5
$\begingroup$

Possible Duplicate:
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.

  • 0
    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.2012-12-09

0 Answers 0