Finite-Dimensional Subspaces are Closed in a Hausdorff LCTVS

# Finite-Dimensional Subspaces are Closed in a Hausdorff LCTVS

Theorem 1: Let $E$ be a Hausdorff locally convex topological vector space. If $M$ is a finite-dimensional subspace of $E$ then $M$ is closed in $E$. |

**Proof:**We will show that if $a \not \in M$ then $a \not \in \overline{M}$ to conclude that $M = \overline{M}$ and hence that $M$ is closed.

- Let $\{ e_1, e_2, ..., e_n \}$ be a basis for $M$ and let $a \not \in M$. Regard $a, e_1, e_2, ..., e_n$ as linear forms on $E' = E^*$ (in particular, make the identification to the linear forms $\hat{a}, \hat{e_1}, \hat{e_2}, ..., \hat{e_n}$ specified for all $f \in E'$ by $\langle f, \hat{a} \rangle = f(a)$ and $\langle f, \hat{e_i} \rangle = f(e_i)$ for each $1 \leq i \leq n$).

- Then by the lemma on The Topological Dual of E Equipped with σ(E, F) is F page, we have that either $a$ is a linear combination of $\{ e_1, e_2, ..., e_n \}$ or there exists an $f \in E'$ such that $\langle a, f \rangle = 1$ and $\langle e_i, f \rangle = 0$ for all $1 \leq i \leq n$.

- Note that $a$ cannot be a linear combination of $\{ e_1, e_2, ..., e_n \}$, for then since $M$ is a subspace of $E$, this would imply that $a \in M$.

- So the latter must be true. Let:

\begin{align} \quad U := \{ x : |\langle x, f \rangle| < 1 \} \end{align}

- Then $U$ is a neighbourhood of the origin. So $U + a$ is a neighbourhood of $a$. Furthermore, observe that $(a + U) \cap M = \emptyset$, for if instead $x \in (a + U) \cap M$, then we can write:

\begin{align} \quad x := a + u \quad \mathrm{and} \quad x := \sum_{i=1}^{n} \lambda_i e_i \end{align}

- where $u \in U$, and $\lambda_1, \lambda_2, ..., \lambda_n \in \mathbf{F}$. But then:

\begin{align} \quad \langle x, f \rangle = \langle a + u, f \rangle = \langle a, f \rangle + \langle u, f \rangle = 1 + \langle u, f \rangle \neq 0 \end{align}

- while:

\begin{align} \quad \langle x, f \rangle = \left \langle \sum_{i=1}^{n} \lambda_i e_i, f \right \rangle = \sum_{i=1}^{n} \lambda_i \langle e_i, f \rangle = 0 \end{align}

- which is a contradiction. Hence indeed, $(a + U) \cap M = \emptyset$ so that $a \not \in \overline{M}$. Thus we must have that $M = \overline{M}$, i.e., $M$ is closed. $\blacksquare$