Finite Unions of Equinumerous Sets

Question

Is the following statement provable in ZF?

Assume that $A_1$ and $A_2$ are two infinite sets. If $A_1$ is equinumerous with $B_1$ and $A_2$ is equinumerous with $B_2$, then $A_1 \cup A_2$ is equinumerous with $B_1 \cup B_2$.

Answer 1

Assuming that the union is disjoint, this is easy, since the union of the bijections is itself a bijection.

If not, then this is easy to see that if we assume that $|A_1\cup A_2|=|B_1\cup B_2|$, for any two pairs as in your assumption, then taking $A_1=A_2$ and $B_1=A_1\times\{0\}$ and $B_2=A_1\times\{1\}$, we get that $|A_1|=|A_1\times 2|=|A_1|+|A_1|$.

This does not follow from $\sf ZF$, for example, if there are infinite Dedekind-finite sets that is easy, and even if we assume $\sf DC$ or its uncountable versions, there can still be counterexamples to this principle, e.g. sets which are $\aleph_1$-amorphous may exist even if we assume $\sf DC$.

On the other hand, assuming $|A|=|A\times 2|$ for any infinite set $A$, we get that $|A_1\cup A_2|=|A_1\times\{0\}\cup A_2\times\{1\}|$, since we have:

$$|A_1\cup A_2|=|(A_1\cup A_2)\times 2|\geq|A_1\times\{0\}\cup A_2\times\{1\}|\geq|A_1\cup A_2|.$$

Which means that every union can be replaced by a disjoint union, and therefore the equality ensues. Finally, Sageev proved that $|A|=|A\times 2|$ for all infinite $A$ does not imply the axiom of choice, indeed, not even the axiom of countable choice.