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. $\mathrm{\square }$

Title	finite dimensional proper subspaces of a normed space are nowhere dense
Canonical name	FiniteDimensionalProperSubspacesOfANormedSpaceAreNowhereDense
Date of creation	2013-03-22 14:58:59
Last modified on	2013-03-22 14:58:59
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	9
Author	asteroid (17536)
Entry type	Result
Classification	msc 15A03
Classification	msc 46B99
Classification	msc 54E52