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. $\\square$

单词	FiniteDimensionalProperSubspacesOfANormedSpaceAreNowhereDense
释义	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. $\\square$
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