# Is a locally compact Hausdorff vectorspace with countable base a topological vector space?

my question is related to the very first part of Chapter 3 of Resnicks book "Extreme Values, Regular Variation and Point Processes. He is claiming that one should think of a locally compact Hausdorff space with countable base as a subset of a compactified finite dimensional euclidean space (which is a topological vector space).

Indeed, if we were in a locally compact Hausdorff topological vector space with countable base this would be an obvious statement since every locally compact topological vector space is finite dimensional. My question is: if we additionally assume that the locally compact Hausdorff space with countable base is a vector space do we automatically get that this space is also a topological vector space?

The mathematical formulation of the problem: Let $$X$$ be a vector space over $$K$$. Assume that $$X$$ is Hausdorff, locally compact and has a countable base. Is $$X$$ a topological vector space?

I tried to tackle the problem using that $$X$$ is metrizable and separable, but this did not lead me anywhere.

Thanks!

• @HennoBrandsma I am a little confused. Why is multiplication not continuous in this case? I assume that $\mathbb{R}$ and $\mathbb{Q}$ are endowed with the standard topology. – Florian Brück Dec 11 '19 at 8:48
• I was confused. The finite dimensionality for locally compact TVS’s only holds for realms and complex scalars. – Henno Brandsma Dec 11 '19 at 9:22

This is obviously false, because the topology could have nothing to do with the vector space structure. For instance, with $$K=\mathbb{R}$$ and $$X=\mathbb{R}$$ with its usual vector space structure, we can pick a bijection between $$X$$ and $$[0,1]$$ and thus give $$X$$ a topology which is homeomorphic to the usual topology on $$[0,1]$$. Or, we could pick some crazy bijection between $$X$$ and $$\mathbb{R}$$ to put a topology on $$X$$ which is homeomorphic to the usual topology but for which addition is not continuous.