Locally Compact Hausdorff Topological Vector Space has Finite Dimension

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

Let $\struct {X, \tau}$ be a locally compact topological vector space over $\GF$.

Then $X$ is a finite dimensional vector space.

Proof
Since $X$ is locally compact, there exists a von Neumann-bounded open neighborhood $V$ of ${\mathbf 0}_X$ such that:
 * $\map \cl V$ is compact.

From Dilations of von Neumann-Bounded Neighborhood of Origin in Topological Vector Space form Local Basis for Origin:
 * $\BB = \set {2^{-n} V : n \in \N}$ is a local basis for ${\mathbf 0}_X$.

We have that:
 * $\ds \map \cl V \subseteq \bigcup_{x \in X} \paren {x + \frac 1 2 V}$

Since $\map \cl V$ is compact, there exists $x_1, \ldots, x_m \in X$ such that:
 * $\ds \map \cl V \subseteq \bigcup_{j \mathop = 1}^m \paren {x_j + \frac 1 2 V}$

Let:
 * $Y = \span \set {x_j : 1 \le j \le m}$

Then $Y$ is finite dimensional.

We show that $Y = X$.

From Finite-Dimensional Subspace of Topological Vector Space is Closed, $Y$ is closed.

Lemma
Recall that $\BB$ is a local basis for ${\mathbf 0}_X$.

From Expression for Closure of Set in Topological Vector Space: Corollary, we obtain:
 * $\ds \bigcap_{n \mathop = 1}^\infty \paren {Y + 2^{-n} V} = \map \cl Y$

Since $Y$ is closed, we have:
 * $\map \cl Y = Y$

from Set is Closed iff Equals Topological Closure.

Hence from the Lemma, we have:
 * $V \subseteq Y$

Hence for each $k \in \N$, we have:
 * $k V \subseteq Y$

From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, we have:
 * $\ds X = \bigcup_{n \mathop = 1}^\infty k V$

From Set Union Preserves Subsets: General Result, we obtain:
 * $\ds \bigcup_{n \mathop = 1}^\infty k V \subseteq Y$

so that:
 * $X \subseteq Y$

Since we have $Y \subseteq X$, we obtain $Y = X$.

So $X$ is a finite dimensional vector space.