# Comparability implies well-orderability?

I am trying to prove a small proposition that got me completely stumped, and I cannot find a single counterexample.

(ZF) Suppose that $E$ is such that for every $A\subseteq\mathcal P(E)$ either $|E|\leq|A|$ or $|A|\leq|E|$, then $E$ can be well-ordered.

It is not a biconditional statement since we have models of ZF (e.g. Solovay's model) where $\omega$ serves as a counterexample to this, but I still make true or false of the above statement.

Is this result known, or known to be false?

If the above is indeed false, how about a stronger requirement:

(ZF) Suppose $E$ is such that the cardinalities below $|\mathcal P(E)|$ are linearly ordered, then $E$ can be well-ordered.

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"can it" must mean "cannot"? –  Lee Mosher Jul 20 '12 at 12:23
Yes. One of the problems of posting from a cellphone. I will edit that in a bit... Thanks! –  Asaf Karagila Jul 20 '12 at 12:34

Hi Asaf. It is open whether the continuum hypothesis for an infinite set $E$ implies the well-orderability of $X$. Of course, if $CH(E)$ holds, then the assumption in your (first) statement holds.

($CH(E)$ is the statement that any subset $A$ of $\mathcal P(E)$, either $A$ injects into $E$, or else $A$ is in bijection with $\mathcal P(E)$.)

This is a question that dates back to Ernst Specker, "Verallgemeinerte Kontinuumshypothese und Auswahlaxiom", Archiv der Mathematik 5 (1954), 332–337. There is a nice presentation in Akihiro Kanamori, David Pincus, "Does GCH imply AC locally?, in "Paul Erdős and his mathematics, II (Budapest, 1999)", Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, (2002), 413–426.

I do not know about the stronger statement you ask for. Part of the difficulty comes from the "bad" cardinal arithmetic we should have below $|{\mathcal P}(E)|$. For example, Specker proved that if CH holds for both $X$ and $\mathcal P(X)$, then $\mathcal P(X)$ is well-orderable.

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Hi Andres, thank you for the answer. I indeed had in mind the case where $CH(E)$ holds, but I had a hunch that would be too hard to attack. I suppose this is as good as it gets... I will leave the question open for a few days and accept your answer if no one else posts anything substantial until then. –  Asaf Karagila Jul 20 '12 at 15:05