proper subspaces of a topological vector space have empty interior

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 $f_x:𝕂⟶V$ given by

$$f_x(\lambda )=\lambda x$$

is also continuous.

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

$$\lambda v\in A-a_0,$$

i.e. we can multiply $v$ by a sufficiently small $\lambda$ such that $\lambda v$ belongs to the open set $A-a_0$.

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. $\mathrm{\square }$

Title	proper subspaces of a topological vector space have empty interior
Canonical name	ProperSubspacesOfATopologicalVectorSpaceHaveEmptyInterior
Date of creation	2013-03-22 17:34:23
Last modified on	2013-03-22 17:34:23
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	6
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 46A99