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Let $X$ and $Y$ be compact metric spaces. Let $f:X \to Y$ be a Borel measurable map and suppose that $T:X \to X$ is a homeomorphism. Can one change the topology on $X$ such that

  1. $X$ is still a compact metrizable space, with the same Borel sets.
  2. The map $T$ remains continuous.
  3. The map $f:X \to Y$ is continuous.

?

Note: If $Y$ is not compact, we can find counterexamples by taking $f$ to be an unbounded function. The answer is YES if we just want complete metrizability of the new topology on $X$, instead of compactness.

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  • $\begingroup$ Should it read "Let $X$ and $Y$ be compact metric.."? $\endgroup$ – Pedro Sánchez Terraf Mar 8 '15 at 12:22
  • $\begingroup$ Yes, thanks. Fixed. $\endgroup$ – Mathemagician Mar 8 '15 at 12:30
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No - This follows from: If $\tau$ properly extends the usual topology on $[0, 1]$, then $\tau$ is not compact. The same is true of any compact Hausdorff space.

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  • $\begingroup$ Interesting, reference? I guess I just need one counterexample, so I will post one in an answer. $\endgroup$ – Mathemagician Mar 9 '15 at 1:16
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The answer is NO even if we forget $T$ (Or equivalently, let $T=Id_X$). I will later post a measure theoretic weakening of this question, which I am more interested in.

Counterexample: Let $X=[0,1]$ and $Y=[0,1] \cup \{-1\}$ and define $f:X \to Y$ by $f(x)=x$ for $x \neq 0$ and $f(0)=-1$. Thus we would require a topology on $X$ such that $\{0\}$ and $(x,1]$ for all $0<x<1$ are open. Thus we have an open cover of $X$ given by $$\{ (x,1] | x \in (0,1) \} \cup \{0\},$$ which has no finite subcover.

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  • $\begingroup$ Maybe an even easier way to see it is that the image $f(X) = \{-1\} \cup (0,1]$ is not compact in $Y$, so it's impossible to have $f$ continuous and $X$ compact. $\endgroup$ – Nate Eldredge Mar 9 '15 at 1:51
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(fubini has a nice answer, I'm just giving a concrete example I thought in the meantime while being offline.)

Take $X=Y=[0,1]$ and $f=\chi_{\{0\}}$, the indicator function of the singleton $\{0\}$. That is, $$ f(x)= \begin{cases} 1 & x=0 \\ 0 & 0<x\leq 1.\end{cases} $$ Then $f$ is Borel, and if you make it continuous, then $\{0\}$ must be open. Then $$ \{0\}, (\tfrac{1}{n},1] $$ with $n\in\mathbb{N}$, is an open cover of $[0,1]$ without a finite subcover.

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  • $\begingroup$ Thanks, didn't see this answer earlier when I posted my own, which is very similair. However technically your example is not a counterexample, since I did not say that the new topology must have more open sets (so $(1/n,1]$ does not have to be open). That is why I defined the function as $f(x)=x$ for $x \neq 0$ instead of just $0$. $\endgroup$ – Mathemagician Mar 9 '15 at 6:56
  • $\begingroup$ @Mathemagician Thanks. Nevertheless, if you are willing to change the topology in a more general way (than simply enlarging it), I guess some restrictrion should me made on this. I expect to see the new question! $\endgroup$ – Pedro Sánchez Terraf Mar 9 '15 at 10:54
  • $\begingroup$ The new question has just been posted :) math.stackexchange.com/questions/1196457/… $\endgroup$ – Mathemagician Mar 19 '15 at 3:07

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