$A_1,A_2,A_3$ forms a partition of $\mathbb N_{>0}$ and $a,b,c \in A_i \implies a+b+c \in A_i$ then at-least one of $A_i$ is closed under addition?

Question

If $A_1,A_2,A_3$ forms a partition of $\mathbb N_{>0}$ such that for any $i=1,2,3$ ; $a,b,c \in A_i \implies a+b+c \in A_i$ then is it true that at-least one of $A_i$ is closed under addition ? I can prove the result to be true for general semigroups when the no. of partitioning set is 2 but in this case the method I applied for the 2 case is not working and I am unable to find a counterexample . Please help

Answer 1

Say that a set $X$ is distinguished if $a+b+c\in X$ whenever $a,b,c\in X$. Obviously, you want each $A_i$ to be non-empty. Now any nonempty distinguished $X$ must be infinite (because for $x\in X$, we have $3x\in X, 3^2x \in X$ etc). So each $A_i$ is infinite.

One of the $A_i$, say $A_1$, contains $1$. Taking $a=b=1$ in the definition, we see that ( * ) : $c+2\in A_1$ for any $c\in A_1$. So $A_1$ contains all the odd numbers. If $A_1$ contained an even number $e$, using (*) again would contain all even numbers larger than $e$, so that ${\mathbb N}\setminus A_1$ would be finite, contradicting the infinitude of $A_2$ and $A_3$. We conclude that $A_1$ is exactly the set of odd natural numbers.

Then $A_2$ and $A_3$ consist of even numbers : there is a partition $B_1\cup B_2$ of ${\mathbb N}_{>0}$ such that $A_2=2B_2$, $A_3=2B_3$. And $B_2$ and $B_3$ are distinguished.

By the same argument as in the second paragraph, one of $B_2$ or $B_3$ (say $B_2$) is the set of all odd natural numbers. Then $B_3=2{\mathbb N}$, so $A_3=4{\mathbb N}$ is stable by addition.

The same argument shows that for any partition of ${\mathbb N}_{>0}$ into distinguished sets $A_1,\ldots,A_k$, up to renaming we have $A_1=2^{k-1}{\mathbb N}_{>0}$, $A_2$ is the set of number of the form $2^{k-2}\times$ odd number, $A_3$ is the set of number of the form $2^{k-3}\times$ odd number etc.