Can Peano arithmetic be completed using only atomic $\omega$-completions? Say we have an axiomatizable theory $T$ extending $Th(A_E)$ where $A_E$ are the axioms of arithmetic. What I call an atomic $\omega$-completion of $T$ (I am not sure if there is a more standard terminology) is a theory of the form $T\cup \lbrace \forall n \phi(n) \rbrace$, where $\phi$ is a formula in one variable such that $T$ proves $\phi(n)$ for every $n\in{\mathbb N}$.
Question. Is there a sequence $(T_n)_{n\geq 1}$ of theories such that for every $n$, $T_{n+1}$ is an atomic $\omega$-completion of $T_n$ and $T_{\infty}=\bigcup_{n\geq 1}T_n$ is a complete theory ?
We assume that Peano arithmetic is consistent, of course, so that each $T_n$ in the sequence will be consistent also.
My thoughts : There might be a formula $\phi$ in two variables such that $T$ proves $\phi(n,m)$ for every $n,m\in {\mathbb N}^2$ but such that $T$ does not prove $\psi_n : \forall m \phi(n,m)$ for any $n$, so that to extend $T$ to a theory that proves $\forall n \psi_n$ we would seem to need an infinite number of statements first, namely all the $\psi_n$'s.
 A: I'm going to assume our starting theory $T$ is sound (= a subtheory of true arithmetic $\mathsf{TA}$). Otherwise, atomic $\omega$-completions need not preserve consistency: $\mathsf{PA}+\neg\mathsf{Con(PA)}+\mathsf{Con(PA)}$ is an inconsistent atomic $\omega$-completion of the consistent theory $\mathsf{PA}+\neg\mathsf{Con(PA)}$.
Say that a sentence $\varphi$ is reachable from a fixed c.e. theory $T$ iff there is some finite sequence of successive atomic $\omega$-completions starting with $T$ which results in a theory containing $\varphi$. The set of reachable sentences is $\Sigma^0_3$ since $\varphi$ is reachable iff there is some finite sequence $(\psi_i(x))_{1\le i\le n}$ of formulas such that $$T\cup\{\forall x\psi_i(x): 1\le i\le n\}\vdash\varphi$$ and for each $i$ and each $k\in\omega$ we have $$T\cup\{\forall x\psi_j(x): j<i\}\vdash \psi_i(\underline{k}).$$ Since true arithmetic is not arithmetic (per Tarski), this means that the set of reachable sentences is not just true arithmetic. But since we started with a subtheory of true arithmetic and the $\omega$-rule is sound, every reachable sentence is true in $\mathbb{N}$. So the set of reachable sentences is not complete.

Note, however, that we can continue this process transfinitely. When we do this, we do in fact wind up with true arithmetic after $\omega^2$-many steps, as long as we start with something reasonably strong (e.g. $\mathsf{PA}$ or even Robinson's $\mathsf{Q}$). Basically, by careful "dovetailing" we can get all the true $\Pi^0_{n+1}$-sentences by stage $\omega\cdot n$, for each finite $n$.
