Compact Subspace of Topological Vector Space is von Neumann-Bounded

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

Let $X$ be a topological vector space over $\Bbb F$.

Let $K$ be a compact subspace (where subspace is meant in the topological sense) of $X$.

Then $K$ is von Neumann-bounded.

Proof
Let $V$ be an open neighborhood of ${\mathbf 0}_X$.

We aim to find $s > 0$ such that $E \subseteq t V$ for $t > s$.

From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists $W \subseteq V$ such that $W$ is a balanced open neighborhood of ${\mathbf 0}_X$.

From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, we have:


 * $\ds X = \bigcup_{n \mathop = 1}^\infty n W$

In particular:


 * $\ds K \subseteq \bigcup_{n \mathop = 1}^\infty n W$

Since $K$ is compact, there exists $n_1, n_2, \ldots, n_k$ such


 * $\ds K \subseteq \bigcup_{i \mathop = 1}^k n_i W$

Relabel these $n_i$ so that:


 * $n_1 < n_2 < \ldots < n_k$

From Finite Union of Dilations of Balanced Set, we then have:


 * $K \subseteq n_k W$

Since $W$ is balanced, for $t > n_k$ we have:


 * $\ds \frac {n_k} t W \subseteq W$

so that:


 * $n_k W \subseteq t W$

Then, we have:


 * $K \subseteq t W$

Since $W \subseteq V$, we obtain:


 * $K \subseteq t V$

for $t > n_k$.

Since $V$ was arbitrary, we have that $K$ is von Neumann-bounded.