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I know that all real, finite-dimensional topological vector spaces are isomorphic to $\mathbb{R}^n$ for some $n$, but are they also homeomorphic?

The reason I'm asking this is because I was wondering whether or not there were any disconnected real topological vector spaces.

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Since sum and scalar product have to be continuous, I guess that the space is in fact path connected, because of $\lambda\mapsto \lambda V+(1-\lambda)W$ is a path connecting any $V$ and $W\in V$. – Leandro Dec 25 '10 at 7:55
Your isomorphism also gives you a homeomorphism – zroslav Dec 25 '10 at 8:39
If you insist that the vector space topology is Hausdorff (otherwise take a seminorm which is not a norm) then it is true that the dimension determines the homeomorphism type, but it is not trivial. It is easier if you require in addition local convexity. André Weil proves the general fact in one of the very first sections of his "Basic Number Theory". – Theo Buehler Dec 25 '10 at 8:45
@Theo: Does he do it using uniform spaces (which were his invention, if I remember correctly)? – Harry Gindi Dec 25 '10 at 11:22
@Pietro: I took that "isomorphic" means "as a linear space". I agree that it is not that difficult (essentially it boils down to compactness of the unit ball in standard $\mathbb{R}^{n}$). @Harry: No, uniform structures do not enter explicitly (but they lurk around, of course). Yes, uniform structures were invented by Weil, they arose in his investigations of topological groups in the late 30's. – Theo Buehler Dec 25 '10 at 11:53
up vote 7 down vote accepted

Any (Hausdorff) topological real vector space of dimension $n<\infty$ is isomorphic to $\mathbf{R}^n$ with the standard topology, see e.g. Rudin, Functional analysis, theorem 1.21.

Here are some comments:

  1. For some reason it is stated there for complex vector spaces, but, as remarked after the theorem, the proof works for real vector spaces as well.

  2. Instead of the Hausdorff axiom Rudin uses the (weaker) $T_1$ axiom in the definition of a topological vector space.

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If a toplogical group is $T_1$ then since {e} is a closed set and $f : (x,y) \mapsto xy^{-1}$ is continuous, the diagonal, in $G$ which is $f^{-1}(e)$ is closed, i.e., $G$ is Hausdorff. – Dick Palais Dec 26 '10 at 7:05
Dick -- that's a nice proof! Rudin does prove that as well, but his proof is a bit longer and more technical. – algori Dec 26 '10 at 7:39
In fact, you only need $T_{0}$ to conclude that a topological group is Hausdorff: If $U$ contains $x$ but not $y$ then $yx^{-1}U$ contains $y$ but not $x$, so a $T_{0}$ group is $T_{1}$. Now Use Dick's argument. – Theo Buehler Dec 26 '10 at 8:01

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