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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?

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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.

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  • $\begingroup$ You mean that I need it to create a map $a\mapsto (i,\bar{a})$ where $a\in A_i$ and $\bar{a}\in |A_i|$? I'm not exactly sure what you mean by equivalence class in this situation. $\endgroup$ – user132940 Mar 4 '14 at 8:11
  • $\begingroup$ Yes, and also a choice of bijections between $a$ and $\bar a$, for each $i$. $\endgroup$ – Asaf Karagila Mar 4 '14 at 9:06

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