finite dimensional proper subspaces of a normed space are nowhere dense

- Let $V$ be a normed space. If $S\subseteq V$ is a finite dimensional proper subspace, then $S$ is nowhere dense.

Proof :

It is known that for any topological vector space (in particular, normed spaces) every proper subspace has empty interior (http://planetmath.org/ProperSubspacesOfATopologicalVectorSpaceHaveEmptyInterior).

From the entry (http://planetmath.org/EveryFiniteDimensionalSubspaceOfANormedSpaceIsClosed) we also know that finite dimensional subspaces of $V$ are closed.

Then, $\mathrm{int}(\overline{S})=\mathrm{int}(S)=\mathrm{\varnothing }$, which shows that $S$ is nowhere dense. $\mathrm{\square }$