Let us denote by **ACC** the axiom of countable choice, namely the assertion that the product of countably many non-empty sets is non-empty, and denote by **UCC** the assertion that a countable union of countable sets is countable.

UCC is a simple theorem of ZF+ACC.

*Proof* Suppose for every $i\in\omega$ we have $X_i$ a countable set, and $X_i\cap X_j=\varnothing$ for $j\neq i$.

Since $X_i$ is countable $O_i=\{f\colon X_i\to\omega\mid f\ \text{ injective}\}$ is non-empty. We can choose $f_i\in O_i$ by the axiom of countable choice, and define: $$F\colon\bigcup X_i\to \omega\times\omega\colon\qquad x\mapsto\langle n,f_n(x)\rangle$$ Where $n$ is the unique $n\in\omega$ such that $x\in X_n$.

The Cantor pairing function shows that $\omega\times\omega$ is countable and we are done.

Is the opposite assertion is true, namely ZF+UCC implies ACC? If the answer is negative, does that imply at least some other weaker form of choice?

As noted by Emil Jeřábek below, UCC implies the axiom of countable choice for countable sets (the latter abbreviated as CCF).

Digging through the paper mentioned by

*godelian*in the comments, I reached [1] in which Howard constructs a model of ZFA in which CCF holds and UCC does not, and by the transfer theorem of Pincus constructs this over ZF. Therefore we have: $$\text{ACC}\Rightarrow\text{UCC}\Rightarrow\text{CCF}$$ The first implication is irreversible in ZFA, by the comment of godelian, and the second irreversible in ZF by [1]. Both papers are two decades old, is there any known progress?

*Bibliography:*

- Howard, P.
**The axiom of choice for countable collections of countable sets does not imply the countable union theorem.***Notre Dame J. Formal Logic Volume 33, Number 2 (1992), 236-243.*

countablesets is nonempty. $\endgroup$ – Emil Jeřábek supports Monica Sep 7 '11 at 12:57