Closure of Linear Subspace of Topological Vector Space is Linear Subspace

Theorem

Let $K$ be a topological field.

Let $X$ be a topological vector space over $K$.

Let $Y$ be a linear subspace of $X$.

Then the closure $Y^-$ of $Y$ is a linear subspace of $X$.

Proof

Since $Y \subseteq Y^-$, we have that $Y^-$ is non-empty.

We now use the One-Step Vector Subspace Test.

We show that for each $\lambda \in K$ and $u, v \in Y^-$ we have:

$u + \lambda v \in Y^-$

That is, we want to show that:

$Y^- + \lambda Y^- \subseteq Y^-$

for each $\lambda \in K$.

Let $\lambda \in K$.

Then:

\(\ds Y^- + \lambda Y^-\)

\(=\)

\(\ds Y^- + \paren {\lambda Y}^-\)

Dilation of Closure of Set in Topological Vector Space is Closure of Dilation

\(\ds \)

\(\subseteq\)

\(\ds \paren {Y + \lambda Y}^-\)

Sum of Closures is Subset of Closure of Sum in Topological Vector Space

Since $Y$ is a linear subspace, for each $u, v \in Y$ we have:

$u + \lambda v \in Y$

So:

$Y + \lambda Y \subseteq Y$

From Topological Closure of Subset is Subset of Topological Closure, we therefore have:

$\paren {Y + \lambda Y}^- \subseteq Y^-$

so that:

$Y^- + \lambda Y^- \subseteq Y^-$

for each $\lambda \in K$ as required.

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.13$: Theorem