Topological Vector Space: $\dim V=n\implies V\cong\mathbb{K}^n$ Let $V$ be a finite dimensional Hausdorff topological vector space.

Prove that it is is isomorphic to the Euclidean vector space of the same dimension:
$$\dim V=n\implies V\cong\mathbb{K}^n$$
I think this works: Assume without loss of generality that $V = \mathbb{F}^n$. Let $\tau$ be the Euclidean topology and $\tau_1$ be any other (Hausdorff) linear topology on $V$.

a) Suppose $x_{\alpha} \to 0$ in $(V,\tau)$, then each component $x_{\alpha}^i \to 0$ in $\mathbb{F}$. Since addition and scalar multiplication are continuous in $(V,\tau_1)$, it follows that
$$
x_{\alpha} = \sum_{i=1}^n x_{\alpha}^i e_i \to \sum_{i=1}^n 0e_i = 0
$$
in $(V,\tau_1)$. Hence, the identity map
$$
I : (V,\tau) \to (V,\tau_1)
$$
is continuous. So,
$$
\tau_1\subset \tau
$$

b) Now consider $B =\{x \in V : \|x\|_2 &lt; 1\}$ and $S = \{x\in V : \|x\|_2 = 1\}$. $S$ is compact in $(V,\tau)$ and since $\tau_1 \subset \tau$, $S$ is compact in $(V,\tau_1)$ as well. Since $(V,\tau_1)$ is Hausdorff, $S$ is closed in $(V,\tau_1)$. Since $0\notin S$, there exists a circled $\tau_1$-neighbourhood $V$ of $0$ such that
$$
V\cap S = \emptyset
$$
I claim that $V\subset B$: If not, then $\exists x \in V\setminus B$. So $\|x\|_2 \geq 1$, whence
$$
\frac{x}{\|x\|_2} \in V\cap S
$$
This is a contradiction, so $V\subset B$. Hence, $B$ is a $\tau_1$-neighbourhood of $0$. Since scalar multiples of $B$ form a $\tau$-base at $0$, it follows that
$$
\tau \subset \tau_1
$$
Proof:

Let $V$ be a finite dimensional Hausdorff topological vector space and $E$ an Euclidean vector space of the same dimension.

Since both dimensions agree there is a linear bijection $L:E\to V$.

Now since the Euclidean space carries the product topology the projections $\pi_k:E\to\mathbb{K}$ are continuous. Similar vector addition and scalar multiplication in $V$ are continuous. But $L(x)=\sum_{k=1}^n\pi_k(x)L(e_k)$. Thus $L$ is continuous as composition of continuous functions.

Next, consider the unit sphere $S$. It is compact in $E$. Thus by continuity $LS$ is compact in $V$. But $V$ is Hausdorff so $LS$ is closed. So its complement is an (open) neighborhood of zero and so is also the balanced core. Lets call it $U$. The crucial observation will be $U\subseteq LB$.

Lemma: $U\subseteq LB$
By construction $U$ is disjoint from $LS$. Since $L$ is injective $L^{-1}U$ is disjoint from $S$. But $L^{-1}U$ is also balanced by linearity of $L$. Thus $L^{-1}U\subseteq B$ and so $U\subseteq LB$ since $L$ was also surjective.

Finally if $N\in\mathcal{N}_0$ then there exists $\delta&gt;0$ such that $\delta B\subseteq N$ and therefore $\delta U\subseteq \delta LB\subseteq N$. But $U$ was a neighborhood of zero so also any of its dilates $\delta U$ and therefore $LN$ is a neighborhood of zero too. That proves that the inverse of $L$ is continuous at zero. But it is linear so it was continuous everywhere.

Concluding that $L$ is an isomorphism. c.q.f.d.