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Frank Drake proves in his 'large cardinals' book, pg 140, that

Theorem: There are $\kappa^+$ ordinals $\alpha$ between $\kappa$ and $\kappa^+$ such that $\alpha+1$ is not the 'constructive order' (i.e. constructive rank) of any subset of $\kappa$.

His proof makes use of the fact that there are $\kappa^+$ models of ZFC in the form $L_\alpha$ for some $\kappa\le\alpha\le\kappa^+$. The proof looks fine to me, but I'm unsure about something. If no new subsets of $\kappa$ are added from $L_\alpha$ to $L_{\alpha+1}$, then how exactly does the constructive hierarchy continue to grow? I mean, we cannot have $L_\alpha = L_{\alpha+1} = L_{\alpha+2}=...$ as then we would not get all of $L$. So when we move up the hierarchy, we must always be getting new sets from somewhere, but where exactly are they coming from?

Furthermore, can we put an upper bound on the length of the gaps? (Drake shows using model theoretic arguments that the gaps must be very large).

Furthermore, has the value of the smallest ordinal $\alpha$ such that there is a gap at $L_\alpha$ been determined?

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  • $\begingroup$ Why shouldn't the hierarchy continue to grow? E.g. $L_\alpha \in L_{\alpha+1}$ always. $\endgroup$ – Jonathan Sep 1 '18 at 15:06
  • $\begingroup$ We define $L_{\alpha+1}$ as $Def(L_\alpha)$ where $Def(A)$ is the collection of sets definable by a formula with parameters in $A$. From this we have if $A = Def(A)$, then certainly $A = Def(A) = Def(Def(A)) = ... $ etc. $\endgroup$ – Elie Bergman Sep 1 '18 at 15:13
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    $\begingroup$ Yes but $Def(A)$ is never $A$ because $A \in Def(A)$ but $A \notin A$. $\endgroup$ – Jonathan Sep 1 '18 at 15:18
  • $\begingroup$ Ah right, so instead of adding subsets of $\kappa$ at the sucessor stage, instead we get sets like $\{\kappa\}$, $\{\{\kappa\}\}$ and the like. (weird stuff basically, but the hierarchy always grows). The other questions are still open of course. $\endgroup$ – Elie Bergman Sep 1 '18 at 15:21
  • $\begingroup$ Well if no new subsets of $\kappa$ get added in stage $\alpha$ then $\{ \kappa\}$ and also $\{ \{ \kappa \}\}$ will already be in $L_\alpha$. New stuff that is added is e.g. the ordinal $\alpha$, the set $L_\alpha$, subsets of $\alpha$... $\endgroup$ – Jonathan Sep 1 '18 at 15:24
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The comments have answered how the $L$ hierarchy continues to grow even in gaps; the point being that even though it may "appear finished" with a certain rank, it will continue to add sets of higher rank. Gaps refer to times when the $L$ hierarchy restricted to a certain level doesn't change.


Now as to the lengths of gaps, basically they are "as big as possible." For example:

If $\kappa$ is infinite and $\lambda<\kappa^+$, then there is an $\alpha\in [\kappa,\kappa^+)$ such that $\alpha$ starts a "$\kappa$-gap" of length at least $\lambda$.

Proof: Let $\alpha$ be the image of $\kappa^+$ under the Mostowski collapse of any cardinality-$\kappa$ elementary submodel $M$ of enough of $L$ ($L_{\kappa^{++}}$ is more than enough) which contains $\lambda+1$ as a subset.

By condensation, the Mostowski collapse of $M$ is a level of $L$ - say, $L_\beta$. Since $\kappa^+$ starts a $\kappa$-gap of length at least $\lambda$ - indeed, of length $Ord$, since $\mathcal{P}(\kappa)\cap L\subseteq L_{\kappa^+}$ by condensation - we have in $L_\beta$ that $\alpha$ starts a $\kappa$-gap of length at least $\mu=$ the image of $\lambda$ under the Mostowski collapse of $M$. But by hypothesis on $M$, $\mu=\lambda$, so we're done. $\Box$

Similarly, we can show that $\kappa$-gaps of arbitrarily great length $<\kappa^+$ occur cofinally in $\kappa^+$.


Now, what about the least $\alpha>\kappa$ which starts a $\kappa$-gap? Here I know virtually nothing - in particular, "the least $\alpha>\kappa$ which starts a $\kappa$-gap" is pretty concrete to me, so improving it seems a hard task - but here's a brief comment:

In the case $\kappa=\omega$, we can try to characterize this $\alpha$ as follows: by finding some appropriate notion of correctness for second-order arithmetic such that $\alpha$ is the least ordinal such that $L_\alpha\cap\mathcal{P}(\omega)$ is correct in this sense.

I'm restricting attention to second-order arithmetic for concreteness; we could look instead at fragments of set theory, and this seems in general more fruitful, but for now let's stay "low-down."

Leeds and Putnam showed that the ordinals which start $\omega$-caps all yield $\beta$-models of second-order arithmetic. However, I don't know if the least such ordinal is also the least gap.

For $\kappa>\omega$, an analogue of the Leeds-Putnam result should hold, but again I don't see that it gives a characteriztion of the corresponding $\alpha$.

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