Can countable ordinals start gaps of every order in the constructible universe?

Define "$$\alpha$$ starts a gap of order $$n+1$$ and length $$\beta$$" iff $$\mathcal P^n(\omega)\cap (L_{\alpha+\beta}\setminus L_\alpha)=\emptyset\land\forall\gamma\in\alpha: L_\alpha\setminus L_\gamma\neq\emptyset$$ where $$\mathcal P^n$$ is the powerset operation iterated $$n$$ times.

Define "$$\alpha$$ is in a gap of order $$n$$" as $$\forall m

Define $$\text{ZFC}^-$$ as $$\text{ZFC}-$$(Powerset axiom). This theory is equiconsistent with $$Z_2$$, second order arithmetic. I believe gaps in the constructible universe were first talked about by Putnam in 1963, or at least that's the oldest source I've read, but that's besides the point. According to Marek and Srebrny, the least ordinal that starts a gap of second order is the minimal model height of $$\text{ZFC}^-$$ and [on p. 372 theorem 3.7] the first ordinal to start a gap of third order is also the minimal model height of $$\text{ZFC}^-+\exists\omega_1$$.

I have four questions associated with this property:

1. Does the pattern hold that the least ordinal to start a gap of order $$n+2$$ is also the minimal model height of $$\text{ZFC}^-+\exists\omega_n$$ or $$Z_{n+2}$$?
2. If so, are all the ordinals $$\alpha$$ that start gaps of order $$n+2$$ exactly those for which $$L_\alpha\models\text{ZFC}^-+\exists\omega_n$$ and $$L_\alpha\cap\mathcal P^{n+1}(\omega)\models Z_{n+2}$$? If the answer to (1) is unknown then how about (2) when $$n=0$$?
3. Does the ordinal $$\sup\{\min\{\alpha~|~\mathcal P^n(\omega)\cap (L_{\alpha+1}\setminus L_\alpha)=\emptyset\}: n\in\mathbb N\}$$ itself start any gaps?
4. Is there a countable ordinal that starts a gap of every order (of order $$n\forall n\in\mathbb N$$) simultaneously?

EDIT: As pointed out by Farmer S, question (4) does not make sense as it is contradicted by page 371 of the very same paper. Thus, I will change it to the following:

Can an ordinal be in a gap of every order (order $$n~\forall n\in\mathbb N$$) simulteneously?

• I think the answer to (4) is yes. Take some countable ordinal $\beta$ and some countable elementary submodel of $L_{\aleph_\omega+1}$ containing $\beta$. The transitive collapse is of the form $L_{\gamma+\delta}$, where $\gamma$ is believed to be $\aleph_\omega$, and $\delta > \beta$. Mar 24, 2022 at 17:32
• With regards to 2. the answer is no. Observe that for all large enough $\alpha$'s, the sets $L_\alpha$ don't have new sets from $\mathcal{P}^n(\omega)$ and hence all large enough ordinals $\alpha$ start a gap of the order $n+1$ (according to your definition). Also we have $L_\alpha\cap \mathcal{P}^{n+1}(\omega\models \mathsf{Z}_{n+2}$ for all large enough $\alpha$'s. However it is easy to see that you don't have this kind of uniformity for all large enough ordinals with regards to the property to be a model of $\mathsf{ZFC}^-+\exists \omega_n$. Mar 25, 2022 at 15:33
• Why is it "easy to see"? Also, I thought that $\alpha$ has to be a limit ordinal in order for $L_\alpha$ to be a model for a theory. Mar 25, 2022 at 19:29
• Your definition of "starts a gap (of order 2)" doesn't agree with the linked paper. On p. 364 of that paper, they define "$\alpha$ is a gap ordinal iff $(L_{\alpha+1}-L_\alpha)\cap\mathcal{P}(\omega)=\emptyset$, and on p. 368, define "$\alpha$ starts a gap iff $\alpha$ is a gap ordinal and $(\beta)_\alpha (L_\alpha-L_\beta)\cap\mathcal{P}(\omega)\neq\emptyset$". Thus, what you are defining as "$\alpha$ starts a gap of order 2 and length 1" is what they define as a "$\alpha$ is a gap ordinal"... Mar 25, 2022 at 21:38
• ...and their "$\alpha$ starts a gap" adds the requirement that "$(\beta)_\alpha (L_\alpha-L_\beta)\cap\mathcal{P}(\omega)\neq\emptyset$". This "$(\beta)_\alpha$" was initially cryptic to me, but comparing it with usage earlier in the paper, and e.g. the notation "$(E\alpha)_{\omega_1^L}$" on p. 365, I think "$(\beta)_\alpha$" must mean "$\forall\beta\in\alpha$" or "$\forall\beta<\alpha$". This then renders the terminology "starts a gap" as more natural (to me at least). Mar 25, 2022 at 21:43

I presume that by "starts a gap (of order $$n$$)" you mean "starts a gap of positive length (of order $$n$$)", since length $$0$$ is trivial.

Note that the answers given by Monroe Eskew and Fedor Pakhomov were written when there was a different definition of "starts a gap" than what is now there.

1. Yes (by $$L_\alpha\models\mathrm{ZFC}^-$$, there are no bounded subsets of $$\alpha$$ in $$L_{\alpha+1}\backslash L_\alpha$$; and if $$L_\alpha\models$$"$$\omega_n$$ is the largest cardinal" but $$L_\alpha\not\models\mathrm{ZFC}^-$$, then $$L_{\alpha+1}\backslash L_\alpha$$ contains some bounded subset of $$\alpha$$, hence some subset of its $$\omega_n$$).

2. No; you need to add the condition that $$L_\alpha\models$$"$$\omega_n$$ is the largest cardinal", and then you get the characterization.

3. No (again assuming that the length of a gap has to be $$>0$$); let $$\alpha$$ be that sup. Then the sequence $$\left<\alpha_n\right>_{n<\omega}$$, where $$\alpha_n$$ is the first start of a gap of order $$n$$, is $$\Sigma_1$$-definable over $$L_\alpha$$ without parameters, and in $$L_\alpha$$, every set is countable. But then it follows that $$L_\alpha$$ is the $$\Sigma_1$$-hull of the empty set in $$L_\alpha$$ (that is, every element of $$L_\alpha$$ is $$\Sigma_1$$-definable over $$L_\alpha$$ from no parameters), which implies that $$\alpha$$ does not start a gap.

4. No (ignoring the fact that "starts a gap of order $$0$$" was not defined); $$\alpha$$ can not simultaneously start a gap of two distinct orders (this is alluded to in the case of orders 2 and 3 on p. 371 of the paper, near the bottom of the page). E.g. for orders 2 and 3, if cofinally many $$\gamma<\alpha$$ project to $$\omega$$, then $$\omega$$ is the largest cardinal in $$L_\alpha$$. (Note Monroe's comment was written when the definition of "starts a gap" was different.)

• On page 371 it is stated, as you said, that an ordinal cannot start both a second and third order gap. The definition on page 370 says that if an ordinal is a "nth order gap" then $\forall m<n: \mathcal P^m(\omega)\cap L_{\alpha+1}\setminus L_\alpha=\emptyset$ So an ordinal can't be the first to start gaps of multiple orders but it can be in gaps of multiple orders? Mar 26, 2022 at 0:47
• And an ordinal can be in a gap of every finite order $n\geq 2$, simulteneously? Mar 26, 2022 at 1:18

Your other questions have been answered, so I provide here only an answer for your last edit, that is, whether an ordinal can be in a gap of every order.

The answer is positive by an argument almost identical to that in Theorem 1.3 of the paper you mention.

Since $$\mathcal{P}^n(\omega)^L \subseteq L_{\omega_n^L}$$, we have $$\mathcal{P}^n(\omega)^L \subseteq L_{\omega_\omega^L}$$ for every $$n$$. So for instance $$L_{\omega_{\omega+1}^L} \models \exists \alpha \forall n \in \omega((L_{\alpha+1}\setminus L_\alpha)\cap \mathcal{P}^n(\omega) = \varnothing)$$ As in the paper, working inside $$L$$ we can apply Downward Löwenheim-Skolem to find an elementary submodel, of for instance countable cardinality. And thus by the absoluteness of the above formula we'll have an ordinal below $$\omega_1^L$$ which is a gap of every finite order.

In fact, an argument as in the paper shows there are arbitrarily big such ordinals below every $$\omega_n^L$$. And like there, this can be generalized to any $$\Sigma_1$$ ordinal operation in $$L$$ replacing the $$+1$$ operation.

These ordinals (gaps of every finite order) can be understood intuitively as $$\omega$$-gaps. And so we can also study the starts of these $$\omega$$-gaps, which as it turns out will have to be limits of limits, and will satisfy a certain set theory. I elaborate on this in my Bachelor's Thesis, "The real numbers in inner models of set theory", which expands on the topics of Marek and Srebrny's paper.

• I, for one, would be interested in seeing that Bachelor's Thesis! And even if I weren't, if it has expositional value, you'd do the community a service by putting it online in a place where it can be googled and where it will be archived. Jun 17, 2022 at 20:41
• $\mathcal P^n(\omega)\subset L_{\omega_n^L}$ is a statement independent of ZFC. You can't just state it as a fact. Do you mean $(\mathcal P^n(\omega))^L\subset L_{\omega_n^L}$? Jun 18, 2022 at 21:07
• But yes, I would love to see your thesis. Jun 18, 2022 at 21:10
• @Gro-Tsen Thank you for your comment! The University of Barcelona will upload it to their digital deposit in some months, but for now I've uploaded it to arxiv and linked it above so you can access it. Jun 23, 2022 at 10:03
• Oh, it's very recent! (I thought you were talking about exhuming a thesis written ages ago.) So, congratulations, and thanks for putting it online, I'm sure it will help make gaps in the constructible universe much more understandable. Jun 23, 2022 at 10:21