Is the axiom of countable choice need for proving the cantor-bernstein theorem?
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$\begingroup$ For the OP: The reason this question was downvoted and put on hold was that it is not appropriate for this site, which is for research mathematics. Math.stackexchange is a better site for such questions (I would not have answered, except that I forgot which site I was on. :P). As a side note, though, you should explain what you've tried and where you got stuck - in particular, for a question like this (where the standard proof goes through in ZF), explain what step(s) seem like they invoke choice. $\endgroup$– Noah SchweberCommented Feb 1, 2016 at 18:01
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1$\begingroup$ @Noah: It might be acceptable to ask this question on this site, I can certainly imagine one of the professors in my department running into me (or any other set theorist) in the hallway and asking this. Not like this, though. And some effort, like typing into the search box "Cantor Bernstein choice" and reading a bunch of the related results is more or less a prerequisite of asking anyway. $\endgroup$– Asaf Karagila ♦Commented Feb 2, 2016 at 0:42
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$\begingroup$ First, I did not believe that any specific part of any proofs I've seen involved choice, but there are many proofs which involve the axiom of countable choice in a way which I would not have realized. For example, the proof for the claim that the countable union of countable sets is countable. Second, if someone can ask whether the Axiom of Infinity is needed in Cantor-Bernstein theorem (on this site), I don't see why I cannot ask this question. Finally, I tried deleting this question but they did not let me. Anyway, sorry. (Also, I looked online for an answer and did not find any.) $\endgroup$– Yitzchak ShmaloCommented Feb 2, 2016 at 1:12
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$\begingroup$ I think this is a reasonable question, for the reasons @AsafKaragila gives, and because for me MO is about being able to consult outside one's own expertise -- this whole "everyone learns this in grad school" trope is in my opinion not so healthy. It could have been asked in a slightly better way, though $\endgroup$– Yemon ChoiCommented Feb 4, 2016 at 0:40
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It is not - the usual proof goes through in ZF without any choice. See https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem.
answered Feb 1, 2016 at 6:33