Prove that $PA$ has a model $M$ with an element $m$, such that $M\models q\mid m$ for ever $q\in Q$ and $M\models \neg q\mid m$ for every $q\in Q-P$

Let $$P\subseteq \mathbb{N}$$ be the set of primes, and suppose $$Q\subseteq P$$ is some subset. Prove that $$PA$$ has a model $$M$$ with an element $$m$$, such that $$M\models q\mid m$$ for ever $$q\in Q$$ and $$M\models \neg p\mid m$$ for every $$p\in Q-P$$

PA is Peano Axioms.

So I believe I want $$\Gamma:= PA\cup \{q\mid m: q\in Q\}$$

Then taking any sub theory of $$\Gamma$$ I have some finite set of $$\{q_1\mid m, q_2\mid m,...,q_k\mid m\}$$

Which I understand can be taken to be prime factors of some $$m$$, but there isn't an $$m$$ which is the product of infinitely many primes. Which makes it seem like there can't be a single $$m$$ which satisfies any finite subset.

• use compactness. Commented Dec 12, 2019 at 5:32
• (Also, what is a context set? A domain? Also, $Q$ might be infinite so there's no product of the primes in $Q.$) Commented Dec 12, 2019 at 5:37
• @spaceisdarkgreen So just take $m$ to be the product of all the primes? Then any finite subset of $Q$ this will be true? Can I make $m$ an infinite product though? Commented Dec 12, 2019 at 5:37
• No. Think about how you use the compactness theorem to show there is a model of PA with an element $m$ such that $m>n$ for every $n\in \mathbb N$. You start by adding a constant symbol, and then prove some collection of new axioms involving that symbol is finitely satisfiable. This works in a similar way. Commented Dec 12, 2019 at 5:40
• @spaceisdarkgreen But I need a collection of $q\mid m$ for every $q$ right? Commented Dec 12, 2019 at 8:05

Add a constant symbol $$c$$ the language and consider the theory consisting of PA plus $$\{\mathbf q\mid c:q\in Q\}\cup\{\mathbf p\not\mid c:p\in P-Q\}$$ where the boldface denotes the numeral (so to be clear about what language we’re in and that we aren’t mixing syntax and semantics).
This theory is finitely satisfiable: use use the usual interpretation in the naturals and interpret $$c$$ as the product of all the $$q\in Q$$ such that $$\mathbf q\mid c$$ appears in the finite subtheory under consideration.
Thus by the compactness theorem, the theory has a model. Then let $$m$$ be the interpretation of $$c$$ in that model.