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Suppose $f$ is a one-to-one function. $\forall{n}: F(A_1)=B_1, F(A_2)=B_2, F(A_3)=B_3, \ldots, F(A_n)=B_n$ (I'm talking about a countable amount of infinite sets, A_1, A_2, A_3...)

$\forall{i} \neq n, A_i \cap A_n = \emptyset$

$\forall{i} \neq n, B_i \cap B_n = \emptyset$

Is there a way to prove, without using the axiom of choice or the axiom of countable choice, that the union of $A_1, A_2, A_3, \ldots, A_n$ is equipotent to the union of $B_1, B_2, B_3, \ldots, B_n$ ?

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Yes, this is true even without any choice.

Since $f(A_i)=B_i$, we can see that $f$ is surjective $\bigcup_i A_i \to \bigcup_i B_i$, and as it is explicitly assumed to be injective, it is actually a bijection between the two unions. Therefore, by definition, they are equipotent.

You don't need the disjointness assumptions.

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  • $\begingroup$ "You don't need the disjointness assumptions." Why not. Suppose B_1=B_2, then F(A 1 ) will hit B_1 and F(A 2 ) will hit B_1 and so the unions will not be equipotent. $\endgroup$ – Yitzchak Shmalo Feb 1 '16 at 17:25
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    $\begingroup$ @YitzchakShmalo But since $f$ is one-to-one, if $B_1=B_2$ then $A_1=A_2$. $\endgroup$ – Noah Schweber Feb 1 '16 at 17:58
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    $\begingroup$ Also, note that no countability assumption is needed either: if $f$ is injective and for each $i\in I$, we have $f(A_i)=B_i$, then $f(\bigcup A_i)=\bigcup B_i$. $\endgroup$ – Noah Schweber Feb 1 '16 at 18:10
  • $\begingroup$ @Noah Schweber You are right. I was thinking about a case in which we have a countable amount of different functions f1, f2, f3... mapping A_1 to B_1 and so on. Sorry. $\endgroup$ – Yitzchak Shmalo Feb 2 '16 at 1:19

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