Finite-Dimensional Vector Spaces Have a Unique LC & H Topology

# Finite-Dimensional Vector Spaces Have a Unique Locally Convex and Hausdorff Topology

Theorem 1: Let $E$ be a finite-dimensional vector space. Then there exists a unique topology for which $E$ becomes a Hausdorff locally convex topological vector space. |

**Proof:**Since $(E, E^*)$ is a dual pair, $\sigma(E, E^*)$ is a Hausdorff locally convex topology on $E$.

- Let $\tau$ be another Hausdorff locally convex topology on $E$. We will show that $\tau = \sigma(E, E^*)$.

- First, observe that since $\sigma(E, E^*)$ is the coarsest topology for which every linear form $f \in E^*$ is continuous, we must have that $\sigma(E, E^*) \subseteq \tau$.

- To show that $\sigma(E, E^*) \supseteq \tau$, we will show that for every absolutely convex and absorbent $\tau$-neighbourhood $U$ of the origin, there exists a $\sigma(E, E^*)$-neighbourhood $V$ of the origin for which $V \subseteq U$.

- Let $U$ be an absolutely convex and absorbent $\tau$-neighbourhood of the origin. Let $\{ e_1, e_2, ..., e_n \}$ be a basis for $E$ and let $\{ e_1^*, e_2^*, ..., e_n^* \}$ be its dual basis (as defined on The Dual Base for E* of a Finite-Dimensional Vector Space E page). By the absorbency of $U$, for each $1 \leq i \leq n$ there exists a $\mu_i > 0$ such that:

\begin{align} \quad e_i \in \mu_i U \end{align}

- Let $\mu = \max \{ \mu_1, \mu_2, ..., \mu_n \}$. Then:

\begin{align} \quad e_i \in \mu U \end{align}

- for all $1 \leq i \leq n$. Let:

\begin{align} \quad V := \left \{ x : \sup_{1 \leq i \leq n} |\langle x, e_i \rangle| \leq (n \mu)^{-1} \right \} \end{align}

- Then $V$ is a $\sigma(E, E^*)$-neighbourhood of the origin. Furthermore, if $x \in V$, then write $\displaystyle{x := \sum_{i=1}^{n} \lambda_i e_i}$. Then:

\begin{align} \quad x = \sum_{i=1}^{n} \lambda_i e_i \in \sum_{i=1}^{n} \lambda_i (\mu U) \overset{(*)} \subseteq \sum_{i=1}^{n} |\lambda_i| (\mu U) = \sum_{i=1}^{n} |\langle x, e_i^* \rangle| (\mu U) \leq n (n \mu)^{-1} \mu U = U \end{align}

- (Where the inclusion at $(*)$ comes from the fact that $U$ is absolutely convex and using a result on the Absolutely Convex Sets of Vectors page).

- Thus $V \subseteq U$, and so $\sigma(E, E^*) \supseteq \tau$. So $\tau = \sigma(E, E^*)$ and so there is only one topology on $E$ which makes $E$ a Hausdorff and locally convex topological vector space. $\blacksquare$