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 $B_1, B_2, B_3, \ldots, B_n$ ?