Proper Linear Subspace of Topological Vector Space has Empty Interior

Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a topological vector space over $\GF$.

Let $L$ be a proper linear subspace of $X$.

Then:

$L^\circ = \O$

where $L^\circ$ denotes the interior of $L$.

Proof

Aiming for a contradiction, suppose that $L^\circ \ne \O$.

Let $x \in L^\circ$.

Then there exists an open neighborhood $U$ of $x$ such that $U \subseteq L$.

From Translation of Open Set in Topological Vector Space is Open, $U - x$ is an open neighborhood of ${\mathbf 0}_X$ with $U - x \subseteq L - x$.

Since $L$ is a linear subspace of $X$, we have $L - x = L$.

Let $V = U - x$.

From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, we have:

$\ds X = \bigcup_{n \mathop = 1}^\infty n V$

On the other hand, since $L$ is a linear subspace of $X$ and $V \subseteq L$, we also have:

$\ds \bigcup_{n \mathop = 1}^\infty n V \subseteq L$

and so $L = X$.

This contradicts that $L$ is a proper linear subspace.

Hence $L^\circ = \O$.

$\blacksquare$