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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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marked as duplicate by Brian M. Scott, JSchlather, Alexander Gruber, Asaf Karagila, draks ... Dec 9 '12 at 10:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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 9 '12 at 4:57