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This is exercise 5.9 Thomas Jech set theory. Let $\{X_i\}_{i \in I}$ and $\{Y_i\}_{i \in I}$ are disjoint families such that $|X_i| = |Y_i|$. Prove that $$\bigg|\bigcup_{i \in I} X_i\bigg| = \bigg|\bigcup_{i \in I} Y_i\bigg|$$ by Axiom of choice.

My attempt: I'm pretty sure it's about the validity of the set $$\{f_i:\text{some bijection between } X_i \text{ and } Y_i \text{ with } i \in I\}$$

To formalize this, for each $i \in I$, the sub-set $$S = \{f : \text{ bijection between } X_i \text{ and } Y_i\} \subset Y_i^{X_i}$$ must not be empty due to $|X_i| = |Y_i|$. Thus, consider $P(S)$ and use the choice function: $$g_i: P(S) \rightarrow \bigcup P(S)$$ on $S$ itself to pick uniquely $g_i(S)$, thus forming a function $i \rightarrow g_i(S)$. Is this the author intent?

Martin Sleziak
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MathInquirer
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1 Answers1

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The last part you wrote about $P(S)$ doesn't make sense to me, but the role of choice is that we use it to choose a bijection $f_i: X_i\to Y_i$ for each $i\in I$. Then $\bigcup_i f_i$ is a bijection $\bigcup_i X_i\to \bigcup_i Y_i$

spaceisdarkgreen
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  • did you spot anything invalid – MathInquirer Dec 25 '24 at 10:56
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    @MathInquirer Your argument doesn't make sense to me, though it seems to have the correct goal of getting a sequence of bijections $f_i:X_i\to Y_i.$ The way to formalize it is to just note that ${S_i: i \in I}$ where $S_i$ is the set of bijections $X_i\to Y_i$ is a family of nonempty sets, so has a choice function $f,$ and then for $i\in I$ let $f_i = f(S_i).$ – spaceisdarkgreen Dec 25 '24 at 18:40