# every subspace of a normed space of finite dimension is closed

Let $(V,\parallel \cdot \parallel )$ be a normed vector space^{}, and $S\subset V$ a finite dimensional subspace^{}. Then $S$ is closed.

Proof

Let $a\in \overline{S}$ and choose a sequence^{} $\{{a}_{n}\}$ with ${a}_{n}\in S$ such that
${a}_{n}$ converges^{} to $a$. Then $\{{a}_{n}\}$ is a Cauchy sequence^{} in $V$ and
is also a Cauchy sequence in $S$. Since a finite dimensional normed
space is a Banach space^{}, $S$ is complete^{}, so $\{{a}_{n}\}$ converges to an
element of $S$. Since limits in a normed space are unique, that limit
must be $a$, so $a\in S$.

Example

The result depends on the field being the real or complex numbers. Suppose the $V=Q\times R$, viewed as a vector space over $Q$ and $S=Q\times Q$ is the finite dimensional subspace. Then clearly $(1,\sqrt{2})$ is in $V$ and is a limit point of $S$ which is not in $S$. So $S$ is not closed.

Example

On the other hand, there is an example where $Q$ is the underlying
field and we can still show a finite dimensional subspace is closed. Suppose
that $V={Q}^{n}$, the set of $n$-tuples of rational numbers, viewed
as vector space over $Q$. Then if $S$ is a finite dimensional subspace
it must be that $S=\{x|Ax=0\}$ for some matrix $A$.
That is, $S$ is the inverse image^{} of the closed set^{} $\{0\}$.
Since the map $x\to Ax$ is continuous^{}, it follows that $S$ is a closed set.

Title | every subspace of a normed space of finite dimension^{} is closed |
---|---|

Canonical name | EverySubspaceOfANormedSpaceOfFiniteDimensionIsClosed |

Date of creation | 2013-03-22 14:56:28 |

Last modified on | 2013-03-22 14:56:28 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 12 |

Author | Mathprof (13753) |

Entry type | Theorem |

Classification | msc 54E52 |

Classification | msc 15A03 |

Classification | msc 46B99 |