Say we have an axiomatizable theory $T$ extending $Th(A_E)$ where $A_E$ are the axioms of arithmetic. Is it possible to extend $T$ such that our extended theory is consistent but not $\omega$-consistent?
Recall that a theory $T$ is defined to be $\omega$-consistent if for each formula $\psi$, if $T \vdash \neg \psi(a)$ for each $a \in \mathbb{N}$, then $T \nvdash \exists b \psi(b)$.
Here are my thoughts: We need to add a formulas to $T$ such that for some $c$, we have $T \vdash \psi(c)$ and for all $a$, $T \vdash \neg \psi(a)$. I am not sure how it is possible to prove these two statemetns while maintaining that $T$ is consistent since we would have that we could prove $\psi(c)$ for some constant $c$ and $\neg \psi(c)$ since $c$ is in our domain. What am I missing here?
PA + "PA is inconsistent" is indeed a consistent theory - this is the conclusion of Gödel's incompleteness theorem .
But there are simpler examples. For instance, start with PA and add a new constant $x$ and an infinite sequence of axioms of the form "$x > 1$", "$x > 2$", "$x > 3$", ... - this theory is consistent by the compactness theorem. It proves $(\exists y) [y = x]$, but it also proves $x \ne n$ for every numeral $n$, because of the extra axioms.