17
$\begingroup$

Let us recall that the Axiom of Countable Choice (denoted by ACC) says that the countable product $\prod_{n\in\omega}X_n$ of nonempty sets $X_n$ is nonempty.

It is easy to see that ACC implies that for any sequence of meager sets $(X_n)_{n\in\omega}$ in a Polish space $X$ the union $\bigcup_{n\in\omega}X_n$ is meager in $X$. Let us denote the latter statement by (UMM), abbreviated from "union of meager is meager".

So, (ACC)$\Rightarrow$(UMM).

On the other hand, it is consistent with (ZF) that the real line can be equal to the union of a countable family of countable sets, in which case (UMM) does not hold. This means that (UMM) cannot be proved in (ZF) alone.

Problem 1. What is the place of (UMM) among other weaker versions of AC?

Denote by (UCC) the statement: the union of a countable family of countable sets is countable.

Problem 2. Does (UMM) imply (UCC)?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Good questions. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 13, 2020 at 7:01
  • $\begingroup$ And what about (UCM), the union of a countable family of countable sets is meager? Does (UCM) imply (UCC)? $\endgroup$
    – bof
    Commented Sep 13, 2020 at 7:26
  • $\begingroup$ @bof: I can't say if the answer is positive or not, but at least in Truss' model, the countable union of countable sets of reals is countable, but $\omega_1$ is singular. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 13, 2020 at 8:48
  • 2
    $\begingroup$ Since (UMM) is only about Polish spaces, it should be preserved from the ground model to symmetric extensions that introduce no new sets of low rank. So I'd be inclined to try violating (UCC) among sets of very high rank and thus answer Problem 2 negatively. $\endgroup$
    – Andreas Blass
    Commented Sep 13, 2020 at 12:41
  • 1
    $\begingroup$ @Andreas: That's a good point. If $\Bbb R$ can be well-ordered, but some strange sets exist above it, it would be a counterexample. I suppose that the right question, then, is restricting these choice principles to sets of reals. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 13, 2020 at 20:06

1 Answer 1

Reset to default
1
$\begingroup$

What is your definition of meager in (ZF)? Being a countable union of nowhere dense sets? In that case, the given argument about the failure of (UMM) in (ZF) is not clear: It is not obvious (to me) that the real line cannot be be meager in (ZF). Recall that the general category theorem of Baire is in (ZF) equivalent to countable dependent choices (DC); perhaps a proof that the real line is not meager already requires more than ZF?

Improve this answer
$\endgroup$
5
  • 4
    $\begingroup$ The Baire category theorem for separable spaces is provable in ZF since the dependent choices one makes can be restricted to a countable dense subset, which is wellordered at the outset so that one can choose canonically. $\endgroup$
    – Gabe Goldberg
    Commented Sep 13, 2020 at 15:54
  • 1
    $\begingroup$ @GabeGoldberg But wouldn't that prove UMM in ZF? $\endgroup$
    – D.S. Lipham
    Commented Sep 13, 2020 at 16:05
  • 1
    $\begingroup$ @D.S. Lipham. Quite the opposite. From GabeGoldberg's (correct) observation, it follows even in ZF that the real line is not meager. Hence the argument in the original question shows indeed that UMM is unprovable in ZF. $\endgroup$
    – Martin Väth
    Commented Sep 13, 2020 at 16:19
  • 1
    $\begingroup$ Martin, ZF proves that $\Bbb R$ is not meager, it does not prove that it is a countable union of countable sets. Therefore it is consistent that the countable union of meager sets is not meager. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 13, 2020 at 20:03
  • $\begingroup$ Asaf, yes that's what I meant exactly: It is consistent (with ZF) that UMM is unprovable. That's why I wrote "quite the opposite" (to the assertion that UMM is provable in ZF). $\endgroup$
    – Martin Väth
    Commented Sep 14, 2020 at 16:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .