Cardinality of union of pairwise disjoint elements needs choice?

Question

If there is an indexed family $(i\mapsto A_i)_{i\in I}$ of pairwise disjoint sets $A_i$, why do we need choice to show that $$ \left|\textstyle{\bigcup_{i\in I}A_i}\right| = \sum_{i\in I}|A_i|? $$ It suffices to give a bijection between $\bigcup_{i\in I}A_i$ and $\{(i,x)\in I\times \bigcup_{i\in I}|A_i| \mid x\in |A_i|\}$. For each $a\in A$, there exists a unique $i\in I$ for which $a\in A_i$. Can't we then send $a\mapsto (i,a)$ to get such a bijection?

Answer 1

Yes, very much. Because when switching to $|A_i|$ you need to effectively choose canonical representatives from each equivalence class, and bijections from $A_i$ to that set. Neither of these processes is well-defined without the axiom of choice.

We can have the following situation:

There exists a set $S$ which can be partitioned into countably many pairs $S_n$, but $S$ does not have a countably infinite subset.

It follows that $|S|=|\bigcup S_n|=\sum |S_n|=\sum |\{0,1\}|=\aleph_0$. But that doesn't make sense, because $S$ doesn't have a countably infinite subset at all.