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Let $E$ be a Hausdorff topological vector space. Then $K\subseteq E$ is compact if and only if $K$ is complete and precompact (i.e., the closure of $K$ in the completion $\hat{E}$ of $E$ is compact).

For the $\Leftarrow$-proof, we can say:$K$ is complete therefore $K=\overline{K}^{\hat{E}}$; $K$ is precompact, therefore $\overline{K}^{\hat{E}}$ is compact. We conclude that $K$ is compact.

Now, why is $K=\overline{K}^{\hat{E}}$? We know that there is an ismorphism $i:E\to\hat{E}$, so we can identify $K$ as a subset of $\hat{E}$, which shows $K\subseteq \overline{K}^{\hat{E}}$. For the other direction, I know that $K=\overline{K}^E$ (i.e., closed in the Hausdorff space $E$, since $K$ is complete). Is $\overline{K}^{\hat{E}}=\overline{K}^E$? Or $K=E\cap \overline{K}^{\hat{E}}$?

Thanks.

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    $\begingroup$ How are you defining the completion without a metric? $\endgroup$ Mar 16 at 18:22
  • $\begingroup$ I've added the theorem that defines the 'completion' of a Hausdorff topological vector space. $\endgroup$
    – Zachary
    Mar 16 at 18:34
  • $\begingroup$ The theorem does not seem to define completion. It only uses it. What is a complete Hausdorff tvs? $\endgroup$ Mar 16 at 18:40
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    $\begingroup$ @almosteverywhere a TVS has an induced structure of uniform space, hence we can define the notion of Cauchy sequences, then a complete TVS is one where every Cauchy net/filter converges. $\endgroup$
    – Alessandro
    Mar 16 at 18:52
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    $\begingroup$ @almosteverywhere - in particular, a sequence $\{v_n\}$ is Cauchy if for every neighborhood $U$ of $0$, there is some $N$ such that $v_n - v_m \in U$ for all $n, m > N$. $\endgroup$ Mar 17 at 14:34

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For general sets $K$, $\overline K^E = \overline K^{\hat E}$ would be false. But because $K$ is complete in $E$, it will also be complete in $\hat E$. And since complete sets are closed, $\overline K^E = K = \overline K^{\hat E}$.

So why does completeness in $E$ imply completeness in $\hat E$? Suppose $\{v_n\} \subset K$ is a Cauchy sequence in $\hat E$. Let $U$ be a neighborhood of $0$ in $E$. Then there is some neighborhood $\hat U$ in $\hat E$ with $U = \hat U\cap E$. Thus there is some $N$ with $v_n - v_m \in \hat U$ for all $m, n > N$. But $v_n, v_m \in K \subset E$, and so $v_n - v_m \in E$, and thus $v_n - v_m \in \hat U \cap E = U$ for all $n,m > N$. Therefore $\{v_n\}$ is a Cauchy sequence in $E$ as well. Since $K$ is complete in $E$, we have $v_n \to v \in K$, and therefore in $\hat E$ as well.

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  • $\begingroup$ Thank you! I was wondering whether $K=E\cap \overline{K}^{\hat{E}}$ is always true? $\endgroup$
    – Zachary
    Mar 18 at 14:24
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    $\begingroup$ Per the embedding, $E$ is a subspace of $\hat E$, i.e., it has the subspace topology inherited from $\hat E$. Thus all open sets in $E$ are the intersection of open sets in $\hat E$ with $E$. And by complementation, all closed sets in $E$ are the intersection of closed sets in $\hat E$ with $E$. $K = E\cap \overline K^{\hat E}$ will be true for any $K$ closed in $E$. But if $K$ is not closed in $E$, it may be true, or it may be false. $\endgroup$ Mar 18 at 14:31

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