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 $\forall m<n:\mathcal P^m(\omega)\cap (L_{\alpha+1}\setminus L_\alpha)=\emptyset$
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:
- 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 $Z_{n+2}$?
- 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 $n=0$? - 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?
- 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?
Best Answer
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.
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 $\omega_n$).
No; you need to add the condition that $L_\alpha\models$"$\omega_n$ is the largest cardinal", and then you get the characterization.
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.
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.)