Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Closed

Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a Hausdorff topological vector space over $\GF$.

Let $n \in \N$.

Let $Y$ be a subspace of $X$ with dimension $n$.

Then $Y$ is closed.

Proof

Let $f : \GF^n \to Y$ be a vector space isomorphism.

From Isomorphism from Cartesian Space to Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Homeomorphism, $f$ is a homeomorphism.

Let $B$ be the unit ball in $\GF^n$.

Let $D$ be the closed unit ball in $\GF^n$.

Let $S$ be the unit sphere in $\GF^n$ and:

$K = f \sqbrk S$

In Isomorphism from Cartesian Space to Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Homeomorphism, it was determined that there exists a balanced open neighborhood $V$ of $\mathbf 0_X$ such that:

$V \cap K = \O$

Now let $\map \cl Y$ be the closure of $Y$.

From Set is Closed iff Equals Topological Closure, we just need to show that $Y^- \subseteq Y$.

From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, there exists $t > 0$ such that:

$p \in t V$

It follows that:

$p \in \map \cl {Y \cap t V}$

In Isomorphism from Cartesian Space to Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Homeomorphism, it was determined that:

$f^{-1} \sqbrk V \subseteq B$

so that:

$Y \cap \paren {t V} \subseteq t f \sqbrk B$

by Image of Preimage under Mapping.

From Image of Dilation of Set under Linear Transformation is Dilation of Image, we have:

$Y \cap \paren {t V} \subseteq f \sqbrk {t B}$

That is:

$Y \cap \paren {t V} \subseteq f \sqbrk {t D}$

Since $D$ is compact, we have:

$t D$ is compact

from Dilation of Compact Set in Topological Vector Space is Compact.

From Continuous Image of Compact Space is Compact, $f \sqbrk {t D}$ is compact.

Since $X$ is a Hausdorff topological vector space, $X$ is in particular Hausdorff.

From Compact Subspace of Hausdorff Space is Closed, $f \sqbrk {t D}$ is therefore closed, and we obtain:

$\map \cl {Y \cap \paren {t V} } \subseteq f \sqbrk {t D}$

So:

$p \in f \sqbrk {t D}$

giving $p \in Y$.

So we have $Y^- = Y$ as desired.

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.21$: Theorem