# Topological vector spaces over field with discrete topology

I know the following theorem on topological vector spaces.

[Theorem]
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.

• 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. Commented Jun 21, 2019 at 2:14

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 $$\Bbb K$$.