# Infinite set of standard primes as the set of standard prime divisors of a nonstandard number

Suppose $(N, +, \cdot, 0, 1, <, =)$ is a proper elementary substructure of $(N^*, +^*, \cdot^*, 0^*, 1^*, =^*, <^*)$. Show that there exists some (infinite) $b$, where $b ∈ N^*$, such that for each prime number $p ∈ N$, $N^* \models p | b$ iff $p ∈ S$, where $S$ is some set containing infinitely many primes.

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Do you mean "for every $S$ and every prime $p\in\mathbb N$, the equivalence holds" or do you mean "there exists some $S$ such that for every prime $p\in\mathbb N$, the equivalence holds"? The former statement is not (always) true, but the latter is. –  Lawrence Wong Oct 28 '13 at 12:43
I think it is for some S. –  max_b Oct 28 '13 at 15:38
The quantifier on $S$ is still not very clear. Is $S$ is given in advance, or are you free to choose it so long as it is infinite? –  user72694 Oct 28 '13 at 15:39
You're free to choose S so long as its infinite. –  max_b Oct 28 '13 at 16:06
@Arthur: Do we need a new tags for non-standard models? –  Asaf Karagila Oct 28 '13 at 17:09

If we choose $S$ to be the collection of all the standard primes, then it suffices to set $b=H!$ for some infinite integer $H$. Since the standard model satisfies the elementary formula expressing the factorial, the same formula evaluated at $H$ will give a non-standard integer divisible by all the primes up to $H$ by elementary equivalence, and in particular by all the finite primes in $S$.