Is there an 'intrinsic' characterization of the usual topology on a finite-dimensional vector space? Let $V$ denote a finite-dimensional vector space. Then $V$ becomes a topological space in a canonical way, by choosing a basis and using this to get an isomorphism to Euclidean space. It turns out that the topology you get is independent of choice of basis, so this makes $V$ into a topological vector space in a canonical way.

Question. Is there a more intrinsic, or abstract, or basis-free approach to characterizing this topology, that gets closer to the heart of why it's important?

 A: There are many characterizations of the usual topology on $V$ that do not involve a choice of a basis.  Probably the most important is:


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*The usual topology on $V$ is the unique topology that makes $V$ a  topological vector space.  That is, it is the unique $T_0$ topology that makes addition $V\times V\to V$ and scalar multiplication $\mathbb{R}\times V\to V$ continuous.


This is a standard theorem in functional analysis.  See How to endow topology on a finite dimensional topological vector space?, for instance.
Here are some other characterizations.  The proofs are easy and I leave them for you to discover.


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*The usual topology on $V$ is the coarsest topology that makes all linear maps $V\to \mathbb{R}$ continuous.

*The usual topology on $V$ is the coarsest topology that makes all linear maps $V\to \mathbb{R}^n$ continuous for all $n\in\mathbb{N}$.

*The usual topology on $V$ is the finest topology that makes all linear maps $\mathbb{R}^n\to V$ continuous for all $n\in\mathbb{N}$.


Note that the "dual" of the first of these three is not correct: the usual topology on $V$ is not the finest topology on $V$ that makes all linear maps $\mathbb{R}\to V$ (or even all affine maps $\mathbb{R}\to V$) continuous, at least not if $\dim V>1$.  Indeed, as every multivariable calculus student learns, there are functions $\mathbb{R}^2\to\mathbb{R}$ whose restrictions to any line are continuous but which are not continuous.
