I'm doing some extra self-exercises on first order logic (I'm taking the course through open university) and I've come across this question:
Let there be a language $L = \{ +, \cdot, 0, 1, < \}\cup \{ c_{n} | n\in \mathbb{N}\}$ and $N$ a structure such that $|N| = \mathbb{N}$ whereby $c^N_{n} = n$ and all the rest of the language is interpreted as the standard symbols in $\mathbb{N}$ ($+$ is addition, $\cdot$ is multiplication etc.). Let us denote $T = Th(N)$
For every model M of T we shall say that $a \in |M|$ is an infinite member if it holds that $c^M_{n} < a$ for every $n\in{N}$
For every model M of T we shall say that $p \in |M|$ is a prime number if it holds that $c^M_1 < p$ and for every $a, b \in |M|$ if $p = a > \cdot^M b$ then $a = p$ or $a = c^M_1$
Show that there exists : $M\vDash T$ such that M has an infinite member.
I don't have a clue what to do here...I've sat on it for a bit but with no luck.
Any help is appreciated, Thx!
HINT: This is a classical use of the compactness theorem. Add an additional constant symbol $c$ and the axioms $c_n<c$ for each $n\in\Bbb N$. Now use the fact that the augmented theory is finitely satisfiable to conclude that it has a model, and it is a model of $T$.