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.
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).