Let $\varphi(x,y)$ be an $\in$-formula which is absolute between transitive models of ZF minus powerset axiom. Then $\exists x\, \varphi(x,y)$ is $H(\kappa)$-absolute, where $H(\kappa)$ is the set $\{x\, |\, card(TC(\{x\}))<\kappa\}$. $\kappa$ is uncountable regular.

This is a homework question and I would prefer a hint instead of a solution.

  • $\begingroup$ Precisely what does "$\psi(y)$ is $H(\kappa)$-absolute" mean? $\endgroup$ – Andrés E. Caicedo Jul 20 '13 at 2:49
  • $\begingroup$ This means that, for any $y\in H(\kappa)$, $\psi(y)$ holds in V iff it holds in $H(\kappa)$. So in this case the task is to show that $x$ can be chosen in $H(\kappa)$ if such a x exists in V. $\endgroup$ – user35359 Jul 20 '13 at 2:52
  • $\begingroup$ Are you assuming that $\kappa$ is regular? $\endgroup$ – Andrés E. Caicedo Jul 20 '13 at 3:08
  • $\begingroup$ No, $\kappa$ is arbitrary. $\endgroup$ – user35359 Jul 20 '13 at 3:59
  • $\begingroup$ I just realized I made a mistake in the question. $\varphi$ should be absolute between models of ZF minus powerset (which includes $H(\kappa)$ for uncountable $\kappa$). Also, $\kappa$ should indeed be uncountable regular. Sorry for all that. $\endgroup$ – user35359 Jul 20 '13 at 4:03

Assuming ZFC in $V$. Since $\kappa$ is uncountable regular, $H(\kappa)$ models all axioms of ZFC except powerset. (*)

The relativization of $\exists x \, \varphi(x,y)$ to $H(\kappa)$ is $\exists x\in H(\kappa) \,\varphi(x,y)$, $\varphi$ is absolute between $V$ and $H(\kappa)$ because of (*). So it needs to be shown that, if for some $y\in H(\kappa)$ there is $x\in V$ with $\varphi(x,y)$, then there is $x\in H(\kappa)$ with $\varphi(x,y)$.

For $y\in H(\kappa)$, let $T_y:=TC(\{y\})\subseteq H(\kappa)$ with $|T_y|<\kappa$. Let $\kappa'\geq \kappa$ regular be so that if there is $x\in V$ with $\varphi(x,y)$, there is $x\in H(\kappa')$ with $\varphi(x,y)$. Let $S_y$ be a Skolem hull of $T_y$ in $H(\kappa')$ and let $M_y$ be the transitive collapse of $S_y$, with $\pi:S_y\to M_y$ being a $\in\!\!-\!\!\in$ isomorphism. $|S_y|<\kappa$ and $|M_y|<\kappa$.

Assume there is a $x'\in V$ with $\varphi(x',y)$. Then there is $x\in S_y$ with $\varphi(x,y)$. Since $S_y$ is an elementary substructure of $H(\kappa')$, it is a model of ZF without powerset axiom and $\varphi^{S_y}(x,y)$ holds, because $\varphi$ is absolute. Because $\pi$ is a isomorphism and its restriction to the transitive set $T_y$ is the identity, $\varphi^{M_y}(\pi(x),y)$ holds. Because $M_y$ is isomorphic to $S_y$, ZF without powerset is true in $M_y$ and therefore $\varphi(\pi(x),y)$.

Now $TC(\{M_y\})=\{M_y\}\cup M_y$ has cardinality less than $\kappa$, so $M_y\in H(\kappa)$. Because $H(\kappa)$ is transitive and $\pi(x)\in M_y$, $\pi(x)\in H(\kappa)$.

  • $\begingroup$ This is "morally correct" but not quite formal. There are several problems. The first is that in $\mathsf{ZFC}$ we cannot really form elementary substructures of the whole universe, so $S_y$ may not be defined. The second is that there may be many $x$ such that $\varphi(x,y)$, so you cannot simply conclude that $x\in S_y$. Besides the mathematical problems, a notational advice: Purely for clarity's sake, avoid using $H_y$, as $H(\kappa)$ already means something else. $\endgroup$ – Andrés E. Caicedo Jul 20 '13 at 6:11
  • $\begingroup$ All this can be fixed easily: Assume $\exists x\,\varphi(x,y)$. First, pick an $H(\lambda)$ with $\lambda>\kappa$ regular, large enough so there is some $x\in H(\lambda)$ with $\varphi(x,y)$ and therefore $\varphi^{H(\lambda)}(x,y)$. Now use this instead of $V$ when defining $S_y$. Note that $H(\lambda)$ satisfies $\exists t\,\varphi(t,y)$, so the same holds in its elementary substructure $S_y$. This means that in $S_y$ there is an $\hat x$ such that $\varphi^{S_y}(\hat x,y)$, etc. That $\hat x$ may be different from the $x$ we began with is irrelevant. $\endgroup$ – Andrés E. Caicedo Jul 20 '13 at 6:17
  • $\begingroup$ I have a question regarding "cannot form elementary substructure of V", if you don't mind. Trying to form a Skolem hull for a set $X$, if there is $x\in V$ with $\varphi(x,y_1,..,y_k)$ for $y_i\in X$, choose $\alpha$ so that x is in $V_{\alpha}$. Since X is a set, there is $\alpha$ so that this works for all $\varphi$ and parameters in $X$ at the same time. Then wellorder $V_{\alpha}$ and choose the smallest element for each such $\varphi$, $y_i$. Take $X_1$ as the set of those elements. Repeat this, get sets $X_i\subset X_{i+1}$, the union ($i<\omega$) should be an elem. substructure of V. $\endgroup$ – user35359 Jul 20 '13 at 6:45
  • $\begingroup$ I see this shouldn't work because it would let ZFC prove its consistency, but I don't see the mistake in the argument at the moment. $\endgroup$ – user35359 Jul 20 '13 at 6:56
  • 1
    $\begingroup$ This is a subtle issue. To formalize it, you need to define a satisfaction predicate, but this is not possible (say, because of Tarski's undefinability of truth theorem). This is discussed in Chapter 12 of Jech's Set theory, or see Kunen's The foundations of mathematics. $\endgroup$ – Andrés E. Caicedo Jul 20 '13 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.