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Assume that the cardinality of the union of two sets is continuum. How to prove that at least one of the sets has the cardinality of a continuum?

I suppose that it's possible to cope with it, using the operations with cardinals (for example, something like $\mathfrak{c}+\mathfrak{c}=\mathfrak{c}$), but i have no meaningful ideas.

Could you give me a hint, please?

Thank you in advance.

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    $\begingroup$ Try to prove by a contradiction argument. $\endgroup$ – DiegoMath Oct 30 '14 at 21:46
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    $\begingroup$ Are you assuming choice or not? With choice, there's a dog-simple argument just based on understanding how cardinality behaves under addition. $\endgroup$ – Steven Stadnicki Oct 30 '14 at 21:55
  • $\begingroup$ And it doesn't sound like something that is even necessarily true without choice. $\endgroup$ – Henning Makholm Oct 30 '14 at 22:11
  • $\begingroup$ @HenningMakholm Since we're talking about a union of a finite number of sets, it's at least possible that choice isn't necessary (or that very limited choice is), but I don't know choiceless arguments well enough. $\endgroup$ – Steven Stadnicki Oct 30 '14 at 22:17
  • $\begingroup$ @Henning: It is consistent without the axiom of choice that this is false. $\endgroup$ – Asaf Karagila Oct 31 '14 at 5:07
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$Card ( A\cup B ) = \max \left(Card(A), Card(B)\right)$ for ordinals greater or equal to $\omega$. For finite ordinals (that is, if $A$ and $B$ are finite) this is obviously false as $Card ( A\cup B ) + Card ( A\cap B ) = Card(A) + Card(B)$ in this case.

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If none of the two sets have cardinality bigger than $\mathbb{N}$, than both of the sets are countable. But the union of two countable sets is again countable. This leads to a contradiction.

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  • $\begingroup$ The problem is not about uncountable cardinality, but about the cardinality of the continuum $\endgroup$ – Hagen von Eitzen Oct 30 '14 at 21:52
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    $\begingroup$ Right, I was assuming the continuum hypothesis... $\endgroup$ – Jef L Oct 30 '14 at 21:53

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