Theorem - Let $V$ be a topological vector space. Every proper subspace $S \subset V$ has empty interior.

Proof : Let $S$ be a subspace of $V$ . Suppose there is a non-empty open set $A \subseteq S$ .

Fix a point $a_0 \in A \subseteq S$ . Since the vector sum operation is continuous, translations of open sets are again open sets. In particular, the set $A - a_0 := \{x-a_0 : x \in A\}$ is an open set of $V$ that contains the origin $0$ .

As $S$ is a vector subspace and $A \subseteq S$ , we see that the translation $A-a_0$ is still contained in $S$ .

Since the scalar multiplication operation is continuous it follows easily that, for every $x \in V$ , the function $f_x: \mathbb{K} \longrightarrow V$ given by

Consider now any vector $v \in V$ . The set $f_v^{-1}(A-a_0)$ is an open set that contains $0$ . Thus, taking a value $\lambda \in f_v^{-1}(A-a_0)$ we see that

Since the set $A- a_0$ is contained in $S$ , we see that $\lambda v \in S$ , and therefore $v \in S$ .

This proves that $V=S$ , i.e. $S$ is not proper.

We conclude that if $S$ is proper then $S$ has empty interior. $\square$