# Axiom of Countable Choice and meager sets

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)?

• Good questions. – Asaf Karagila Sep 13 at 7:01
• And what about (UCM), the union of a countable family of countable sets is meager? Does (UCM) imply (UCC)? – bof Sep 13 at 7:26
• @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. – Asaf Karagila Sep 13 at 8:48
• 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. – Andreas Blass Sep 13 at 12:41
• @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. – Asaf Karagila Sep 13 at 20:06

• 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. – Asaf Karagila Sep 13 at 20:03