I'm working on the following problem:
Let $p_0,p_1,...$ be a list of the prime numbers in increasing order. Show that for any set $X\subseteq\mathbb{N}$, there is a model of Th($\mathbb{N})$ which contains a number $a$ such that $a$ is divisible by $p_i$ iff $i\in X$.
So I have defined, primes, divisibility, "<". I'm aware that I can do this in one swoop but I need to use the technique of adding a constant $c$ and sentences $p_i$ divides $c$ if $i\in X$ and $p_i$ does not divide $c$ if $i\notin X$ and finding a way to apply compactness but I'm having trouble seeing how.
Let $\mathcal{L}'=\{+,\cdot,0,1\}\cup \{c\}$ where $c$ is a constant symbol not originally in the language of $\mathbb{N}$. Given $X\subseteq \mathbb{N}$, consider the $\mathcal{L}'$-theory given by $$\Gamma_X=\operatorname{Th}(\mathbb{N})\cup \{\underbrace{\exists x(p_i\cdot x=c)}_{\text{"$p_i$ divides $c$"}}:i\in X\}\cup \{\underbrace{\forall x(p_i\cdot x\neq c)}_{\text{"$p_i$ does not divide $c$"}}:i\in X\}$$
First we show that $\Gamma$ is finitely consistent. If $\Gamma_0\subset^{\text{fin}}\Gamma$, then it contains finitely many sentences of the form $\varphi_i:=\exists x(p_i\cdot x=c)$ (for $i=i_1,\ldots,i_k$) and $\psi_j:=\forall x(p_j\cdot x\neq c)$ (for $j=j_1,\ldots,j_\ell$). Notice that $\{i_1,\ldots,i_k\}\cap\{j_1,\ldots,j_\ell\}=\emptyset$.
Take $m_0=\displaystyle{\prod_{t=1}^{k}p_{i_t}}$. Then we have that $p_{i}$ divides $m_0$ for all $i=i_1,\ldots,i_k$, and $p_j$ does not divide $m_0$ for $j=j_1,\ldots,j_\ell$. Thus, the $\mathcal{L}'$-structure given by $\mathcal{N}=(\mathbb{N},+,\cdot,0,1c^{\mathcal{N}}:=m_0)$ is a model of $\Gamma_0$. (This is the standard natural numbers, interpreting the new constant as the number $m_0$)
Therefore, since every finite subset of $\Gamma$ is consistent, then by compactness we conclude that $\Gamma$ is consistent. By completeness theorem, there is an $\mathcal{L}'$-structure $\mathcal{M}\models \Gamma$, and by the construction of $\Gamma$ it would be a model of $\operatorname{Th}(\mathbb{N})$ with an element $c=c^\mathcal{M}$ such that $p_i$ divides $c$ if and only if $i\in X$
Note: Each sentence $\varphi_i=\exists x\left(p_i\cdot x=c\right)$ is explicitly written as: $$\exists x\left((\underbrace{1+\cdots +1}_{\text{$p_i$ times}})\cdot x=c \right)$$