I have been trying to show that if $\gamma$ is such that there is a real (an element of $2^\omega$) in $L_{\gamma+1} \setminus L_\gamma$, then there are countable $M$ and a surjection $f : \omega \rightarrow M$ such that $(M, \in) \cong (L_\gamma, \in)$ and $f \in L_{\gamma + \alpha}$ for some "small" ordinal $\alpha$, a result I recall is by Enderton and Putnam.

The proof I have in mind goes like this: let $\bar a \in L_\gamma$ be the parameters used to define a real $z \subseteq L_\gamma$ and let $M := \{x \in L_\gamma : \text{$x$ is definable with parameters $\bar a$ as a singleton that is a subset of $L_\gamma$ }\}$. Then $M \prec L_\gamma$, and the Mostowski collapse $\bar M$ of $M$ models $V=L$ because $\bar M \cong M \equiv L_\gamma$ and thus $\bar M \models V=L$, meaning $\bar M = L_\beta$ for some $\beta$. Since $ z$ is definable in $\bar M$, we conclude $\beta = \gamma$.

To take the Mostowski collapse of $M$ we need to show that it is extensional. An obvious thing to try is to appeal to a definable well-order $<_L$ in $L$ to get a definition of a counterexample of extensionality as a singleton. (By a counterexample I mean $y$ such that $y \in x$ and $y \not \in x'$, where $x \cap M = x' \cap M$ and $x, x' \in M$). This does not work, since $M$ is not definable in $L_\gamma$.

How can one show the extensionality?


1 Answer 1


I am assuming $\prec$ refer to elementary embedding.

At first glance, I am not sure why your $M$ should be an elementary substructure.

However, if you define $M$ to be the Skolem hull of $z$, then you should be fine.

But then extensionality is first order definable:

$(\forall x)(\forall y)(((\forall z)(z \in x \Leftrightarrow z \in y)) \Rightarrow x = y)$

Since $L_\gamma$ obviously satisfies extensionality and $M$ is an elementary substructure, $M$ satisfies extensionality as well.

  • $\begingroup$ My $M$ is the Skolem hull, so to speak. $\endgroup$
    – Pteromys
    Oct 15, 2015 at 22:33

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