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Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice?
Showing that $\mathbb{R}$ and $\mathbb{R}\backslash\mathbb{Q}$ are equinumerous using Cantor-Bernstein

What can be said about $|\mathbb R\setminus\mathbb Q|$ under weaker versions of AC or alternative axioms (Martin's axiom, for example)? thank you.

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  • Related: http://math.stackexchange.com/a/72140/ and http://math.stackexchange.com/questions/105990/showing-that-mathbbr-and-mathbbr-backslash-mathbbq-are-equinumerous. – Jonas Meyer Dec 09 '12 at 04:50
  • The second link given by @Jonas contains a complete answer: one can use the Cantor-Schröder-Bernstein theorem to show that $|\Bbb R\setminus\Bbb Q|=|\Bbb R|$ in ZF. – Brian M. Scott Dec 09 '12 at 04:57

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