Theorem 1   In a topological vector space the closure of a vector subspace is a vector subspace.

Proof. Let $X$ be the topological vector space over where is either or , 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 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.