Why is AC needed for $|\bigcup X_i|=|\bigcup Y_i|$, $\forall i$ $|X_i|=|Y_i|$, $\{X_i\}_{ i\in I}$, $\{Y_i\}_{i\in I}$ pairwise disjoint?

On Page 60, Set Theory, Jech(2006),

5.9 If $\{X_i : i \in I\}$ and $\{Y_i : i \in I\}$ are two disjoint families such that $|X_i| = |Y_i|$ for each $i \in I$, then $|\cup_{i \in I}X_i| = |\cup_{i \in I}Y_i|$ [Use AC]

Here's how far I goes:

$|X_i| = |Y_i|$ implies there exists a bijective function $f_i: X_i \to Y_i$ for each $i \in I$ ex ante. Since $\{X_i : i \in I\}$ is a disjoint family, for each $x \in \cup_{i \in I}X_i$, there exists exactly one $i \in I$, such that $x \in X_i$. So we could define a bijective function $f:\cup_{i \in I}X_i \to \cup_{i \in I}Y_i$, by $f(x)=f_i(x)$, if $x \in X_i$.

I don't see any usefulness of AC in problem 5.9, as opposed to problem 5.10, in which $|\cup_{i \in I}X_i| = |\cup_{i \in I}Y_i|$ is replaced by $|\prod_{i \in I}X_i| = |\prod_{i \in I}Y_i|$. The reason is that without AC, the cardinality of a cartesan product of non-empty sets is arbitary.

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Try to give meaningful titles. 'Why Bother Axiom of Choice' tells me nothing about the question inside... –  Asaf Karagila Dec 1 '12 at 11:30
@AsafKaragila: No problem. I should have thought of that. –  Metta World Peace Dec 1 '12 at 11:35
The title actually made me think the content is more philosophical in nature (e.g. why do we need the axiom of choice, or why do we bother to mention/verify when it is used). To these questions I have some long answers posted already and I was actually setting my mind to search for them in order to point the duplicate... :-) –  Asaf Karagila Dec 1 '12 at 12:05

You have to choose bijections for every $i$. It is consistent that there is a family $\{P_i\mid i\in\omega\}$ of disjoint pairs which does not have a choice function on any infinite subfamily.

One can show (quite easily too) that $\bigcup P_n$ is uncountable and in fact Dedekind-finite.

However $|P_n|=|\{2n,2n+1\}|$ whereas $|\bigcup P_n|\neq|\bigcup_{n\in\omega}\{2n,2n+1\}|=|\omega|=\aleph_0$.

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Russell sets are so wonderful. –  Asaf Karagila Dec 1 '12 at 11:38
Thank you for counterexample and the little excercise which I have to spend some time to figure out why. They are so nice. –  Metta World Peace Dec 1 '12 at 12:00
@Metta: I actually wrote something incorrect. It is consistent to have partial choice in the sense that there is an infinite subfamily which do have a choice function, in which case the union of the pairs is Dedekind-infinite, but the corrected version is correct, if no infinite family of pairs have a choice function then the union is Dedekind-finite. –  Asaf Karagila Dec 1 '12 at 12:01

For each $i\in I$ there are in general many bijections from $X_i$ to $Y_i$, so you’re using the axiom of choice when you pick a specific bijection $f_i$ for each $i\in I$.

Specifically, for each $i\in I$ let $B_i$ be the set of bijections from $X_i$ to $Y_i$. Then $\{B_i:i\in I\}$ is a non-empty family of non-empty sets, and you want to pick a set $\{f_i:i\in I\}$ such that $f_i\in B_i$ for each $i\in I$. In order to do that with no further information, you need the axiom of choice.

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Thanks, I got it. –  Metta World Peace Dec 1 '12 at 11:18
@MettaWorldPeace: You’re welcome. (That kind of use of AC is easy to overlook, especially when you’re just learning.) –  Brian M. Scott Dec 1 '12 at 11:21