Cofinality of $\omega^{CK}_{\omega_1+1}$

What is the cofinality of this ordinal (say in ZFC)? Is it countable or not?

Edit: Based on reason for close, I have added some informal motivation for question. Based on the computability-theoretic definition, whenever $$\omega^{CK}_\alpha$$ is countable then so is $$\omega^{CK}_{\alpha+1}$$. But, formally speaking, it doesn't apply directly when $$\alpha$$ isn't countable.

Denoting $$\beta=\omega^{CK}_{\omega_1+1}$$, intuitively there are several "interesting" and large enough points $$p$$ (such that $$p<\beta$$) so that $$p$$ will have countable finality (as one example when $$p$$ is the first fixed point of a normal function such as $$x \mapsto (\omega_1)^x$$).

But this informal intuition doesn't help in giving a formal answer to cofinality of $$\omega^{CK}_{\omega_1+1}$$. Hence the reason for asking the question (I want to know how well the informal intuition does or doesn't transfer into formal reasoning).

• What is $\omega_{\omega_1 + 1}^{CK}$? I can make sense of $\omega_\alpha^{CK}$ for countable $\alpha$ but for uncountable ones, I'm not sure what the intended meaning is. If $\omega_\alpha^{CK}$ is the $\alpha$-th admissible ordinal, then $\omega^{CK}_{\omega_1 + 1}$ has countable cofinality. – Stefan Mesken Mar 4 '19 at 9:20
• @StefanMesken I would be out of depth in trying to explain $\omega_\alpha^{CK}$ (for uncountable $\alpha$) myself. But you might look at the answer of a question I asked number of months ago: mathoverflow.net/questions/310244. So given that answer, yes it seems that "$\omega_\alpha^{CK}$ as the $\alpha$-th admissible ordinal" should be the intended meaning. – SSequence Mar 4 '19 at 11:06
• Given your comment (as I took it), $\omega^{CK}_{\omega_1 + 1}$ has (provably) countable cofinality. In that case, you might convert your comment as an answer. – SSequence Mar 4 '19 at 11:09
• The accepted answer is incorrect - see my answer and Stefan's comment. – Noah Schweber Mar 4 '19 at 14:23
• @NoahSchweber Thanks for the effort on the answer. I was definitely not expecting $cf(\theta)>\omega$. I certainly can't make heads or tails out of this. I will try to explain the brief reasoning (in a paragraph or two) why I am feeling very confused by this by bumping the relevant MO question (which was the reason for this question) a bit later. – SSequence Mar 4 '19 at 15:10

I disagree with Stefan Mesken's answer. First, let me explain why I think the general claim made there is false. Consider the theory $$T$$ = KP + V=L + "There is some ordinal $$\alpha$$ such that $$\alpha+\alpha$$ doesn't exist but $$\alpha+\beta$$ exists for all $$\beta<\alpha$$." We have $$L_{\omega_1+\omega_1}\models T$$, but no $$\alpha\in (\omega_1,\omega_1+\omega_1)$$ models $$T$$; so $$L_{\omega_1+\omega_1}$$ is the first level of $$L$$ satisfying $$T$$ above $$\omega_1$$, yet $$cf(\omega_1+\omega_1)\not=\omega$$.

Now here's an argument showing that the cofinality of $$\omega_{\omega_1+1}^{CK}$$ - which latter ordinal I'll call "$$\theta$$" for simplicity - is $$\omega_1$$.

The key is projecta. If $$\alpha$$ is the next admissible above $$\beta$$, then there is an injection $$\pi$$ of $$\alpha$$ into $$\beta$$ which is $$\Sigma_1$$-definable (in $$L_\alpha$$). This doesn't contradict admissibility of $$\alpha$$ since in order to turn this into a $$\Sigma_1$$ cofinal map from $$\beta$$ to $$\alpha$$ we would need $$ran(\pi)$$ to be $$\Delta_1$$, and it's only $$\Sigma_1$$.

Indeed, $$ran(\pi)$$ has the following strong property: for any cofinal set $$A\subseteq \alpha$$, $$\pi[A]$$ is not $$\Delta_1$$. Otherwise we could indeed get a $$\Sigma_1$$ map from $$\beta$$ to $$\alpha$$ with cofinal image - namely, send $$x\in A$$ to $$\pi^{-1}(x)$$ and $$x\not\in A$$ to $$17$$. This is in fact $$\Sigma_1$$, since if we know ahead of time that $$x\in ran(\pi)$$ we can search for $$x$$'s preimage without risk of looping forever.

OK, now suppose $$cf(\theta)=\omega$$. Fix$$^1$$ a constructible cofinal $$\omega$$-sequence $$(\eta_i)_{\eta<\omega}$$, and let $$\gamma_i=\pi(\eta_i)$$. Then $$S=(\gamma_i)_{i\in\omega}$$ is an $$\omega$$-sequence of countable ordinals, so by condensation$$^1$$ we get $$S\in L_{\omega_1}$$. But now take $$S$$ - or rather, the range of $$S$$ - to be our $$A$$.

(OK fine, strictly speaking all I've shown is that $$cf(\theta)>\omega$$. But since $$\vert\theta\vert=\omega_1$$, this does mean $$cf(\theta)=\omega_1$$.)

$$^1$$Why does this exist? On the face of things, after all, I seem to be using V=L (or at least that $$L$$ correctly computes cofinalities) here. This isn't necessary, but at present I'm forgetting how the argument goes; I'll add it when I have time to recall it.

• I agree that my argument doesn't work -- I incorrectly assumed that closing a given level under Skolem terms would land me strictly below the model of $T$ of uncountable cofinality in the hierarchy. – Stefan Mesken Mar 4 '19 at 14:21
• Btw, you should mention that you work in $L$, i.e. $\omega_1 = \omega_1^L$ in your post. – Stefan Mesken Mar 4 '19 at 14:27
• @StefanMesken Whoops, good point - addressed. – Noah Schweber Mar 4 '19 at 14:48
• I was having a look at this answer. I have a number of questions. Firstly, is $\pi$ supposed to be a total function or not? The word "injection" doesn't seem to imply that directly. If $\pi$ is supposed to be total then I think I can make sense of the first half of the answer (by converting the corresponding notions to computation related ones). If not, then I might probably be misinterpreting and/or misunderstanding something significant. – SSequence Sep 16 '19 at 19:12
• The map $\pi$ is indeed total (and when we speak of an injection of one set into another we do intend totality unless otherwise stated specifically). The details of the construction of $\pi$ are covered in e.g. Sacks' book. – Noah Schweber Sep 16 '19 at 19:20