When I was reading my notes, I encountered a problem that I don't know how to start with.
Prove that if $V$ is a finite-dimensional vector space over $\mathbb{F}$ there is a unique topology on $V$ making it a Hausdorff topological vector space.
My guess is that we can construct a topology on $V$ and show that it is Hausdorff and unique but what kind of topology should I start with? I was thinking to start with a topology that every points are closed s.t. it is Hausdorff...
There is an isomorphism $T: V \to F^{n}$ for some $n$ and we can declare $U$ to be open in $V$ if $T(U)$ is open in $F^{n}$. This makes $V$ a Hausdorff TVS.
It is a well known fact that all linear maps on finite dimensional TVS's are continuous. Hence, if $\tau_1$ and $\tau_2$ are two Hausdorff vector topologies on $V$ then the identity map $: (V,\tau_1) \to (V,\tau_2)$ is a homeomorphism which proves that $\tau_1=\tau_2$.