7
$\begingroup$

A topological space $X$ is called (strong) Choquet if the player II has a winning strategy in the (strong) Choquet game.

It is known that a metrizable space $X$ is

$\bullet$ Choquet if and only if it contains a dense complete-metrizable subspace;

$\bullet$ strong Choquet if and only if it is complete-metrizable.

Since a topological group is complete-metrizable if and only if it contains a dense complete-metrizable subspace, we obtain the following characterization.

Theorem. A metrizable topological group is Choquet if and only if it is strong Choquet.

Corollary. A cosmic topological group is Choquet if and only if it is strong Choquet.

Let us recall that a regular topological space is cosmic if it has a countable network of the topology. It is easy to show that a topological group is second-countable if it is cosmic and Baire. That is why Corollary follows from Theorem.

Theorem and Corollary suggest the following intriguing

Problem. Is each Choquet topological group strong Choquet?

Improve this question
$\endgroup$
6
  • $\begingroup$ Hello, excuse me, do you have any reference for the characterization of the Choquet spaces in a metrizable space? I only knew that a Choquet space is productively Baire. Also if $X$ is a Choquet space without isolated points then $X$ contains a Cantor set. Thanks $\endgroup$
    – Gabriel Medina
    Commented Oct 16, 2019 at 2:35
  • 1
    $\begingroup$ Choquet spaces in the class of metrizable separable spaces were considered in Section 8.C of Kechris' book "Classical Descriptive Set Theory". $\endgroup$
    – Taras Banakh
    Commented Oct 17, 2019 at 3:35
  • $\begingroup$ Thanks Professor Taras, in the hypothesis of the Theorem (8.17) of Kechris's book says that " if $X$ is a nonempty separable metrizable space and $\tilde{X}$ a Polish space in which $X$ is dense. Then $X$ is Choquet iff $X$ is comeager in $\tilde{X}$ ". How can you conclude that the result is valid for any metrizable space? Thanks $\endgroup$
    – Gabriel Medina
    Commented Oct 17, 2019 at 3:41
  • $\begingroup$ Also for the part " $X$ is comeager in $\tilde{X}$ " we don't need the hypothesis about "separability". $\endgroup$
    – Gabriel Medina
    Commented Oct 17, 2019 at 3:45
  • 1
    $\begingroup$ I think one should just repeat the classical proof: first embed the space into a complete metric space, then given a winning strategy for the Non-empty player and using the presence of a metric, to construct a tree of open sets such that the union of open sets at each level of the tree is dense and intersection of open sets at each branch is a singleton. Then the union of those singletons is a dense $G_\delta$-set (so completely-metrizable). The same proof should work for any submetrizable space (i.e. a topological space admitting a continuous metric). $\endgroup$
    – Taras Banakh
    Commented Oct 17, 2019 at 3:51

0

Reset to default

You must log in to answer this question.