# Understanding proof: compact iff complete and precompact

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.

• How are you defining the completion without a metric? Mar 16 at 18:22
• I've added the theorem that defines the 'completion' of a Hausdorff topological vector space. Mar 16 at 18:34
• The theorem does not seem to define completion. It only uses it. What is a complete Hausdorff tvs? Mar 16 at 18:40
• @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. Mar 16 at 18:52
• @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 $n, m > N$. Mar 17 at 14:34

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.
• Thank you! I was wondering whether $K=E\cap \overline{K}^{\hat{E}}$ is always true? Mar 18 at 14:24
• 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. Mar 18 at 14:31