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I have a question with the following theorem,

Every finite dimensional topological separated vector space, over the $\mathbb{K}$(=$\mathbb{R}$ o $\mathbb{C}$), is isomorphic and homeomorphic to $\mathbb{K}^{n}$.

In class, we saw the prove of this theorem, but we never use the condition that the space is separated. And I really don't see, why is a necessary condition of this theorem the separation? Thanks!

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    Without that condition you could give the space the indiscrete topology. In the proof at some point you ought to be using the fact that a continuous bijection between compact Hausdorff spaces is a homeomorphism, and so you'll need to know that stuff is Hausdorff. – Qiaochu Yuan Oct 09 '17 at 21:22
  • @RobArthan: No, that's not correct. Scalar multiplication is not continuous for the discrete topology (remember that the $\mathbb{K}$ appearing in the scalar multiplication map is still given its usual topology). – Eric Wofsey Oct 09 '17 at 21:44
  • @EricWofsey: thanks for the correction. I've been working with structures like TVSs where the scalars are an arbitrary topological field, so I'd forgotten that the topology on the scalars will be a given for the OP. – Rob Arthan Oct 09 '17 at 21:48

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For any vector space $V$, the indiscrete topology makes it a topological vector space, since then every map to $V$ (in particular, addition and scalar multiplication) is continuous. So you definitely need to assume some separation axiom to prove that a finite-dimensional topological vector space is homeomorphic to $\mathbb{K}^n$.

I don't know what proof you saw, but probably at some step in your proof you tacitly used the fact that a compact subset of your topological vector space is automatically closed. This is only true if you assume the space is Hausdorff.

Eric Wofsey
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  • Thanks for your answer. In the proof, I used the fact that a compact subset of the topological vector space is closed... Thanks! –  Oct 09 '17 at 21:50