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I have a question regarding the following passage in Just/Weese (p 192), $\mathbf{V}$ denoting the cumulative hierarchy:

"But consider the following situation:

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Where in "($\beta$)" does the $\mathbf{V}$ come from? As I understand it, "($\alpha$)" is the statement that from $ZF$ we can prove $(AC)_{/\mathbf X} \land (GCH)_{/\mathbf X} \land \varphi_{0/\mathbf X} \land \dots \land \varphi_{n/\mathbf X}$ for every $n$. Rewriting "($\alpha$)" in symbols gives me $\forall n \in \omega( ZF \vdash (AC)_{/\mathbf X} \land (GCH)_{/\mathbf X} \land \varphi_{0/\mathbf X} \land \dots \land \varphi_{n/\mathbf X})$. How is this the same as "($\beta$)"? Thanks for your help.

Adding some context:

  • Given a class $\mathbf{X}$ and a formula $\varphi$ in the language of set theory ($L_S$), Just-Weese use $\varphi_{/\mathbf{X}}$ to denote the relativization of $\varphi$ to $\mathbf{X}$; what is perhaps more commonly denoted $\varphi^{\mathbf{X}}$.
  • $\{ \varphi_n : n \in \omega \}$ is an enumeration of the axioms of ZF.

The question boils down to "Why are ($\alpha$) and ($\beta$) logically equivalent?", that is, "Why $ZF \vdash (\alpha) \leftrightarrow (\beta)$?".

$ZF \vdash (\alpha) \leftarrow (\beta)$ is clear: If we assume $ZF + (\beta)$ and fix $n \in \omega$ then we want to show that $ZF \vdash (AC)_{/\mathbf X} \land (GCH)_{/\mathbf X} \land \varphi_{0/\mathbf X} \land \dots \land \varphi_{n/\mathbf X}$. If $ZF$ is inconsistent then we're done. If $ZF$ is consistent then it has a model. Let's call it $\mathbf{V}$. We then use $(\beta)$ to obtain $\mathbf X \models ZF + AC + GCH$ which is the same as $(ZF)_{/\mathbf X} + (AC)_{/\mathbf X} + (GCH)_{/\mathbf X}$ from which $(\alpha)$ immediately follows.

To see $ZF \vdash (\alpha) \rightarrow (\beta)$, assume $ZF + (\alpha) + (\mathbf{V}\models ZF)$. The goal is to show $\mathbf{X} \models ZF + AC + GCH$ which by definition 13(b) is the same as saying that $(ZF)_{/\mathbf X}$ and $(AC)_{/\mathbf X}$ and $(GCH)_{/\mathbf X}$ hold in $\mathbf V$. Since we assumed $ZF$ we get $(AC)_{/\mathbf X}$ and $(GCH)_{/\mathbf X}$ by applying $(\alpha)$.

But how to prove $(ZF)_{/ \mathbf X}$? One cannot argue as follows: For every $n \in \omega$ we have $ZF \vdash \varphi_{0/\mathbf X} \land \dots \land \varphi_{n/\mathbf X}$ by $(\alpha)$ and since we assumed $\mathbf V \models ZF$ it follows that $\mathbf X \models \{\varphi_{0}, \dots ,\varphi_{n} \}$. From $\mathbf X \models \{\varphi_{0}, \dots ,\varphi_{n} \}$ for every $n \in \omega$ it follows that $\mathbf X \models ZF$. This does not work since "$\varphi_0 \land \varphi_1 \land \varphi_2 \land \dots$" is not a well-formed formula.

Question: How to prove $(ZF)_{/ \mathbf X}$ from $(\alpha)$, ZF and $\mathbf V \models ZF$?

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Could you remind me what the $/X$ notation means? Also, for any theory $T$ saying that $V \models T$ is just saying that $T$ holds (I don't know if that's relevant to what you're asking.) – Trevor Wilson Dec 18 '12 at 22:30
Isn't this just the completeness theorem within $\mathbf{V}$? – Quinn Culver Dec 19 '12 at 6:34
What are the $\varphi_n$'s? – Trevor Wilson Dec 19 '12 at 6:39
Thank you, @Arthur! – Rudy the Reindeer Dec 19 '12 at 8:36
I've been thinking about this question for some time now. I feel the authors get a bit confused between treating $\vDash$ as a metatheoretic assertion and a formal one. If this is treated metatheoretically, the argument you give should be fine. If, on the other hand, $\mathbf{X}\vDash\mathrm{ZFC+GCH}$ stands for $\forall x\in\mathrm{ZFC+GCH}:\mathit{Bew}_{\mathrm{ZF}}(x_{/\mathbf{X}})$ or something similarly horrible, the same argument, but formalized, should still work. – Miha Habič Dec 25 '12 at 11:53

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