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 $f_{x}:\mathbb{K}\longrightarrow V$ 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. $\square$

Title proper subspaces of a topological vector space have empty interior ProperSubspacesOfATopologicalVectorSpaceHaveEmptyInterior 2013-03-22 17:34:23 2013-03-22 17:34:23 asteroid (17536) asteroid (17536) 6 asteroid (17536) Theorem msc 46A99