Find all sets $A$ that satisfy condition $a+b \in A \implies ab\in A$. Find all subsets from $ \mathbb N$ such that:
We say that a subset from $\mathbb N$ that it is complete (let’s call it $A$) if :
$a+b \in A \implies ab\in A$.
With $a,b \in \mathbb N$
Find all complete subsets from $\mathbb N$.
So far I’ve found some:
$A=\{0,1,2\}, \{0,1\}$
 A: If
$1+(x-1)\in A \Rightarrow 1×(x-1)\in A$
$1+(x-2)\in A \Rightarrow 1×(x-2)\in A$
$1+(x-3)\in A \Rightarrow 1×(x-3)\in A$
Etc.
By infinite descent if $x\in A$ then all natural numbers less than $x$ is also in $A$.
If
$2+(y-2)\in A \Rightarrow 2×(y-2)=2y-4$
$2y-4>y\iff y>4$
This means if one element greater than 4 is in $A$ then there is another element greater than that one.
$2+(2y-6)\in A \Rightarrow 2×(2y-6)=4y-12$
$2+(4y-14)\in A \Rightarrow 2×(4y-14)=8y-28$
By induction if there is an element in $A$ that is greater than 4 then there is no greatest element in $A$ (infinitely many elements). This combined with the decent rule implies that if there is an element greater than 4 in $A$ then all elements of $\Bbb{N}$ are in $A$.
What's left is the null set and sets that have all the natural numbers less than or equal to $k$. Where $k$ is $0\le k\le 4$. It can be easily verified that all five sets from each the five values of $k$ satisfies the condition that $a+b\in A\Rightarrow ab\in A$
This means there are seven subsets that satisfy the criterion.
$\emptyset$
$\lbrace 0\rbrace$
$\lbrace 0,1\rbrace$
$\lbrace 0,1,2\rbrace$
$\lbrace 0,1,2,3\rbrace$
$\lbrace 0,1,2,3,4\rbrace$
$\Bbb{N}$
