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?



When is a linear subspace to be closed in all compatible topologies
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 $\|x\_N-b\_0\|\le 2^{-(N+1)} \to0$.


