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Let $G$ be a pseudometrizable compact abelian topological group, $X$ a compact metric space and $f:X\rightarrow G$ a continuous bijective function.

Suppose there exists $g\in G$ such that if $d_{G}(g_{1},g_{2})\leq\epsilon$ then there exists $n$ such that $d_{X}(f^{-1}g^{n}g_{1},f^{-1}g^{n}g_{2})\leq\epsilon.$

If $G$ is not metrizable then in general $f^{-1}$ is not continuous but can we conclude $f^{-1}$ is Borel?

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    $\begingroup$ Could you please add a few words of background and information? Otherwise it looks "too localized". $\endgroup$
    – Goldstern
    Commented Apr 21, 2013 at 17:36
  • $\begingroup$ Sure. The multiplication of g is a dynamical system, a rotation in a compact abelian group. I want to construct a dynamical isomorphism from a dynamical system in X to G. $\endgroup$
    – FelipeG
    Commented Apr 22, 2013 at 18:13

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Let's try this. I'm not using the hypothesis in your second paragraph, so maybe I am missing something.

Suppose $G$ is pseudometrizable but not metrizable. The the closure of $\{e\}$, (the identity), is a closed subgroup $N$ of $G$. And every open set in $G$ either contains $N$ or is disjoint from $N$. Then this same thing is true for every Borel set. On the other hand, for any two points of $X$, there is an open set that contains one but not the other. So, whatever bijection $f$ we choose, it is not Borel.

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  • $\begingroup$ Thanks for your comment. There is something I don't understand. Are you assuming the closure of the identity is not a single point? Is this is necessarily true? $\endgroup$
    – FelipeG
    Commented Apr 22, 2013 at 18:27
  • $\begingroup$ Pseudo-metrizable but not metrizable means not Hausdorff, right? In a topological group that is not Hausdorff, the closure of a single point is more than the point. And the closure of a subgroup is a subgroup. $\endgroup$
    – Gerald Edgar
    Commented Apr 26, 2013 at 14:18

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