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In an answer to this question it was shown that for each finite $k \geq 1$ it is possible to partition $\mathbb{R}$ into exactly $k$ parts, each of which is additively closed (i.e., is a subsemigroup of $(\mathbb{R},+)$). The argument given there is an induction argument relying on the fact that $\mathbb{R} \cong \mathbb{R}^2$ as groups, and so the partitions obtained for $k \geq 4$ are unlikely to be very nice.

My question is: For a given $k \geq 4$, is it possible to partition $\mathbb{R}$ into $k$ Borel sets, each of which is closed under addition? If so, I'd also be interested to know how complicated the Borel sets are, and in particular if that complexity changes with $k$.

Chris Eagle
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    Possibly related: https://math.stackexchange.com/questions/3048432/partition-of-positive-reals-with-each-part-closed-under-addition-without-choice – Asaf Karagila Aug 24 '20 at 17:44
  • I bet there are very strong restrictions on what a closed under addition set with positive measure look like (I would guess only the obvious ones) and then that would be enough to say you can't do more than 3. Admittedly this is just a guess. (edit of earlier comment since I forgot monoids have identity) –  Aug 24 '20 at 17:45

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Here is a proof that you can not have a partition into four Borel sets with the desired properties.

First, if an additively closed set, $A$, contains an interval of positive numbers then it contains $(x,\infty)$ for some $x$ since there is an $n$ where $(na,nb)$ overlaps with $a+(na,nb)$. So, for large enough $n$, you get that $(na,\infty)$ is a subset of $A$. There is a corresponding statement for containing an interval of negative numbers to get a $(-\infty,x)$ subset.

A consequence is that if $A$ contains both a negative interval and a postive interval then it is $\mathbb{R}$. If we have a partition $A,B,C$ where $A$ has a positive interval and $B$ has a negative interval then the only other set $C$ can be is $\{0\}$ since any other additively closed set would intersect one of $(x,\infty)$ or $(-\infty,x)$. This means we can not partition $\mathbb{R}$ into $4$ additevly closed sets where one of the sets has a postive interval and one has a negative interval.

Now we need to show any Borel paritions we end up with a set that contain $(x,\infty)$ and a set that contains $(-\infty,y)$. Any Borel set is measurable so at least one of these additively closed sets has positive measure. It is a fact that if $A$ has positive measure then $A+A$ contains an interval. By the previous paragraph we get that a partition into additively closed Borel sets has at most cardinality three.

It turns out that in ZF it is consistent that every set of reals is measurable. So, there are models of ZF where there is no partition into four additively closed subsets.

C.F.G
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