How many layers of consistency can PA recognize?

Question

$\newcommand{\Cons}{\operatorname{Cons}}\newcommand{\PA}{\mathsf{PA}}$Assuming $\PA$ is consistent, we know that $\PA \nvdash \Cons(\PA)$ and $\PA + \Cons(\PA) \nvdash \Cons(\PA + \Cons(\PA))$, and so on.

Let $T_0 := \PA$ and $T_{\alpha^+} := T_\alpha + \Cons(T_\alpha)$ and $T_\beta := \bigcup_{\alpha < \beta} T_\alpha$.

What is the minimal ordinal $\alpha$ such that $T_\alpha$ cannot be written as a sentence in the language of Peano arithmetic?

Answer 1

Well, first of all it doesn't really make sense to talk about $T_\alpha$ for $\alpha$ an ordinal; instead we need to talk about ordinal notations (these are basically just "particularly nice" computable well-orderings of $\mathbb{N}$ - specific copies of ordinals rather than the more abstract ordinals themselves). Two notations for the same ordinal may wind up behaving quite differently in terms of the "iterated consistency extension" they yield - see e.g. here.

That said, it turns out that there is a rather sharp answer to your question. Ordinal notations go up through the computable ordinals, that is, up to $\omega_1^{CK}$. Meanwhile, every arithmetically definable well ordering is in fact isomorphic to a computable one (indeed much more is true: "hyperarithmetic = computable" for ordinals, this is a theorem of Spector).

So briefly:

We can iterate consistency principles along computable well-orderings (not computable ordinals per se) without trouble, but we can't even arithmetically refer to $\omega_1^{CK}$ or above.