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 $𝔽$ where $𝔽$ 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 $\lambda ,\mu \in 𝔽$ 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