FALSE: countable unions of countable sets are countable is a theorem of ZF

FALSE. The statement

a union of countably many at most countable sets is at most countable

is a theorem of ZF: it can be proved from the Zermelo-Fraenkel axioms with no appeal to any choice principle.

The claim is plausible because the proof looks like pure bookkeeping. One writes the elements of $A_n$ as $a_{n,0}, a_{n,1}, a_{n,2}, \dots$, reads off the array by diagonals, and every single step of that argument is elementary, with the countability of $\mathbb{N} \times \mathbb{N}$ ($\mathbb{N} \times \mathbb{N} \approx \mathbb{N}$) doing the real work and needing no choice. What is easy to miss is the very first move: writing the elements of $A_n$ as a list means choosing one enumeration of $A_n$, for every $n$ at once, out of the many that each $A_n$ admits. That is exactly the Axiom of Countable Choice (The Axiom of Countable Choice ($\mathrm{AC}_\omega$)), and Countable unions of at most countable sets, assuming $\mathrm{AC}_\omega$ flags it at the step where it is spent.

• What is and is not proved here. The refutation is a correct argument given the cited independence result, but that result is not proved in this library: the Feferman-Levy model is built by forcing, which is deferred. The honest reading is conditional, namely that $U$ is a theorem of ZF only if ZF is inconsistent. It is recorded this way deliberately rather than presented as fully derived, exactly as in FALSE: Zorn's lemma is a theorem of ZF.

• The correct reading of the true theorem. Countable unions of at most countable sets, assuming $\mathrm{AC}_\omega$ proves $U$ from ZF together with $\mathrm{AC}_\omega$ and is not weakened by this item; what this item says is that the choice assumption is doing real work and cannot be dropped. The library's habit of naming the exact step that spends a choice principle is what makes the difference visible.

• How strange the Feferman-Levy model is. In it (The Feferman-Levy model: the reals as a countable union of countable sets ‡) $\mathbb{R}$ is a countable union of countable sets, yet $\mathbb{R}$ is still uncountable, since $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874) is a ZF theorem. There is no contradiction: countably many countable sets can have an uncountable union when no single function enumerates them all simultaneously. What fails is not any statement about $\mathbb{R}$ but the ability to assemble the enumerations.

• The same phenomenon is why "a countable union of countable sets of reals" arguments in analysis, for instance in measure theory, are usually stated over ZFC or at least ZF plus $\mathrm{AC}_\omega$. The choice ledger is not a formality there either.