$ for $k>N$. Hence $(a_n)_{n>N}$ is a sequence in $X\setminus U$, and $X\setminus U$ is a proper closed non-complete subspace of $X$, as shown by this non-convergent Cauchy sequence.
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It suffices to consider a Cauchy sequence that is not convergent and come to a contradiction. The set $S$ consisting of the terms of the sequence is closed; otherwise some subsequence of it would be convergent, and for Cauchy sequences this would imply that the whole sequence is convergent. To make sure that $S$ is not the whole space, delete from it one point. Then you get a closed, proper, but not complete subset of the space.
LeoTheKub
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