4
$\begingroup$

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 $ZFC \vdash CON(\ulcorner ZFC-P \urcorner)$.

$\endgroup$
3
  • 1
    $\begingroup$ 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

6
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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