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Consider a compact separable Hausdorff space $X$ endowed with a finite Radon measure $\mu$ of full support and a continuous measure-preserving ergodic transformation $T$. Is there a dense orbit for the transformation?

Recall that a topological space is separable if it admits a dense countable subset. It is second countable if there is a countable family of open sets such that any open sets is the union of members of this family. A Hausdorff compact space is metrizable if and only if it is second countable.

The result is well known if the space is second countable (or metrizable). An example of compact group that is separable but not second countable is given by a product of uncountably (cardinal of the continuum) many circles. It is easily checked that ergodic translations on such a group have a dense orbit.

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  • $\begingroup$ are you assuming $\mu$ is Borel? $\endgroup$
    – Asaf
    May 20, 2017 at 18:01
  • $\begingroup$ Yes. Radon means Borel together with inner regularity. $\endgroup$
    – coudy
    May 20, 2017 at 18:02
  • $\begingroup$ So why can't you take the usual construction of $T$-generic points (due to Furstenberg)? Uryshon's lemma will give you countably-many functions on $C(X)$ through which you can deduce that the orbit closure of a generic point contains each pt from the dense subset, and hence the orbit closure is the whole space. $\endgroup$
    – Asaf
    May 20, 2017 at 18:06
  • $\begingroup$ @Asaf I think that you need second countability to use this approach. Separability is not enough. $\endgroup$
    – coudy
    May 20, 2017 at 18:10
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    $\begingroup$ @Asaf Indeed, separability + first countability is enough. Note that for a compact group, first countability implies second countability (Birkhoff Kakutani). To my surprise, there exist compact, first countable, separable spaces that are not second countable, e.g. the Helly space. So this is an improvement over second countability. $\endgroup$
    – coudy
    May 24, 2017 at 20:26

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