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).

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    $\begingroup$ 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. $\endgroup$ – Stefan Mesken Mar 4 '19 at 9:20
  • $\begingroup$ @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. $\endgroup$ – SSequence Mar 4 '19 at 11:06
  • $\begingroup$ 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. $\endgroup$ – SSequence Mar 4 '19 at 11:09
  • $\begingroup$ The accepted answer is incorrect - see my answer and Stefan's comment. $\endgroup$ – Noah Schweber Mar 4 '19 at 14:23
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Stefan Mesken Mar 4 '19 at 14:21
  • $\begingroup$ Btw, you should mention that you work in $L$, i.e. $\omega_1 = \omega_1^L$ in your post. $\endgroup$ – Stefan Mesken Mar 4 '19 at 14:27
  • $\begingroup$ @StefanMesken Whoops, good point - addressed. $\endgroup$ – Noah Schweber Mar 4 '19 at 14:48
  • $\begingroup$ 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. $\endgroup$ – SSequence Sep 16 '19 at 19:12
  • $\begingroup$ 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. $\endgroup$ – Noah Schweber Sep 16 '19 at 19:20

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