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!
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.