4
$\begingroup$

A homework(ish) problem from models of set theory:

Define $\varphi(x) :\leftrightarrow Lim(x) \land \forall y\in x \, (Lim(y)\rightarrow y=0)$ where $Lim(x)$ means that $x$ is a limit ordinal. $\varphi$ says that $x=\omega$. I can prove that $\varphi(x)$ is a definite formula : For any transitive model $M\subseteq V$ of $ZF$ and $x\in M$, $\varphi(x)$ is true iff the relativation $\varphi(x)^M$ is true. (because all quantifiers used in the definition of $\varphi(x)$ are bounded)

Now, being asked to prove that the term $\omega= \bigcap \{ x \, |\, Ind(x) \,\}$ is definite (where $Ind(x)\leftrightarrow 0\in x\land \forall y\in x : y\cup \{y\}\in x$), I want to argue as follows.

ZF proves $\exists x \varphi(x)$ and $\forall x (\varphi (x) \leftrightarrow x=\omega)$. So for a model $M$ of ZF, the relativations $\exists x\in M \varphi(x)$ and $\forall x\in M (\varphi(x)\leftrightarrow x=\omega^M)$ should be true. The first relativation implies $\omega\in M$ and the second implies $\omega^M=\omega$.

I have seen a proof that $\omega^M=\omega$ (for transitive models M of ZF) using a more direct argument, but is this also correct?

$\endgroup$
  • $\begingroup$ Are you talking about transitive models? $\endgroup$ – Asaf Karagila Jun 29 '13 at 17:45
  • $\begingroup$ Yes, M is supposed to be transitive. Will edit.. $\endgroup$ – user35359 Jun 29 '13 at 17:57
  • $\begingroup$ Nitpick: Since you wrote $\forall y\in x(\operatorname{Lim}(y)\to y=0)$ instead of $\forall y\in x(\neg\operatorname{Lim}(y))$, it appears that you really intended $\operatorname{Lim}(x)$ to mean that $x$ is not a successor ordinal. $\endgroup$ – Brian M. Scott Jun 29 '13 at 20:32
  • $\begingroup$ @Brian: I met (on this site) several people who were given definitions that $0$ is a limit ordinal. Not everyone work with the definition that $0$ is the unique non-successor and non-limit ordinal. $\endgroup$ – Asaf Karagila Jun 29 '13 at 21:11
  • $\begingroup$ @Asaf: They may be stuck using those definitions, but as far as I’m concerned, calling $0$ a limit ordinal is seriously abusing the term limit. At the very least they should be aware that for many of us $0$ is not a limit ordinal. (Mind you, there’s certainly nothing wrong in defining the predicate $mathrm{Lim}$ to include $0$.) $\endgroup$ – Brian M. Scott Jun 29 '13 at 21:14
0
$\begingroup$

Your proof is essentially right. Someone might object that, where you wrote "So for a model $M$ of ZF, the relativations $\dots$ should be true," you should write the actual relativizations, which involve $\varphi^M$, and then invoke the first paragraph of your question to replace $\varphi^M$ with $\varphi$. But this is just a question of how much detail to include.

As long as people are pointing out in the comments that it is customary not to call $0$ a limit ordinal, I might as well also point out the usual terminology "relativizations" (instead of "relativations") and "absolute" (instead of "definite").

$\endgroup$

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.