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?

  • $\begingroup$ It works if $\alpha$ is admissible, that is, if $L_\alpha$ is a model of $\mathsf{KP}$. $\endgroup$
    – Hanul Jeon
    Feb 8, 2021 at 16:53
  • 1
    $\begingroup$ @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 $80$). $\endgroup$ Feb 8, 2021 at 16:59
  • $\begingroup$ @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. $\endgroup$
    – Hanul Jeon
    Feb 8, 2021 at 17:05
  • $\begingroup$ @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? $\endgroup$ Feb 8, 2021 at 17:34
  • $\begingroup$ @spaceisdarkgreen I believe it does, but since I don't have a source on-hand I didn't want to claim it. $\endgroup$ Feb 8, 2021 at 17:38

1 Answer 1


In fact, condensation is very strong:

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.

  • $\begingroup$ 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. $\endgroup$ Feb 8, 2021 at 17:09

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