enter image description here

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 $K=E\cap \overline{K}^{\hat{E}}$?


  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

1 Answer 1


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.

  • $\begingroup$ Thank you! I was wondering whether $K=E\cap \overline{K}^{\hat{E}}$ is always true? $\endgroup$
    – Zachary
    Mar 18 at 14:24
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.