# closure of a vector subspace is a vector subspace

###### Theorem 1.

In a topological vector space^{}
the closure (http://planetmath.org/Closure) of a vector subspace is a vector subspace.

###### Proof.

Let $X$ be the topological vector space over $\mathbb{F}$ where $\mathbb{F}$ is either $\mathbb{R}$ or $\u2102$, let $V$ be a vector subspace in $X$, and let $\overline{V}$ be the closure of $V$. To prove that $\overline{V}$ is a vector subspace of $X$, it suffices to prove that $\overline{V}$ is non-empty, and

$$\lambda x+\mu y\in \overline{V}$$ |

whenever $\lambda ,\mu \in \mathbb{F}$ and $x,y\in \overline{V}$.

First, as $V\subseteq \overline{V}$, $\overline{V}$ contains the zero vector^{},
and $\overline{V}$ is non-empty.
Suppose $\lambda ,\mu ,x,y$ are as above.
Then there are nets ${({x}_{i})}_{i\in I}$, ${({y}_{j})}_{j\in J}$ in $V$ converging to
$x,y$, respectively.
In a topological vector space, addition^{} and multiplication are continuous^{}
operations^{}. It follows that there is a net ${(\lambda {x}_{k}+\mu {y}_{k})}_{k\in K}$ that converges to $\lambda x+\mu y$.

We have proven that $\lambda x+\mu y\in \overline{V}$, so $\overline{V}$ is a vector subspace. ∎

Title | closure of a vector subspace is a vector subspace |
---|---|

Canonical name | ClosureOfAVectorSubspaceIsAVectorSubspace |

Date of creation | 2013-03-22 15:00:19 |

Last modified on | 2013-03-22 15:00:19 |

Owner | loner (106) |

Last modified by | loner (106) |

Numerical id | 8 |

Author | loner (106) |

Entry type | Theorem |

Classification | msc 46B99 |

Classification | msc 15A03 |

Classification | msc 54A05 |

Related topic | ClosureOfAVectorSubspaceIsAVectorSubspace |

Related topic | ClosureOfSetsClosedUnderAFinitaryOperation |