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It is an easy exercise using transfinite recursion to prove the following (in ZFC):

There exists sets $S,T$ that partition $\mathbb{R}_{>0}$ such that each of $S$ and $T$ is closed under addition.

It is equally easy, using a transcendence base of $\mathbb{R}$ over $\mathbb{Q}$, to prove the following generalization:

For any cardinal $k < \#(\mathbb{R})$, it is possible to partition $\mathbb{R}_{>0}$ into $k$ parts each of which is closed under addition.

The problem, of course, is that the transfinite recursion depends on AC. My question is, can the first weak version be proven in ZF (no choice)? If so, I would suspect that ZF can prove:

For any finite $k$, it is possible to partition $\mathbb{R}_{>0}$ into $k$ parts each of which is closed under addition.

I am also curious to know whether ZF can prove any of the following:

$\mathbb{R}_{>0}$ can be partitioned into countably many parts each of which is closed under addition.

$\mathbb{R}_{>0}$ can be partitioned into uncountably many parts each of which is closed under addition.

I am unable to think of any algebraic way to split the positive reals in the desired fashion. Clearly, one of them must be uncountable, but I do not even see an obvious uncountable subset that is closed under addition. I would guess that ZF cannot prove any of them, but maybe that is just because I cannot see how to do it.

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    $\begingroup$ I have a feeling this was discussed here before. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 21, 2018 at 11:50
  • $\begingroup$ I couldn't find it. Please link any that you find? =) $\endgroup$
    – user21820
    Commented Dec 21, 2018 at 11:51
  • $\begingroup$ Why do you need a transcendence basis? Wouldn't you need a Hamel basis? $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 21, 2018 at 11:51
  • $\begingroup$ Yes algebraic independence is unnecessary; I was just whacking with the usual hammers I wield. $\endgroup$
    – user21820
    Commented Dec 21, 2018 at 11:52
  • $\begingroup$ I am guessing that math.stackexchange.com/a/360828/622 can be extended to countably many as well. Or in general to "less than the additivity of null/meager ideals". $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 21, 2018 at 11:55

1 Answer 1

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Suppose that $(A_i\mid i<\alpha)$ is such partition. I claim that $\alpha$ is at least the size of the additivity of the null ideal, or one of the $A_i$'s is not Lebesgue measurable.

To see that this is indeed the case, recall that $A+A=\{a+b\mid a,b\in A\}$ contains an interval for any $A$ of positive measure. If each $A_i$ is null, and $\alpha$ is less than the additivity of the null ideal, then $\bigcup A_i$ is null, in which case we have a contradiction. At the same time, if one of the $A_i$'s got positive measure, then $A_i+A_i$ contains an interval, which is impossible.

The same can be said about the meager ideal and the Baire Property.

(The above is based on https://math.stackexchange.com/a/360828/622)

In particular, at least under $\sf ZF+DC$, it is consistent that there is no finite or countable partition of this sort (e.g. if all sets are Lebesgue measurable or have the Baire Property).

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  • $\begingroup$ To clarify, what assumptions do you need over ZF to prove that there is a model of ZF plus DC plus no countable partition of this sort? Is Con(ZF) enough, or do we need an inaccessible cardinal? Also, do you know the answer for whether ZF proves an uncountable partition of this sort exists? $\endgroup$
    – user21820
    Commented Dec 21, 2018 at 12:41
  • $\begingroup$ Lebesgue = inaccessible; Baire = ZF. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 21, 2018 at 12:42
  • $\begingroup$ Holo has given the obvious simple answer for the uncountable partition. I was blur. So I guess your answer completely addresses the non-trivial part of my question. Thanks! =) $\endgroup$
    – user21820
    Commented Dec 21, 2018 at 14:54
  • $\begingroup$ Interestingly, $\Bbb{R/Q}$ might have strictly larger cardinality in models of ZF+DC+BP/LM, which raises the interesting question about exactly $2^{\aleph_0}$ sets, and requiring them all to have size continuum too. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 21, 2018 at 18:25
  • $\begingroup$ Hmm interesting. ZFC can construct a partition of size $\#(\mathbb{R})$ each of whose parts has size $\#(\mathbb{R})$, by dovetailing. But can ZF+DC do it even if we drop the size requirement, and simply want a partition of size $\#(\mathbb{R})$? $\endgroup$
    – user21820
    Commented Dec 22, 2018 at 7:42

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