This is something I couldn't figure out while answering this question. I know that $\aleph_1\lt2^\mathfrak c$ and $\mathfrak c\lt2^\mathfrak c$, whence it follows that $\aleph_1\cdot\mathfrak c\le2^\mathfrak c$, but I didn't see how to prove the strict inequality $\aleph_1\cdot\mathfrak c\lt2^\mathfrak c$ without the axiom of choice.
Is $\aleph_1\cdot\mathfrak c=2^\mathfrak c$ consistent with ZF?
Best Answer
Not quite literally.
Since there is a surjection from $\Bbb R$ onto $\omega_1$, there is one from $\Bbb R$ onto $\omega_1\times\Bbb R$.
In particular, $\omega_1\times\Bbb R$ cannot be in bijection with $\mathcal P(\Bbb R)$, as the latter has no surjection from $\Bbb R$ itself.