# Cardinality of $\mathbb R\setminus\mathbb Q$ without AC [duplicate]

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

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