Hausdorff LCTVS: Precompact Nbhd Implies Finite-Dimensional

# For Hausdorff LCTVS, Precompact Neighbourhood Implies Finite-Dimensional

Theorem 1: Let $E$ be a Hausdorff locally convex topological vector space. If $E$ has a precompact neighbourhood of the origin then $E$ is finite-dimensional. |

**Proof:**We will show that if $E$ has a precompact neighbourhood of the origin then $E$ is a subspace of a finite-dimensional space and is thus itself, finite-dimensional.

- Suppose that $E$ has a precompact neighbourhood $U$. By the proposition on the Boundedness of Precompact Sets in a LCTVS page, $U$ is bounded. Since $E$ is a Hausdorff locally convex topological vector space possessing a bounded neighbourhood of the origin, $E$ must be normable by the theorem on the Bounded Neighbourhood Criterion for a Hausdorff LCTVS to be Normable page.

- Let $\| \cdot \|$ denote the norm on $E$, and let $\overline{B}(o, 1) := \{ x : \| x \| \leq 1 \}$. Since $\{ \lambda \overline{B}(o, 1) : \lambda > 0 \}$ is a base of neighbourhoods of the origin, there exists a $\lambda > 0$ such that:

\begin{align} \quad \lambda \overline{B}(o, 1) \subseteq U \end{align}

- But since $U$ is precompact, the subset $\lambda \overline{B}(o, 1)$ is precompact, and so $\overline{B}(o, 1) = \lambda^{-1} (\lambda \overline{B}(o, 1))$ is precompact (as scalar multiples of precompact sets are precompact).

- Since $\overline{B}(o, 1)$ is precompact and since $\frac{1}{2} \overline{B}(o, 1)$ is an absolutely convex neighbourhood of the origin, there exists $a_1, a_2, ..., a_n \in E$ such that:

\begin{align} \quad \overline{B}(o, 1) \subseteq \bigcup_{i=1}^{n} \left ( a_i + \frac{1}{2} \overline{B}(o, 1) \right ) \end{align}

- Let $M := \mathrm{span} (a_1, a_2, ..., a_n)$. Then $M$ is a finite-dimensional space and:

\begin{align} \quad \overline{B}(o, 1) \subseteq \bigcup_{i=1}^{n} \left ( a_i + \frac{1}{2} \overline{B}(o, 1) \right ) & \subseteq M + \frac{1}{2} \overline{B}(o, 1) \\ & \subseteq M + \frac{1}{2} \left ( M + \frac{1}{2} \overline{B}(o, 1) \right ) \\ & \subseteq ... \\ & \subseteq M + \frac{1}{2^k} \overline{B}(o, 1) \end{align}

- for all $k \in \mathbb{N}$. (Note that $M + \frac{1}{2}M + \frac{1}{4}M + ... + \frac{1}{2^k}M = M$ since $M$ is a subspace).

- Let $x \in U$. Then for each $k \in \mathbb{N}$ we have that $x = m_k + \frac{1}{2^k} n_k$ where $m_k \in M$ and $\frac{1}{2^k}n_k \in \frac{1}{2^k} \overline{B}(o, 1)$. Clearly $\frac{1}{2^k}n_k \to 0$, and thus $(m_k) \subset M$ converges to $x$. So $x \in \overline{M}$ and thus $\overline{B}(o, 1) \subseteq \overline{M}$. But $M$ is a finite-dimensional subspace of $E$ and is thus closed as seen on the Finite-Dimensional Subspaces are Closed in a Hausdorff LCTVS page. So $\overline{M} = M$ and $\overline{B}(o, 1) \subseteq M$. Finally:

\begin{align} \quad E = \bigcup_{\lambda > 0} \lambda \overline{B}(o, 1) \subseteq \bigcup_{\lambda >0} \lambda M = M \end{align}

- So $E$