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).

Problem. Is (UMM) true in ZF? Or (UMM)$\Rightarrow$(UCC)?

Here (UCC) denotes the statement: the union of a countable family of countable sets is countable.

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)?