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I am wondering that if one can show $ZFC \vdash CON(\ulcorner ZFC-P \urcorner)$. There is an argument in Set Theory, An Introduction to Independence Proofs by Kunen (page 145), but I am confused about the proof.

Let $\phi$ be the formula for the coding of $ZFC-P$ in natural numbers, and $X_{ZFC-P}=\{n\in \omega :\phi(n)\}$.

By Godel Completeness Theorem as the formal sentence $\forall X (CON(X) \leftrightarrow \exists \mathfrak{M}(\mathfrak{M} \models X) )$, it is enough to prove that: $ZFC \vdash H(\omega_1) \models X_{ZFC-P} $, or say $ZFC \vdash \forall x \in X_{ZFC-P} (H(\omega_1) \models x)$. By Completeness and Soundness Theorems, it is enough to show that whenever $M$ is model of $ZFC$, $M$ models $\forall x \in X_{ZFC-P} (H(\omega_1) \models x)$. This amounts to showing for all $x \in X_{ZFC-P}$, $H(\omega_1) \models x$ is true in $M$.

However, if $M$ is a nonstandard model which has nonstandard natural numbers, $X_{ZFC-P}$ may be strictly larger than the actual collection of codings of $ZFC-P$. Let $x_0$ be the coding of a nonstandard axiom $\psi$ which has infinite length looking from outside while we have $\phi(x_0)$. In Kunen's book, they showed $H(\omega_1) \models x$ for actual axioms of $ZFC-P$, but not for infinite sentences like $\psi$.

In fact, $CON(\ulcorner ZFC-P \urcorner)$ as a formal sentence also includes possible nonstandard axioms. I am wondering that if there is a way to deal with these nonstandard axioms, or if one can show $ZFC \vdash CON(\ulcorner ZFC-P \urcorner)$.

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    $\begingroup$ What is P here? Power set axiom? $\endgroup$
    – Wojowu
    Jun 10, 2020 at 23:41
  • $\begingroup$ Yes, P is Power Set axiom. $\endgroup$
    – Feng Liang
    Jun 11, 2020 at 0:11
  • $\begingroup$ What's the meaning of ZFC $\Rightarrow$ Con(X) (where X is a set of axioms)? I don't understand "ZFC" as an assertion (while "$M$ is a model of ZFC" or "Con(ZFC)") are logical assertions). (Oh, it's just been edited, thanks) $\endgroup$
    – YCor
    Jun 12, 2020 at 12:53

1 Answer 1

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You can directly show from $ZFC$ that $\forall n \in X_{ZFC-P}\, \colon \, ( H(\omega_1) \vDash n)$. To see this remind yourself that $ \vDash$ is expressible by a single formula $\psi$, so that $H(\omega_1) \vDash \varphi(z_1,...,z_m)$ iff $\psi(H(\omega_1), \ulcorner \varphi \urcorner, \vec{z},1)$. Now for $n \in X_{ZFC-P}$ you have a finite case distincition what kind of axiom $n$ is. E.g. if $n$ is ` $\forall A \, \forall \vec{z} $ replacement for the formula $\varphi_n(x,y, \vec{z})$ with respect to $A$ holds ', let $A \in H(\omega_1)$ and $\vec{z} \in H(\omega_1)^{<\omega}$ be arbitrary and define the set $$B:=\{y \in H(\omega_1) \, \colon \, \exists x \in A \, \, H(\omega_1) \vDash \varphi_n(x,y, \vec{z})\}.$$ Finally prove that $B \in H(\omega_1)$, and so $H(\omega_1) \vDash n$. The subtlety here is that you are only using replacement with respect to $\psi$. Therefore, this is a finite proof.

On the meta-level, for every finite $\Delta \subseteq ZFC$ you can prove $CON(\Delta)$. But $ZFC$ does not prove $\forall \Delta \subseteq ZFC \, \text{finite}\, \colon CON(\Delta)$ as this would contradict Gödel's second Incompleteness Theorem.

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  • $\begingroup$ Actually I wanted to post this as a comment, but it was too long... so I put it in an answer. $\endgroup$ Jun 10, 2020 at 22:33
  • $\begingroup$ I agree with the part that if $n \in X_{ZFC-P}$ represents some real axiom of $ZFC-P$, then $H_{\omega_1} \models n$. However, $ZFC$ allows nonstandard models which have nonstandard natural numbers. Such natural numbers have infinite predecessors looking from outside but are regarded as finite numbers inside the model. When we define $X_{ZFC-P}$, we are using some formula $\phi$ such that $X_{ZFC-P}=\{ n : \phi (n) \}$. It is not clear if $X_{ZFC-P}$ contains nonstandard naturals. If $n_0 \in X_{ZFC-P}$ is nonstandard, we don't have a corresponding axiom in ZFC-P to $n_0$. - $\endgroup$
    – Feng Liang
    Jun 11, 2020 at 0:01
  • $\begingroup$ -In this case it is not clear if $H_{\omega_1} \models n_0$. $\endgroup$
    – Feng Liang
    Jun 11, 2020 at 0:02
  • $\begingroup$ @FengLiang In any model of ZFC that has nonstandard natural numbers, some of them will be in $X_{ZFC-P}$, but that makes no difference. It is a theorem of ZFC (and therefore true in all models of ZFC, even nonstandard ones) that $H_{\omega_1}$ satisfies all axioms of ZFC-P (and "all" here means all in the sense of the model, including nonstandard axioms). $\endgroup$ Jun 11, 2020 at 2:29
  • $\begingroup$ Do you know where I can find a proof of it? To be clear, we need to show: $ZFC \vdash H_{\omega_1} \models X_{ZFC-P}$, which is stronger than the statement: assuming $ZFC$, $H_{\omega_1}$ is a model of $ZFC-P$. For the latter one, it is proved as Theorem 6.5 in Chapter IV of Kunen's book. I agree with that proof, but all it proved is only all standard axioms of $ZFC-P$ are true in $H(\omega_1)$. Besides that, I don't know a proof of $ZFC \vdash H_{\omega_1} \models X_{ZFC-P}$. $\endgroup$
    – Feng Liang
    Jun 11, 2020 at 2:51

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