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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?

goblin GONE
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    Isn't it just the smallest topology, such that all elements of $V^*$ are continuous? – MooS May 05 '17 at 09:16
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    By the way you should add that you work over $\mathbb R$ or $\mathbb C$ and fix the usual topology on the base field. – MooS May 05 '17 at 09:16
  • @MooS, woah nice idea! Arguably you're right about $\mathbb{R}$ or $\mathbb{C}$. However for me the notion of a "vector space over a field" is a bit unnecessary because they're just modules. So I tend to use vector space to mean $\mathbb{R}$-module. If I'm working over anything other than $\mathbb{R}$, I just start saying $R$-module or $k$-module or whatever. – goblin GONE May 05 '17 at 09:16
  • To prove it, the only thing you have to do is check it for the case $V=\mathbb R^n$, i.e. you have to check whether all inverse images of open sets in $\mathbb R$ w.r.t. linear maps $\mathbb R^n \to \mathbb R$ generate the euclidean topology. – MooS May 05 '17 at 09:20
  • @MooS, just a thought, but perhaps the fact that every closed convex set is an intersection of halfspaces is relevant. – goblin GONE May 05 '17 at 09:29
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    This topology is unique. If you have a finite dimensional TVS then there is only one topology which is Hausdorff. – user60589 May 05 '17 at 11:41
  • Nice, this seems pretty intrinsic to me :) And this of course also automatically proves my claim that you can define this topology by saying that linear functionals are continuous. – MooS May 05 '17 at 12:04

1 Answers1

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

  • 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.

  • 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.

Eric Wofsey
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  • The linked answer proves your claim for the $T_2$ (and therefore $T_1$) case, but not $T_0$. So I asked a question about this here. – WillG Sep 12 '22 at 00:00