# 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

It is open whether the continuum hypothesis for an infinite set $E$ implies the well-orderability of $E$. 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 $\mathcal P(E)$.)
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.
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