I am thinking about the Axiom of Choice and I am trying to understand the Axiom with some but a little progress. Many questions are arising in my head. So, I know that there exists a model of ZF set theory in which the set of real numbers, which is provably uncountable, is a countable union of countable sets.

**Question:** does there exist a model of ZF set theory for which there exists a collection $A_n$, $n\in\mathbb{N}$, of pairwise disjoint **two-element** sets such that their union is not countable?

Some thoughts. Let $A_n$, $n\in\mathbb{N}$, be a collection of pairwise disjoint two-element sets. Then for every $n\in\mathbb{N}$ there exists a bijection $f:\{1,2\}\to A_n$. But when we want to prove that $\bigcup_{n\in\mathbb{N}}A_n$ is countable, we have to choose a countable number of bijections $f_n:\{1,2\}\to A_n$, $n\in\mathbb{N}$, at once (simultaneously). After this we plainly define the bijection $f:\mathbb{N}\to\bigcup_{n\in\mathbb{N}}A_n$ by $f(1):=f_1(1)$, $f(2):=f_1(2)$, $f(3):=f_2(1)$, $f(4):= f_2(2)$, and so on. Rigorously, we write $f(k)=f_l(1)$ if $k=2l-1$ and $f(k)=f_l(2)$ if $k=2l$. Clearly, $f$ is a bijection and we are done. But without the Axiom of Countable Choice we can not choose $f_n$, $n\in\mathbb{N}$, simultaneously and the argument does not work.

It is worth mentioning that if $A_n$ are subsets of $\mathbb{R}$, then we **can** choose $f_n$, $n\in\mathbb{N}$, simultaneously.
Indeed, we can define $f_n(1):=\min A_n$ and $f_n(2):=\max A_n$, $n\in\mathbb{N}$,
and the natural proof given above works.
So if a counterexample exists, the sets $A_n$, $n\in\mathbb{N}$, have to be "abstract", say pairs of socks.

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