# closure of a vector subspace is a vector subspace

###### Proof.

Let $X$ be the topological vector space over $\mathbbmss{F}$ where $\mathbbmss{F}$ is either $\mathbbmss{R}$ or $\mathbbmss{C}$, 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 $\lambda,\mu\in\mathbbmss{F}$ 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 ClosureOfAVectorSubspaceIsAVectorSubspace 2013-03-22 15:00:19 2013-03-22 15:00:19 loner (106) loner (106) 8 loner (106) Theorem msc 46B99 msc 15A03 msc 54A05 ClosureOfAVectorSubspaceIsAVectorSubspace ClosureOfSetsClosedUnderAFinitaryOperation