closure of a vector subspace is a vector subspace

Theorem 1.

In a topological vector spacethe closure (http://planetmath.org/Closure) of a vector subspace is a vector subspace.

Proof.

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

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

whenever $\lambda ,\mu \in 𝔽$ 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 continuousoperations. 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.∎