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

Title finite dimensional proper subspaces of a normed space are nowhere dense FiniteDimensionalProperSubspacesOfANormedSpaceAreNowhereDense 2013-03-22 14:58:59 2013-03-22 14:58:59 asteroid (17536) asteroid (17536) 9 asteroid (17536) Result msc 15A03 msc 46B99 msc 54E52