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I want to show the following: Let $V$ be a finite dimensional topological vector space and Hausdroff. Suppose $f: \mathbb{K}^n \rightarrow V$ to be an vector space isomorphism. Show that $f$ is a homeomorphism.

Since $f$ is a vector space isomorphism, we know that it is bijective. So the only thing left to show is that $f$ and $f^{-1}$ is continuous. Further, since $f$ is an isomorphism between $V$ and $\mathbb{K}^n$, we can conclude that $dim V=n$.

Let $B \subseteq V$ be an open subset. I want to show that $f^{-1}(B)$ is open in $\mathbb{K}^n$. What do I also know: The addition $+:V \times V \rightarrow V$ and scalar multiplication $\cdot:\mathbb{K} \times V \rightarrow V$ are continuous.

I do not really know how to continue from here on. Hints/Solution would be appreciated.

Philip
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    What is the relation between $X$ and $f$? – Keen-ameteur Mar 10 '24 at 10:22
  • @Keen-ameteur sorry that was a typo $X=V$, I corrected it. – Philip Mar 10 '24 at 10:43
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    There's a solution to an equivalent problem here. It's also discussed here but the answer makes a somewhat opaque assumption about the topology being induced by a norm. – Izaak van Dongen Mar 10 '24 at 11:31
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    Also Proposition 3.1 in here is essentially your problem. – Tobi Mar 10 '24 at 11:40
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    @IzaakvanDongen the proof of this relies on compact sets, but if anything there should be a new answer added to that post from 2012 than a new answer here, since its still a duplicate – Jakobian Mar 10 '24 at 13:08
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    @Jakobian, yes I agree that this question is a duplicate and should probably be closed as such. The compact sets argument is included in the first link, but the fact in the second link also answers the question and is more highly upvoted. I was a bit unsure which is a better dupe target/if OP would be satisfied with those questions/if there is a better dupe target, so I thought I'd wait and see what OP/others think. – Izaak van Dongen Mar 10 '24 at 13:21

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I am not sure if it is correct, but here are my thoughts:

Let $\{b_1,...,b_n\}$ be a basis of $\mathbb{K}^n$, a vector space isomorphism is determined by how it maps $\{b_1,...,b_n\}$ to $V$. Let $x \in \mathbb{K}^n$, i.e $x=\sum_{k=1}^{n}\lambda_k b_k$. Then $T(x)=\sum_{k=1}^{n}\lambda_k T(b_k)$ and since the addition in $V$ and scalar multiplication is continuous (V is a topological vector space) we can conclude that $T$ is continuous.

A vector space isomorphism is a linear bijection, thus can be inverted. By the same argumentation, $T^{-1} : V \rightarrow \mathbb{K}^n$ is continuous.

Peter
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  • An argument roughly like this can work, but I'm not sure this is complete. It seems that the difficulty is in proving that each of those components is continuous. Indeed the argument must rely in some way on the Hausdorffness, since there are counterexamples without that condition (take the topology induced by any seminorm). – Izaak van Dongen Mar 10 '24 at 12:48