Let us say that two sets $A$ and $B$ are comparable if there is an injection from $A$ to $B$ or there is an injection from $B$ to $A$. Obviously, in a model of ${\rm ZFC}$ any two sets are comparable by comparing their cardinalities. But this is not necessarily the case in a model of ${\rm ZF}$. For example in Cohen's classical symmetric submodel showing the independence of the Axiom of Choice from ${\rm ZF}$, there is a set of reals which is incomparable with $\omega$.

Is there a known construction of a model of ${\rm ZF}$ in which the reals are not well-ordered, but any two sets of reals are comparable?

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    $\begingroup$ If every set of reals is either countable or has a perfect subset, then any two sets of reals are comparable (by Cantor-Schroeder-Bernstein) but there cannot be a wellorder of the reals. $\endgroup$ Feb 10, 2018 at 1:00

1 Answer 1


Yes. The perfect set property will ensure every set of reals is countable or has size continuum.

Solovay, R.M., A model of set-theory in which every set of reals is Lebesgue measurable, Ann. Math. (2) 92, 1-56 (1970). ZBL0207.00905.

In the paper Solovay proved that from the consistency of an inaccessible cardinal it is consistent that every set of reals is countable or has the perfect set property. Of course, there you need the inaccessible cardinal if you want Dependent Choice. If you're willing to forego that, you can forego the inaccessible cardinal, as shown by Truss.

Truss, John, Models of set theory containing many perfect sets, Ann. Math. Logic 7, 197-219 (1974). ZBL0302.02024.

There he shows that you can always push this up in the sense that you can have that every set of reals is well-orderable or has a perfect subset and you can pretty much have no limit on what "well-orderable" means in that case. (Although I believe Solovay also does that at the end of his paper, but I do not recall at the moment.)

  • $\begingroup$ Thanks, Asaf! This is very nice, somehow I didn't think about the perfect set property! But I realized that what I really need is that any two subsets of $H_{\omega_1}$ are comparable, and the Solovay model doesn't have this property. So I am still stuck. $\endgroup$ Feb 10, 2018 at 14:18
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    $\begingroup$ Well. That's harder. Since that implies $\aleph_1\leq2^{\aleph_0}$. So if anything, then Truss' model might have a chance where every set of reals can be well ordered or has a perfect subset. $\endgroup$
    – Asaf Karagila
    Feb 10, 2018 at 14:23
  • $\begingroup$ How about the easy symmetric models? Are there obvious incomparable sets in those? Say, force with ${\rm Add}(\omega,\omega_2)$ and symmetric names are fixed by automorphisms fixing initial segments. $\endgroup$ Feb 10, 2018 at 15:14
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    $\begingroup$ I'm not entirely sure. I would need you to describe this extension more precisely, but if it's anything like what I have in mind, it's probably the case that $\omega_2$ is incomparable with the set of the generic reals (or some other "relatively canonical set of reals"). $\endgroup$
    – Asaf Karagila
    Feb 10, 2018 at 18:33

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