# 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 |
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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 |