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    $\begingroup$ If every set of reals is Lebesgue measurable then $\omega_1 \nleq 2^{\omega}$, but then you can partition $2^{\omega}$, or rather $\mathcal{P}(\omega\times\omega)$, into $\omega_1+2^{\omega} > 2^{\omega}$ pieces by putting two wellorderings of $\omega$ in the same piece iff they have the same order type, and all non-wellorderings into singleton pieces. $\endgroup$ Commented Apr 29, 2010 at 5:07
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    $\begingroup$ @Anton: This is not at all obvious, it's a theorem of Jean Raisonnier from the 1980's. $\endgroup$ Commented Apr 29, 2010 at 14:34
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    $\begingroup$ The statement that either there is a nonmeasurable set or R has a surjection to a larger set, is Sierpinski: Sur une proposition qui ..., Fundamenta Math, 34(1947), 157-162. It is also in my book (with Vilmos Totik) Problems and Theorems in Classical Set Theory, Springer, 2006, page 127. $\endgroup$ Commented Jul 2, 2010 at 15:57
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    $\begingroup$ Did you set up the account just to post this? $\endgroup$ Commented Oct 10, 2013 at 16:19
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    $\begingroup$ As already observed in a comment here, this argument shows only that the definition of "more" (in the sense of existence of an injection and lack of a bijection) does not make much sense without choice. Something which is not really surprising given that many sets have no cardinality (in the sense of cardinal numbers) without choice. If looked at it this way, the argument only boils down to the fact that $\mathbb R$ is such a set if all sets are Lebesgue measurable. $\endgroup$ Commented Sep 26, 2020 at 18:44