closure of a vector subspace is a vector subspaceClosureOfAVectorSubspaceIsAVectorSubspace22005-02-04 13:20:262005-03-01 10:18:14Theorem% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{thm}{Theorem}
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\newtheorem{cor}{Corollary}\begin{thm} In a topological vector space
the \PMlinkname{closure}{Closure} of a vector subspace is a vector subspace.
\end{thm}
\begin{proof}
Let $X$ be the topological vector space over $\mathbbmss{F}$ where
$\mathbbmss{F}$ is either $\R$ or $\C$, 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.
\end{proof}