# Hypotheses for Gödel's condensation lemma

Gödel's condensation lemma is that given a suitable $$L_{\alpha}$$, and given $$X \prec L_{\alpha}$$, then upon taking the Mostowski collapse on $$X$$ we are returned with $$L_{\beta}$$ for some $$\beta < \alpha$$.

I'm concerned about the exact requirements on $$\alpha$$ or $$L_{\alpha}$$. The way I learnt it was that $$L_{\alpha} \vDash T$$ is required, where $$T$$ is the finite fragment of $$ZF-P$$ that is used to formulate model theory, definability and construct the $$L$$-hierarchy. Then $$L_{\alpha} \vDash V = L$$, and is correct about the levels of the $$L$$ hierarchy so that $$(L)^{L_{\alpha}} = L_{\alpha}$$ and $$(L_{\gamma})^{L_{\alpha}} = (L_{\gamma})$$ for $$\gamma < \alpha$$. Then by elementarity + isomorphism the same holds for $$\pi(X)$$ ($$\pi$$ being the Mostowski collapse) and from this the result easily follows.

However it seems that we can loosen the requirements beyond $$L_{\alpha} \vDash T$$. I've seen weakenings to $$\alpha$$ being a limit or $$\alpha$$ being such that $$\alpha = \omega \cdot \alpha$$. I'm not sure how this is supposed to work. If $$\alpha$$ is simply a limit, it seems to me that it might not satisfy a good chunk of $$ZF-P$$ and potentially $$T$$. For instance the proof of $$(Comprehension)^L$$ uses the reflection theorem scheme, and simply assuming $$\alpha$$ a limit, it seems that $$(Comprehension)^{L_{\alpha}}$$ may not work. Is Comprehension then not needed to formulate model theory, the $$L$$-hierarchy, etc needed in the proof above? If $$L_{\alpha} \not \vDash T$$, then it seems that the traditional argument (or the one I used above) doesn't follow through.

So how far can the hypotheses on $$\alpha$$ be weakened? Can $$\alpha$$ really just be a limit ordinal or that $$\alpha = \omega \cdot \alpha$$? Does this follow simply from being very careful in observing what exactly is used in $$T$$ above and coming to the conclusion that limits are enough? If that was the case, then I was hoping to see/understand this analysis and find out what exactly is needed, and what is the complexity of this when put as a sentence, etc. Or is it a different proof entirely?

• It works if $\alpha$ is admissible, that is, if $L_\alpha$ is a model of $\mathsf{KP}$. Feb 8, 2021 at 16:53
• @HanulJeon It works for arbitrary limit $\alpha$ in fact, and with only the hypothesis of $\prec_1$ in place of $\prec$ - see Devlin's proof here (page $80$). Feb 8, 2021 at 16:59
• @NoahSchweber Thank you for your reply. I did know that condensation holds for Jensen's $J$-hierarchy with $\prec_1$, and $J_\alpha=L_\alpha$ if $\alpha$ is admissible. I did not know that condensation holds for every limit ordinal, however. Feb 8, 2021 at 17:05
• @Noah I thought it held for successor levels too (though with much more difficult proof and maybe not for just $\Sigma_1$). Or am I making that up? Feb 8, 2021 at 17:34
• @spaceisdarkgreen I believe it does, but since I don't have a source on-hand I didn't want to claim it. Feb 8, 2021 at 17:38

Suppose $$\alpha$$ is a limit ordinal and $$X\prec_1L_\alpha$$. Then $$X\cong L_\beta$$ for a unique ordinal $$\beta$$, and the Mostowski collapse provides the unique isomorphism.
Here "$$\prec_1$$" refers to elementarity for $$\Sigma_1$$ formulas only: $$A\prec_1 B$$ iff $$A$$ is a substructure of $$B$$ and for every $$\Sigma_1$$ formula $$\varphi(\overline{x})$$ and every $$\overline{a}\in A$$ we have $$A\models\varphi(\overline{a})\iff B\models\varphi(\overline{a})$$.
The proof isn't really different from the proof of the weaker argument that this holds with $$\prec$$ in place of $$\prec_1$$ and restricting attention to $$\alpha$$s such that $$L_\alpha$$ satisfies a reasonably strong set theory, it's just much more tedious. Part of this is essentially the realization that arbitrary limit levels of $$L$$ satisfy a larger fragment of set theory than one might expect; the other part is showing that the construction of the $$L$$-hierarchy is a $$\Sigma_0$$ proccess in a precise sense. These fine-grained analyses are moot if we assume that $$L_\alpha$$ satisfies a reasonable fragment of set theory.
Devlin's presentation (page $$80$$) of this is quite good in my opinion. Devlin's book has serious flaws elsewhere, but this material is solid.
• Thanks for the link. So basically by a painful analysis of the complexity of $T$ you can conclude that being a limit is enough and that the sentence capturing $T$ is $\Sigma_1$! I vastly overestimated how complex $T$ is it seems. Feb 8, 2021 at 17:09