0
$\begingroup$

Let $(X,d)$ be a metric space. I can't find confirmation that $X$ is complete if and only if its closed subsets are exactly its complete subsets. Although it seems like it would be a useful thing to know. So I'm in doubt. Is this true ?

Here's my proof.

If $X$ is complete, then every Cauchy sequence of a closed subset $S$ converges. In metric spaces, closed subsets are sequentially closed hence the sequence converges in $S$ and $S$ is complete. If $S$, now, is assumed complete instead of closed, any of its convergent sequences is Cauchy, hence converges in $S$ by completeness, therefore $S$ is closed.

Now let's assume that $X$'s closed subsets are its complete subsets. Any Cauchy sequence is bounded, meaning it lies within a closed ball, therefore a complete ball, so it converges.

$\endgroup$
3
$\begingroup$

It is true simply because $X$ is a closed subspace of itself. Since we are assuming that the closed subsets of $X$ are exactly its complete subsets, it follows that $X$ is complete.

Your proof seems correct, though.

| cite | improve this answer | |
$\endgroup$

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.