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

Question

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?

Answer 1

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

Answer 2

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 $V$.

Answer 3

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 $\sigma(V,W)$.