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Question: In ZF (so AC does not necessarily hold) does the following claim hold?

$\aleph_0<2^{\aleph_0}$ implies $\aleph_1\le 2^{\aleph_0}$

This question arose to me when reading the top response in https://mathoverflow.net/questions/173090/how-much-of-gch-do-we-need-to-guarantee-well-ordering-of-continuum

The link, in short, seeks to prove that the continuum ($2^{\aleph_0}$) is well-orderable by assuming $\mathsf{CH}$ holds for $\aleph_1$ and $2^{\aleph_0}$ (notice AC is not being assumed and GCH is not being assumed)

Knowing that when AC doesn't hold, cardinal arithmetic becomes very tricky, I suspected that the above claim implicitly assumes $2^{\aleph_0}$ is well-orderable to being with. Am I wrong?

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Alberto Takase
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1 Answers1

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You cannot prove that $\aleph_1\leq2^{\aleph_0}$ without assuming some choice, but there is no need to assume the continuum can be well-ordered.

In some models (most famously Solovay's model and the Feferman-Levy model) it is not true that there is an injection from $\omega_1$ into the real numbers. On the other hands if you just destroy the well-orderability of the reals (e.g. Cohen's first model) without collapsing cardinals, then you will have an injection from the ground model witnessing that $\aleph_1\leq2^{\aleph_0}$.

(Note, by the way, that it is always the case that $|X|\leq2^{|X|}$, so the question amounts to asking whether or not $\aleph_1\leq2^{\aleph_0}$ without choice.)

You may also want to read about the different formulations of $\sf CH$ without the axiom of choice:

(and there are probably a few more)

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Asaf Karagila
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