Prove that there is a maximum subset of $\mathbb{N}$ who is closed under summation and does not contain any prime number.

Question

Prove that there is a maximum subset of $\mathbb{N}$ who is closed under summation and does not contain any prime number.

My work:
Idea is to use Zorn lemma. Let $F \subseteq \mathbb{N}$ be set of all such set with given properties. We also must see that $F$ isn't an empty set - for example in my set theory class we include $0\in \mathbb{N}$ so $F=\{0\}$ is one example.
Let $L$ be an arbitrary nonempty chain in $F$. We know that upper bound of chain $L$ is union of all elements in $L$.
But Zorn lemma told us that upper bound has to be in $F$. So we need to check this property first. Let $L_1$ and $L_2$ be arbitrary set from our chain L.
L is chain and that implies that $L_1 \subseteq L_2$ or $L_2 \subseteq L_1$.
Without loss of generality we can assume that $L_1 \subseteq L_2$.
Ok, but I stuck here because I have no idea how to prove that union holds given properties.

Answer 1

You are almost there. The final blow is a classic in applying Zorn's lemma: pick any two elements from the union and argue that there has to be one of the $L_i$'s in which they both lie. Therefore so does their sum.

As for the primes, that is easier to see.

Answer 2

I do not think Zorn's lemma is really needed here. Assume that $E\subseteq\mathbb{N}$ is closed with respect to $+$ and let $m=\min E$. Then $E$ contains $m,2m,3m,4m,\ldots$ and if $\mathcal{P}\cap E=\emptyset$ we have that $m$ is a composite number. If $E$ also contains some number which is coprime with $m$, then $[N,+\infty)\subset E$ for some $N\in\mathbb{N}$ by Bézout's lemma, contradicting $\mathcal{P}\cap E=\emptyset$. In particular any two elements of $E$ share a common divisor greater than $1$ and any $E$ fulfilling the given constraints is an AP (arithmetic progression) of the $k\mathbb{N}$ form with some elements dropped. The actual meaning of maximum in the original question is not crystal clear to me, but in any case it is not difficult to deal with lacunary APs.

Answer 3

The maximal subsets of $\mathbb N$ that are closed under addition and contain no primes are exactly $$ p\mathbb N\setminus\{p\} = \{ 0, 2p, 3p, 4p, 5p, \ldots \} $$ for prime $p$.

To see that these are maximal, note that if an addition-closed set contains two coprime numbers $a$ and $b$, it will contain every number greater than $ab-a-b$, which means that it can't be prime free.

On the other hand, suppose we have a prime-free addition-closed set. The gcd of this set must (by the above argument) be greater than $1$ and therefore all elements of the set have a prime factor $p$ in common, and therefore it is contained in $p\mathbb N\setminus\{p\}$.