6 Removed comma from reference.

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:

1. 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.
5 Corrected the link of [1]

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:

1. 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.
4 added 878 characters in body

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:

1. 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.
3 Corrected a minor LaTeX typo.