TITLE: Consistent sentences with no arithmetically definable models
QUESTION [5 upvotes]: I've seen a construction of a sentence of first order logic that is consistent, but has no models with underlying set $\mathbb{N}$ and recursive functions and relations. Do there also exist consistent sentences with no model on $\mathbb{N}$ with arithmetically definable functions and relations? If no, at which levels of the arithmetical hierarchy are there consistent sentences with no model at that level? If yes, what about the same question for analytically definable models?

REPLY [6 votes]: In this edit Propositions A and B are fine-tuned (thanks to input from Alex Mennen and David Marker).

This note complements the answer by Joel Hamkins. The question was posed for consistent sentences and asked for the arithmetical definability of the open diagram of some model of the given sentence, but as pointed out by Joel, a much stronger result is true, namely: every consistent arithmetically definable theory has a model whose elementary diagram is arithmetically definable. 
In contrast, there are extensions $T$ of PA (Peano arithmetic) with the property that the open diagram of no model of $T$ is arithmetically definable; indeed something much stronger is true:
Proposition A. For every $X \subseteq \omega$ there is a consistent theory $T$ in the language of arithmetic extending PA  such that $X$ is Turing reducible to the open diagram of every model of $T$ whose universe is $\omega$.
Proposition A can be readily established as an immediate corollary of the following classical result:
Theorem B (Scott). Every countable Scott set can be realized as the standard system of some pointwise definable model of PA.
Proof sketch of the proposition. It is easy to see that any $X\subseteq\omega$ is included in a countable Scott set (for example the Scott set can be chosen as all subsets of $\omega$ that are arithmetical in $X$), so by Theorem B there is a pointwise definable model $\mathcal{M}_0$ of PA whose standard system $\mathrm{SSy}(\mathcal{M}_0)$ includes $X$. Let $T$ be the first order theory of $\mathcal{M}_0$, and note that $X$ is in the standard system of every model of $T$.
On the other hand, it is well-known that if $\mathcal{M}$ is a model of PA whose universe is $\omega$, then every member of $\mathrm{SSy} (\mathcal{M}$) is Turing reducible to the graph of both the addition and the multiplication operation of $\mathcal{M}$ [this fact plays a key role in the proof of Tennenbaum's theorem that states that there is no nonstandard model of Peano arithmetic whose open diagram is computable]. This implies that if $\mathcal{M}$ is additionally a model of $T$, then every member of
$\mathrm{SSy}(\mathcal{M}$), and in particular $X$, is Turing reducible to the graph of the addition operation of $\mathcal{M}$.