I recently asked a question on a french forum about the proof that any norms on a finite dimensional (real or complex) vector space are equivalent. Someone showed me the proof of a more powerful statement:
Let $T_{1}$ and $T_{2}$ be two Hausdorff topologies on an $n$-dimensional (real or complex) vector space $E$. Then, $T_{1}=T_{2}$
It was so interesting, in particular because it uses only one powerful result (Tychonoff's theorem), that I thought it could be a good thing to share it here. I rewrote the proof below.
I did not have the idea of this proof on my own. I tried to rewrite it here to be sure I understood it correctly and because I found it interesting to share. I tried to write the more details I could, so that anyone with a basic understanding of topology could understand the proof. If there is any mistake, please let me know. The idea is the following:
1) We consider some separated topology $T$ on a finite dimensional real vector space $E$ and the usual product topology $T_{0}$ on $\mathbb{R}^{n}$ and we proof that $T=T_{1}$ (when $T_{1}$ is the particular topology induced by $T_{0}$ on $E$) on compact neighbourhood of $0_{E}$.
2) Since neighbourhoods of $0_{E}$ determines the entire topology on $E$, we can conclude.
As long as a complex vector space of dimension $m$ can be seen as a real vector space of dimension $2m$, it is whithout loss of generality that we only consider the real case. Let $E$ be an $n$-dimensional real vector space. Let $T_{0}$ be the usual product topology on $\mathbb{R}^{n}$.
Let $(e_{1},\dots,e_{n})$ be a basis of $E$ and $T$ be a separated topology of topological vector space (t.v.s.) on $E$, i.e. vector addition $E\times E\to E:(u,v)\mapsto u+v$ and scalar multiplication $\mathbb{R}\times E\to E:(\lambda,u)\mapsto \lambda u$ are continuous functions with respect to $T$.
Let $U$ be a $T$-open set such that $0_{E}\in U$. The mapping $(\mathbb{R}^{n},T_{0})\to (E,T):x\mapsto \sum_{i=1}^{n}x_{i}e_{i}$ is continuous in $0$ (and bijective) as a composite function of continuous functions. Then, there exists a $T_{0}$-neighbourhood $V$ of $0_{\mathbb{R}^{n}}$ such that $\forall x\in \text{int}(V):\sum_{i=1}^{n}x_{i}e_{i}\in U$ by definition of a continuous function between topological spaces. Now, by definition of a $T_{0}$-neighbourhood in $\mathbb{R}^{n}$, there exists $\epsilon>0$ such that $]-\epsilon,\epsilon[^{n}\subset V$, so that $\left\{\sum_{i=1}^{n}x_{i}e_{i}\,\left.\right\vert\,\forall i=1,\dots,n\,:\vert x_{i}\vert\le\epsilon\right\}\subset U$. (*)
If we define $T_{1}$ as the topology obtained by mapping a $T_{0}$-open set $W$ to the $T_{1}$-open set $X:=\{\sum_{i=1}^{n}x_{i}e_{i}\,\vert\,x\in W\}$, then the previous inclusion (*) shows that $T\subset T_{1}$, i.e. any $T$-open set is a $T_{1}$-open set. This is dued to the fact that neighbourhoods of $0_{E}$ unambiguously determines the entire topology on $E$ because translations are continuous (by definition of a t.v.s.) and bijective, so that we know all neighbourhoods of $y$ for any $y\in E$ and since an open set is the neighbourhood of all his points, we know all open sets. Here, we have determined that any open $T$-neighbourhood of $0_{E}$ is an open $T_{1}$-neighbourhood, hence $T\subset T_{1}$.
We now know that we can restrict to neighbourhoods (not especially open) of $0_{E}$ to determine the entire topology. Consider $C:=\left\{\sum_{i=1}^{n}x_{i}e_{i}\,\left.\right\vert\,\forall i=1,\dots,n\,:\vert x_{i}\vert\le\epsilon\right\}$. It is compact for $T_{1}$ because it is compact for $T_{0}$ by Tychonoff's theorem (the product of compact sets is a compact set w.r.t. product topology). Since $T\subset T_{1}$ and $T$ is seperated by construction, $C$ is compact for $T$.
We prove that if $T\subset T_{1}$ are both compact topologies on $C$, then $T=T_{1}$ on $C$ (w.r.t. the respective induced topologies). By reductio ad absurdum, suppose that there exists some $Y\in T_{1}\setminus T$. Then, $Y^{c}$, the complementary of $Y$, is closed for $T_{1}$. A closed set in a compact set is compact (w.r.t. $T_{1}$), hence $Y^{c}$ is also $T$-compact because $T\subset T_{1}$. Since every compact is closed (in a separated space), $Y^{c}$ is $T$-closed, i.e. $Y$ is $T$-open, which is a contradiction. Hence, $T=T_{1}$.
Since $T=T_{1}$ on any $T_{1}$-compact neighbourhood of $0_{E}$, $T=T_{1}$ on any $T_{1}$-compact neighbourhood of $y\in E$, so that $T=T_{1}$ on $E$.
Remark: one could ask why it is not a strict inequality in (*). As long as it holds for the strict inequality, we can take $\epsilon/2$ to have the inequality.