# When is a linear subspace to be closed in all compatible topologies

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?

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

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 $$\|x_N-b_0\|\le 2^{-(N+1)} \to0$$.

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 $$V$$.

• 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. Apr 18, 2021 at 15:40

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 $$\sigma(V,W)$$.

• 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? Apr 20, 2021 at 19:34