3
$\begingroup$

Let $V$ be a real vectors space, and $W$ be a linear subspace.

Say $W$ is obviously closed if, for every topology on $V$ that makes $V$ a Hausdorff locally convex topological vector space, the subspace $W$ is closed in $V$.

We know $V$ is obviously closed, and any finite-dimensional subspace of $V$ is obviously closed. Is there a known characterization of which subspaces are obviously closed? Are there other sufficient conditions for a subspace to be obviously closed? Are there other known examples?

$\endgroup$
1
  • 3
    $\begingroup$ Your proof will clearly involve the Axiom of Choice. So why not start immediately with a Hamel basis? $\endgroup$ Commented Apr 18, 2021 at 14:22

3 Answers 3

6
$\begingroup$

Only the subspaces mentioned by the OP are obviously closed.

Let $V$ be a real vector space and $W\subset V$ a proper infinite-dimensional linear subspace. We shall endow $V$ with a norm so that $W$ will not be closed in $(V, \|\;.\;\|)$. For this we consider a Hamel basis for $W$ that we partition into a countable part $\{b_1, b_2, \dots\}$ and some (possibly empty) rest $B_1$. We augment this basis to a basis $B$ for $V$ by adding a distinguished vector $b_0$ and a (possibly empty) set $B_2$ (since $V\neq W$ there is such a $b_0$). For an element $x=\sum_{b\in B} b'(x)b$ (the $b'$ are the corresponding coefficient functionals) we put $$ \|x\| = \sup_{b\in B_1} |b'(x)| + \sup_{b\in B_2} |b'(x)| + \sup_n |b_n'(x) + 2^{-n} b_0'(x)|. $$ (This imitates the sup norm on the linear span of $c_{00}$ and $(2^{-n})$.)

Let $x_N= \sum_{k=1}^N 2^{-k}b_k$; we shall argue that $x_N\to b_0$. Since $x_N\in W$ and $b_0\notin W$ this shows that $W$ is not closed. Now $$ \|x_N-b_0\| = \sup_n |b_n'(x_N-b_0) +2^{-n} b_0'(x_N-b_0)|, $$ and for $n\le N$ this term is $=0$, whereas it is $=2^{-n}$ for $n> N$; consequently $\|x_N-b_0\|\le 2^{-(N+1)} \to0$.

$\endgroup$
3
$\begingroup$

This is a comment but I am not entitled. I don’t know if this the sort of thing you are looking for but I would suggest those subspaces which are closed in the weak topology $\sigma(V,W)$ where $W$ is the algebraic dual of $V$.

$\endgroup$
1
  • 1
    $\begingroup$ Welcome to MO! I can convert your answer to a comment if you would really prefer, but for now I'll leave it as an answer for two reasons: 1.) I think it really is at least a partial answer and 2.) As a new user, you are able to comment on your own posts. So if anybody has any follow-up questions about this answer, they can comment here and you can comment in response. $\endgroup$
    – Tim Campion
    Commented Apr 18, 2021 at 15:40
-2
$\begingroup$

This is an addendum to my answer above. It seemed so obvious that this was a characterisation that I didn’t fill in the details which are much simpler than the above. Firstly, if a subspace is closed for every l.c. topology, then obviously for $\sigma(V,W)$. On the other hand, if a subspace is closed for some l.c. topology then also for the weak topology induced by the corresponding dual and so also for the stronger topology $\sigma(V,W)$.

$\endgroup$
2
  • 2
    $\begingroup$ Please note that you can edit your answer to add this remark instead of posting a new answer. $\endgroup$
    – gmvh
    Commented Apr 20, 2021 at 13:09
  • $\begingroup$ You seem to say that a subspace that is closed for some lc topology is closed for the weak topology $\sigma(V,V')$ you are mentioning; but every linear subspace is closed for $\sigma(V,V')$. So what is the consequence of this line of reasoning? Maybe I'm missing something? $\endgroup$ Commented Apr 20, 2021 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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