# Subsets of Real Numbers (Edited & Revised Version)

Question 1: Is it consistent with $\text{ZF}$ that only countable subsets of $\mathbb{R}$ are well-orderable?

Question 2: Is it consistent that for some $\lambda$, $\aleph_0 < \lambda < 2^{\aleph_0}$, only those subsets of $\mathbb{R}$ are well-orderable which have size $\leq \lambda$?

Question 3: Is it consistent with $\text{ZF}$ that there is a partition of $\mathbb{R}$ into two sets $A$ and $B$ such that $|A|=|B|<|\mathbb{R}|$?.

• My answer, was, it seems (and I agree with Emil) to a slightly different question. About what is provably true about $\{X\subseteq\Bbb R\mid X\text{ can be w.o.}\}$, and about its complement. Note, to your edit, that sets are the semantical objects in set theory. If $A$ is a set of reals in a model $M$ either it can or cannot be well-ordered, and the axiom of choice says nothing about it. If we want to ask whether or not every definable (with real parameters?) set of real numbers can be well-ordered, that's another question (whose answer is similar to mine), which admits consistency results. – Asaf Karagila Dec 1 '13 at 14:46
• @EmilJeřábek: I edited the question. Thanks for your notifications. – user42090 Dec 2 '13 at 11:42
• Several comments have been deleted from this thread; the full thread can be found here: tea.mathoverflow.net/discussion/1628/… – François G. Dorais Dec 3 '13 at 3:14
• – François G. Dorais Dec 3 '13 at 3:30
• My above comment refers to my previous answer and to the original version of the question. Of course that my current answer refers to the current version of the question. – Asaf Karagila Dec 3 '13 at 21:38

For the first question, yes. It is consistent to have that.

1. Solovay's model, or any model of $\sf AD$ for example.
2. Truss' models, which are similar to Solovay's model, only we start with a general limit cardinal (rather than an inaccessible). The result has that $\aleph_1$ is singular, but still cannot be embedded into $\Bbb R$.
3. The Feferman-Levy model where $\Bbb R$ is a countable union of countable sets. This again has a striking similarity to the previous two models with this and the Truss model generated by collapsing $\aleph_\omega$ being very different despite having very similar constructions.

For your second question it is also easily achievable for every $\lambda$ whose cofinality is not $\omega$. Simply start with a model where $2^{\aleph_0}=\lambda$, then add any infinite number of Cohen reals and make them a non-well orderable set. This is a simple variation of Cohen's first model which one can find in many places in the literature.

Since the Cohen forcing doesn't collapse cardinals, it is easy to show that any subset of the real numbers in the full generic extension has size $\leq\lambda$, and that in the symmetric extension itself not all the sets of reals can be well-ordered. The ground model reals witness a set of size $\lambda$.

Finally, the third question is an interesting one. But the answer is no. This is a nice application of the "Division by three" paper by Conway and Doyle. There they show that without the axiom of choice it holds that if $|A\times 3|=|B\times 3|$ then $|A|=|B|$. They remark that the claim is also true when replacing $3$ by $2$. Now we will have: $$|A|+|A|=|\Bbb R|=|\Bbb R|+|\Bbb R|\implies |A|=|\Bbb R|.$$

I will add the remark, though, that $\Bbb R$ can consistently be partitioned into two sets of strictly smaller cardinality. But they will not be equipotent.

• In the first paragraph, I think you should say that $B$ contains every subset of $\mathbb R$ which provably (in ZF) has size continuum. For example, ZF proves that the set of constructible reals is well-orderable. Yet in some models, that set has size continuum. The essential point is that a set cannot be simultaneously provably well-orderable and provably of size continuum, but it can be provably either one of these and (unprovably) just happen to be the other. – Andreas Blass Dec 1 '13 at 22:55
• Yes, you are right Andreas. In either case, however, $B$ has size $2^\frak c$ (just note that taking any subset containing $[0,1]$ does the trick). – Asaf Karagila Dec 1 '13 at 22:56
• I did not understand what St.Georg meant with his definition of $B$, and I do not see which set (or formula?) $B$ this answer refers to. Could you please add your definition of $B$ to your answer? – Goldstern Dec 1 '13 at 23:18
• @Goldstern: My understanding of the question, as I pointed out in the comments to the main question, was that $A$ and $B$ are definable sets of sets of real numbers which form a partition between well-orderable and non-well-orderable sets of real numbers. The question, in my eyes, was (despite being written otherwise) about the consistency of the cardinalities of these two sets in $\sf ZF$. In models where $\Bbb R$ cannot be well-ordered (perhaps with a "nice" well-order?), it's clear that $B$ is equipotent with $2^\Bbb R$; of course in $V=L$ it's clear that $B$ is empty. – Asaf Karagila Dec 1 '13 at 23:22