0
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ "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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.