Locally Bounded Hausdorff Topological Vector Space with Heine-Borel Property has Finite Dimension

Theorem

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

Let $\struct {X, \tau}$ be a locally bounded Hausdorff topological vector space over $\GF$ with the Heine-Borel property.

Then $X$ is a finite dimensional vector space.

Proof

Since $\struct {X, \tau}$ is locally bounded, there exists a von Neumann-bounded open neighborhood $V$ of ${\mathbf 0}_X$.

From Closure of von Neumann-Bounded Subset of Topological Vector Space is von Neumann-Bounded, $\map \cl V$ is von Neumann-bounded.

From Topological Closure is Closed, $\map \cl V$ is closed.

So $\map \cl V$ is von Neumann-bounded and closed.

From the Heine-Borel property, $\map \cl V$ is compact.

So $V$ is an open neighborhood of ${\mathbf 0}_X$ with compact closure.

Hence $X$ is locally compact.

Hence from Locally Compact Hausdorff Topological Vector Space has Finite Dimension, $X$ is a finite dimensional vector space.

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.24$: Theorem