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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?

user132940
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1 Answers1

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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.

Asaf Karagila
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  • 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. – user132940 Mar 04 '14 at 08:11
  • Yes, and also a choice of bijections between $a$ and $\bar a$, for each $i$. – Asaf Karagila Mar 04 '14 at 09:06
  • @Asaf Karagila Struggling to understand that. Why do you need choose canonical representatives for the $|A_i|$ when you can use the $A_i$ themselves and the identity bijection. And your set $S$ seems impossible if you mean a countably infinite number of pairs (or otherwise your final line doesn't make sense to me). – Martin Rattigan Mar 20 '24 at 21:38
  • @Martin: I don't understand. (I mean, I do, but I'm not sure you're pinpointing your difficulty, so we're going to do the Socratic method here for a moment.) Why would that help? The final line is an exact reference to the quoted fact. – Asaf Karagila Mar 20 '24 at 21:54
  • Let's take it one at a time. Your quoted fact would seem to imply ... finitely (and hence countably) many pairs... , otherwise the set of smallest elements in each pair would constitute a countably infinite subset of $S$. @user132940 specifies that his $A_i$ and $A_j$ $i\neq j$ are disjoint. If this applies in your final line then the sums must be finite and the final $=\aleph_0$ incorrect, But if it doesn't apply it would also appear not to be a counterexample. – Martin Rattigan Mar 20 '24 at 22:13
  • @Asaf Karagila Sorry - I forgot to address the above comment. – Martin Rattigan Mar 20 '24 at 23:14
  • @Martin: Finitely? Why? I used countable here to mean countably infinite. – Asaf Karagila Mar 20 '24 at 23:23
  • @Asaf Karagila But then, as I said, you can't say in the same sentence that there exists a set $S$ partitioned into a countably infinite set of pairs that doesn't have a countably infinite subset, because the function from the partition that maps each pair to it's smallest element is a bijection from the partition into $S$ whose converse domain is a countably infinite subset of $S$. – Martin Rattigan Mar 20 '24 at 23:36
  • @Martin: But that's the whole point. There is no smallest element. This is not a set of numbers or anything. It's just a crazy oddity that is consistent with the failure of choice. – Asaf Karagila Mar 20 '24 at 23:58
  • @Asaf Karagila OK got it. Without AC you can't prove the situation impossible. So that brings us to the second point. In user132940's example why do you need to choose canonical representatives for the $|A_i|$ when you can use the $A_i$ themselves and the identity bijection apparently bypassing any use of AC? – Martin Rattigan Mar 21 '24 at 00:15
  • @Asaf Karagila Rather than the identity bijection, I should have quoted user132940 and said $a$ maps to $(i,a)$. – Martin Rattigan Mar 21 '24 at 00:41
  • Let us continue this discussion in chat. – Martin Rattigan Mar 21 '24 at 01:48