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prove or disprove:

If two infinite sets $A$,$B$ have the same cardinality, then $A\cup B$ and $A$ have the same cardinality.

I even cannot make a judgement.

P.S: Can this be done without using cardinals? This concept has not been introduced in class yet.

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    For infinite cardinals $\kappa, \lambda,$ $\kappa + \lambda = \max\{\kappa + \lambda\}$.2012-09-19
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    @tomcuchta Do you mean $\max\{\kappa,\lambda\}$?2012-09-19
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    have you learned about bijective functions?2012-09-19
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    bijective makes sense~2012-09-19
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    I believe this requires the Axiom of Choice (or some fragment thereof.)2012-09-19
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    Recall that one definition of an infinite set $S$ is that there exists a set $S'\subset S$ for which a bijection exists between $S'$ and $S$. That should help.2012-09-19
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    Also, a very useful tool for constructing bijections is the Cantor–Bernstein–Schroeder theorem (https://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem)2012-09-19
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    @crf Assuming you mean "proper subset," this is not the usual definition of "infinite," although it is equivalent to it assuming the Axiom of Choice2012-09-19
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    @yunone Yes I do mean that, whoops! Too late to edit now :( Also irrelevant.2012-09-19
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    @TrevorWilson http://mathworld.wolfram.com/InfiniteSet.html disagrees. I'm no expert; are they wrong?2012-09-19
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    I would not say "wrong" because the two definitions are equivalent in ZFC. In this case I do think Wikipedia (https://en.wikipedia.org/wiki/Infinite_set) is a better resource than Wolfram.com, however.2012-09-19
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    "If the axiom of choice is also true, infinite sets are precisely the Dedekind-infinite sets [which are sets which can be put into bijection with some of their proper subsets]." Huh, learn something new every day. That axiom of choice sure does pop up a lot.2012-09-19

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