Is there a sequence of extensions of ZFC where the corresponding sequence of proof theoretic ordinals has $\omega_1^{CK}$ as least upper-bound I was reading this question on MO  where they define an infinite sequence of extensions of ZF by creating iteratively  a new theory which includes the consistency of the previous ones.  The definition in this question is ambiguous.
Independently from this particular definition I wonder if it possible  to build a sequence of extensions of ZFC (or ZF), in such a way that for any  recursive ordinal $\alpha$ there exists $\kappa$ such that $\text{pto}(\text{ZFC}_{\kappa})>\alpha$, where  $\text{pto}$ stays for proof theoretic ordinal and $\text{ZFC}_{\kappa}$ is the $\kappa$-th element of the sequence.
I know that the proof theoretic ordinal of ZFC is a recursive ordinal, but it is unknown  and the proof theoretic ordinals are known for theories up to $\Pi^1_2 -\text{CA}_0$. Anyway I am not asking for the explicit proof theoretic ordinals of that sequence, but only if they satisfy the property I have mentioned.
 A: I'll assume that $\mathsf{ZFC}$ is at least $\Pi^1_1\vee\Sigma^1_1$-sound, and use the following definition of the proof-theoretic ordinal:

We let $pto(T)$ be the supremum of the lengths of primitive recursive well-orderings which $T$ proves are in fact well-ordered.

Here $T$ is an "appropriate" theory - for simplicity, let's say it's a computably axiomatizable $(\Pi^1_1\vee\Sigma^1_1)$-sound extension of $\mathsf{ZFC}$.
Then we can produce a uniformly primitive recursive family of theories $(T_i)_{i\in I}$ which provably in $\mathsf{ZFC}$ has the following nice properties:

*

*Each $T_a$ is a consistent extension of $\mathsf{ZFC}$ by a single additional axiom.


*$I$ is a linear order, and the $T_i$s are ordered by strength appropriately: if $a<_Ib$ then $T_b$ proves every axiom of $T_a$.


*For each $\alpha<\omega_1^{CK}$ there is some $a\in I$ such that $T_a$ is $\Pi^1_1\vee\Sigma^1_1$-sound and proves the well-foundedness of some primitive recursive ordering of type $\alpha$.
This uses a couple tricks. First, we apply Barwise-Kreisel Compactness to get a primitive recursive linear ordering $(I,<_I)$ with no hyperarithmetic descending sequences whose well-founded part has type $\omega_1^{CK}$ (these are called Harrison orders). Now for $a\in I$ let $T_a$ be $\mathsf{ZFC}$ + "The initial segment of $I$ up to $a$ is well-founded." The only way $T_a$ could be inconsistent is if $\mathsf{ZFC}$ disproves the well-foundedness of $I_{<a}$. Since $\mathsf{ZFC}$ is $\Pi^1_1$-sound, this can only happen if $a$ is in the illfounded part of $I$; since $I$ has no hyperarithmetic descending sequences, there must be some $b$ in the illfounded part of $I$ such that $T_b$ is consistent (and hence each $T_c$ is consistent for $c<_Ib$). WLOG then $T_a$ is consistent for each $a\in I$ (otherwise replace $I$ with $I_{<b}$).
The only nontrivial point now is to prove that if $a$ is in the well-founded part of $I$ (and remember that has ordertype $\omega_1^{CK}$) then $T_a$ is $\Pi^1_1\vee\Sigma^1_1$-sound. To see this, suppose $T_a$ proves some false $\Pi^1_1\vee\Sigma^1_1$ sentence $\varphi$. Then $\mathsf{ZFC}$ proves the sentence "If $I_{<a}$ is well-founded then $\varphi$," which is a false $\Sigma^1_1\vee(\Pi^1_1\vee\Sigma_1^1)$-sentence, or a false $\Pi^1_1\vee\Sigma^1_1$-sentence. But this contradicts the $\Pi^1_1\vee\Sigma^1_1$-soundness of $\mathsf{ZFC}$.
This is an old argument, but I don't know who it's due to - it may well be folklore.
