I want to prove that -
(Axiom of Choice) Let $I$ be a set and for each $\alpha\in I$ let $X_\alpha$ be a non-empty set. Suppose that $X_\alpha \cap X_\beta=\emptyset$ for all $\alpha\ne\beta$. Then there is a set $Y$ such that $|Y\cap X_\alpha|=1\ \forall \alpha\in I$.
Here I take $A=\bigcup_{\alpha\in I} X_\alpha$ and $B=I$ and define $g:A\to B$ be $g(x)=\alpha$ if $x_\alpha\in X_\alpha$. This map $g$ is well defined from the fact that $X_\alpha\cap X_\beta=\emptyset$. Again $g$ is onto because $X_\alpha\ne \emptyset \ \forall \alpha\in I$.
So by our given hypothesis $\exists$ an injective map $f:I\to\bigcup_{\alpha\in I}X_\alpha$. But it doesn't imply that $f(\alpha)\in X_\alpha\ \forall \alpha\in I$.
Can anyone help me to complete the proof?
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1The statement should maybe be that $f$ obeys $f \circ g = \textrm{id}_A$ as well, instead of the existence of a 1-1 $f$. Or use that comparability of cardinals implies AC. See e.g. this and its answers – Henno Brandsma Sep 13 '20 at 17:21
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1You are trying to prove that the Partition Principle implies the Axiom of Choice; this is not known to be true. – Brian M. Scott Sep 13 '20 at 17:57
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2@Brian: Thanks for the link. One may also be interested in a more recent post, http://karagila.org/2020/countable-sets-of-reals/ – Asaf Karagila Sep 13 '20 at 18:04
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Trichotomy of cardinals implies AC, see here or here e.g.
Your statement implies that trichotomy, I think. So the proof is not as direct as you'd like...
Henno Brandsma
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