To simplify the notation, assume $V=L$. We have $\lvert V_{\omega_{1}} \rvert=\aleph_{\omega_{1}}$ and $\lvert H(\aleph_{1})\rvert=\aleph_{1}$, so in particular $V_{\omega_{1}} \models \exists x \forall \alpha\; x \not\in L(\alpha) $ since $L(\omega_{1})=H(\aleph_{1})$.

Using the Löwenheim–Skolem theorem we have a transitive countable set $M\prec V_{\omega_{1}}$. In particular, $M\in H(\aleph_{1})$. We have $H(\aleph_{1})\prec_{1}V_{\omega_{1}}$, so $M\prec_{1}H(\aleph_{1})$. By the condensation lemma $M=L(\alpha)$. $\alpha$ must be a limit ordinal, but $L(\beta)\in L(\beta+1)$, so $L(\alpha)\models \forall x\exists \beta \; x\in L(\beta)$ for all ordinal limit $\alpha$. This is a contradiction with $M\prec V_{\omega_{1}}$.

Where is the mistake? I didn't find it.

Edit: I think the Mostowski collapse lemma is not the core of problem. We have that exist a $\alpha$ limit with $V_{\omega_{1}}\in L(\alpha)$, and we have $V_{\omega_{1}}\prec_{1}L(\alpha)$, but as $V_{\omega_{1}}$ is transitive his transitive collapse is himself, but is not possible.

The heart of problem is the relation $V_{\omega_{1}}\prec_{1} L(\alpha)$, but why is it wrong?

Edit: thanks the comentaries the solution is, despite $V_{\omega_{1}}$ and $L(\alpha)$ satisfies the same $\Sigma_{1}$ sentences $\varphi$, we need to consider all formulas $\varphi(x)$ that are $\Sigma_{1}$

  • 2
    $\begingroup$ I don't think LS theorem guarantees $M$ is transitive. $\endgroup$
    – Wojowu
    Mar 3, 2022 at 23:33
  • $\begingroup$ no, but we can take his transitive collapse and we have $M\in H(\aleph_{1})$ $\endgroup$ Mar 3, 2022 at 23:34
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    $\begingroup$ @LSpice respectfully disagree. He isn't asking us to check his argument, he knows it's wrong. It seems like a reasonable mathematical question, and more importantly, may stimulate an answer from an expert that will be educational for us. $\endgroup$
    – Nik Weaver
    Mar 4, 2022 at 1:26
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    $\begingroup$ @NoahSchweber already showed where the mistake is. But regarding the edit, the relation $V_{\omega_1}\preccurlyeq_1 L_\alpha$ is not true (for any $\alpha$). For it requires that $V_{\omega_1}\subseteq L_\alpha$, hence $\alpha\geq\aleph_{\omega_1}$. But then for each $x\in V_{\omega_1}$, $L_\alpha\models$"$x$ is constructible" (a $\Sigma_1$ assertion about $x$), but there are $x\in V_{\omega+2}$ such that $V_{\omega_1}\models$"$x$ is not constructible". $\endgroup$
    – Farmer S
    Mar 4, 2022 at 12:16
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    $\begingroup$ ...I'm not sure what your point is. We have a $\Sigma_1(x)$ assertion, about some $x\in V_{\omega_1}$, which is false over $V_{\omega_1}$ but true over $L_\alpha$. So $V_{\omega_1}\not\preccurlyeq_1 L_\alpha$. $\endgroup$
    – Farmer S
    Mar 4, 2022 at 12:28

1 Answer 1


Your mistake is that taking the Mostowski collapse does not preserve elementarity.

We do have a countable transitive $A$ and a countable $M$ with $$A\cong M\preccurlyeq V_{\omega_1},$$ where $M$ comes from downward Lowenheim-Skolem applied to $V_{\omega_1}$ and $A$ is the collapse of $M$, but that does not imply $A\preccurlyeq V_{\omega_1}$. Elementary substructurehood is more than just agreement of theories, it also takes into account exactly how the smaller structure sits inside the larger one. Moving the smaller structure around (e.g. via the Mostowski collapse) may break elementarity.

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    $\begingroup$ And in this case, any elementary submodel will contain many $V_\alpha$s which are uncountable, which witness its intransitivity. $\endgroup$
    – Asaf Karagila
    Mar 4, 2022 at 6:23
  • $\begingroup$ I don't think that is the heart of problem $\endgroup$ Mar 4, 2022 at 12:03
  • $\begingroup$ @AsafKaragila the sentences absolute over transitive models are $\Delta_{1}$, but be uncountable is $\Pi_{1}$, so i din't understand your point $\endgroup$ Mar 4, 2022 at 12:08
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    $\begingroup$ @Ândsonjosé: If $M\prec V_{\omega_1}$, then $V_{\omega+1}\in M$, but $M$ is countable, so $V_{\omega+1}$ is not a subset of $M$. When you collapse $M$ to be transitive, there is some $x\in M$ such that $M$ 'thinks' that $x$ is $V_{\omega+1}$, but that $x$ is not, itself, $V_{\omega+1}$. $\endgroup$
    – Asaf Karagila
    Mar 4, 2022 at 12:13

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