I know the following theorem on topological vector spaces.

On every finite dimensional vector space over field $k$, there is a unique topological vector space structure.
(The proof is here https://www.math.ksu.edu/~nagy/func-an-2007-2008/top-vs-3.pdf)

Let $k=\mathbb{R}$ with discrete topology and $V=\mathbb{R}$ with usual topology.
Both $k$ and $V$ are topological vector spaces over $k$ which have the dimension $\mathrm{dim}_k (k) = \mathrm{dim}_k (V)=1$.

When I apply above theorem to $\mathbb{R}$, I get $k \cong V$ as topological vector space.
Is this correct $??$ Please give me opinions.

  • 1
    $\begingroup$ When $k$ is discrete and $V$ isn't, you've likely violated one of the axioms requiring the algebraic operations to be continuous. So, $V$ is NOT a topological vector space anymore. A TVS is more than a topological space that is also a vector space. $\endgroup$ – Randall Jun 21 at 2:14

The theorem you reference assumes the field "k" is either R or C equipped with the standard topology. It doesn't say anything about a vector space over R with R given the discrete topology.


Quoting the first line of those notes:

Throughout this note $\Bbb K$ will be one of the fields $\Bbb R$ or $\Bbb C$, equipped with the standard topology. All vector spaces mentioned here are over $\Bbb K$.

So it doesn't apply to the field in the discrete topology.

  • $\begingroup$ @EricWofsey so it is. I’ll remove it. Thx. $\endgroup$ – Henno Brandsma Jun 21 at 7:01

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