# every finite dimensional subspace of a normed space is closed

###### Theorem 1

Any finite dimensional subspace of a normed vector space^{} is closed.

Proof. Let $(V,\parallel \cdot \parallel )$ be such a normed vector space, and $S\subset V$ a finite dimensional vector subspace.

Let $x\in V$, and let ${({s}_{n})}_{n}$ be a sequence in $S$ which converges to $x$. We want to prove that $x\in S$. Because $S$ has finite dimension, we have a basis $\{{x}_{1},\mathrm{\dots},{x}_{k}\}$ of $S$. Also, $x\in \mathrm{span}({x}_{1},\mathrm{\dots},{x}_{k},x)$. But, as proved in the case when $V$ is finite dimensional (see this parent (http://planetmath.org/EverySubspaceOfANormedSpaceOfFiniteDimensionIsClosed)), we have that $S$ is closed in $\mathrm{span}({x}_{1},\mathrm{\dots},{x}_{k},x)$ (taken with the norm induced by $(V,\parallel \cdot \parallel )$) with ${s}_{n}\to x$, and then $x\in S$. QED.

## 0.0.1 Notes

The definition of a normed vector space requires the ground field to be the real or complex numbers. Indeed, consider the following counterexample if that condition doesn’t hold:

$V=\mathbb{R}$ is a $\mathbb{Q}$ - vector space, and $S=\mathbb{Q}$ is a vector subspace of $V$. It is easy to see that $dim(S)=1$ (while $dim(V)$ is infinite), but $S$ is not closed on $V$.

Title | every finite dimensional subspace of a normed space is closed |
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Canonical name | EveryFiniteDimensionalSubspaceOfANormedSpaceIsClosed |

Date of creation | 2013-03-22 14:58:56 |

Last modified on | 2013-03-22 14:58:56 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 14 |

Author | Mathprof (13753) |

Entry type | Corollary |

Classification | msc 46B99 |

Classification | msc 15A03 |

Classification | msc 54E52 |