Let $\Sigma = \left\{ 0,1,+,\times , < \right\}$ be a signature, where the last 3 symbols have arity 2 and all except the last symbol are functions symbols. How can I prove that the set $X=\left\{ \ulcorner A \urcorner \ : \mathbb{N} \vDash A \right\}$, where $ \ulcorner A \urcorner $ is the code of the $\Sigma$ formula $A$, and "$\vDash$" means "is a model of", is not arithmetic ?
I can use the fact that the set $B=\left\{ (c,n)\in \mathbb{N}^2\left| \begin{array}{l} c= \ \ulcorner T \urcorner\text{, where } T \text{ is a } \Sigma \text{ formula that}\\ \text{ has only } v_0 \text{ as a free variable, and }\mathbb{N} \vDash T_{v_0 \rightarrow \underline{n}} \end{array}\right.\right\},$ where $A_{v_0 \rightarrow \underline{n} }$ is the $\Sigma$ formula were the variable $v_0$ in $T$ is replaced with the term $ \underline{n}$ (that corresponds to the number $n$ meaning $ \underline{n}=++\ldots+011\ldots1$, where there are $n$ additional symbols) is not arithmetic.
What I got until now is the following: I supposed, that $X\subseteq \mathbb{N}$ were arithmetic. Thus there exists a $\Sigma$ formula $\psi$ which has exactly one free variable $v$ such that $\psi_\mathbb{N}=X$ (we have defined $\psi_\mathbb{N} =$ {$ m \in \mathbb{N}: \mathbb{N} \vDash \psi_{v \rightarrow \underline{m}} $} - and if the formula contains 2 free variables, the definition changes accordingly). Thus I can plug $\psi$ in the set $B$. But this is the point from which on I don't kniw how to continue - and I haven't yet used the fact that $B$ is not arithmetic as well.
(For more info, as to how $\Sigma$ formulas etc. were defined in my course, see this post)