# Finite fragments of ZFC

This is a question regarding exercise (II.2.15) in Kunen's Set Theory (2011):

"We shall see later that $ZFC\vdash$Con$(\Gamma)$ whenever $\Gamma$ is a finite subset of $ZFC$. Use this to define explicitly, in $ZFC$, a binary relation $E$ on $\omega$ such that $ZFC\vdash\varphi^{\omega,E}$ for each axiom $\varphi$ of $ZFC$".

My understanding is that I am proving a scheme in the metatheory. ie given an axiom $\varphi$ of $ZFC$, I want to prove that $ZFC\vdash[(\omega;E)\models\varphi$]. I understand that this does not necessarily imply $ZFC\vdash[(\omega;E)\models ZFC]$ since $(\omega;E)\models ZFC$ is a sentence of $ZFC$ and if this were the case then we would have $ZFC\vdash$Con$(ZFC)$.

Is my understanding correct thus far?

"List the axioms of $ZFC$ in some computable way, as {$\xi_i:i<\omega$}; let $ZFC_n=${$\xi_i:i<\omega$}. Working in $ZFC$, we can define $\Gamma$ to be $ZFC$ if Con$(ZFC)$; if $\neg$Con$(ZFC)$, let $\Gamma=ZFC_n$ where $n$ is largest such that Con$(ZFC_n)$. Then clearly Con$(\Gamma)$, and observe that the proof of the Completeness Theorem yields an explicit $E$ such that $(\omega;E)\models \Gamma$".

Now my understanding is that for each axiom $\varphi$ of $ZFC$ I want to show:

1) $ZFC+$Con$(ZFC)\vdash[(\omega;E)\models \varphi]$

2) $ZFC+\neg$Con$(ZFC)\vdash[(\omega;E)\models \varphi]$

and after showing 1) and 2), it will then follow that $ZFC\vdash[(\omega;E)\models\varphi$].

1) seems straightforward, since arguing in $ZFC+$Con$(ZFC)$, I get that $ZFC$ has a model by the Completeness theorem, and so I can get a countable structure $(\omega,E)$ to model $ZFC$ with a bit of work and by using the proof of the Completeness Theorem. So since I've shown that $ZFC+$Con$(ZFC)\vdash[(\omega;E)\models ZFC]$, then it follows that $ZFC+$Con$(ZFC)\vdash[(\omega;E)\models \varphi]$ for each axiom $\varphi$ of $ZFC$.

For 2), I can argue in $ZFC+\neg$Con$(ZFC)$ that some finite fragment of $ZFC$ must be inconsistent by Compactness. So then I get:

$ZFC+\neg$Con$(ZFC)\vdash\exists n\in\omega [$Con$(ZFC_n)\wedge \forall m\in\omega[m>n \implies \neg$Con$(ZFC_m)]]$. So $ZFC+\neg$Con$(ZFC)\vdash$Con$(ZFC_i)\wedge \forall m\in\omega[m>i \implies \neg$Con$(ZFC_m)]]$ for some $i<\omega$

Now I want to use the fact that $ZFC\vdash$Con$(\Gamma)$ whenever $\Gamma$ is a finite subset of $ZFC$, to say that $ZFC\vdash$Con$(ZFC_{i+1})$, which means my reasoning is definitely flawed.

Assuming I am approaching this exercise correctly, how do I go about showing 2)?

I apologize for the long length in explaining my question.

Any help is appreciated, thanks.

We want to define $E$ over $\mathrm{ZFC}$ such that $\mathrm{ZFC}\vdash\varphi^{\omega,E}$ for every $\varphi\in\mathrm{ZFC}$. If $\mathrm{ZFC}$ is inconsistent in the real world, then it proves everything and we are done. So we may assume $\mathrm{ZFC}$ is consistent in the real world. I think this is the key point of the argument.

Let $\varphi\in\mathrm{ZFC}$. To show that $\mathrm{ZFC}\vdash\varphi^{\omega,E}$, we take an arbitrary model $V\models\mathrm{ZFC}$, and attempt to prove $V\models\varphi^{\omega,E}$. If $V\models\mathrm{Con}(\mathrm{ZFC})$, then we are in the ‘straightforward’ case as you observed. So we may assume $V\models\neg\mathrm{Con}(\mathrm{ZFC})$.

Now $V\models\mathrm{ZFC}+\neg\mathrm{Con}(\mathrm{ZFC})$. Following Kunen's hint, let $n$ be the largest $n\in\omega^V$ such that $V\models\mathrm{Con}(\mathrm{ZFC}_n)$. This $n$ must be bigger than all natural numbers $i$ in the real world because of what Kunen says at the beginning. As a result, if we apply the completeness theorem in $V$ to find a (countable) model $M$ of $\mathrm{ZFC}_n$ in $V$, then $M\models\mathrm{ZFC}_i$ for every natural number $i$ in the real world, which is the same as saying $M\models\mathrm{ZFC}$ in the real world. Notice $V\not\models(M\models\mathrm{ZFC})$ in this case.

Sorry for referring to the ‘real world’ so many times. I cannot find a better phrase for it. Suggestions are welcome.

• Thanks for the reply, but I'm still a bit confused about your explanation. – user52534 Aug 11 '13 at 0:10
• By "we are assuming $ZFC$ is consistent in the real world" do you mean that we are arguing in $ZFC+$Con$(ZFC)$? If we are arguing in $ZFC+$Con$(ZFC)$, then don't we get a countable set model $M$ s.t. $M\models ZFC$ by the proof of the Completeness theorem? Your argument seems to argue in $ZFC+$Con$(ZFC)$ the following: By the 2nd Incompleteness Thm, $ZFC\nvdash$Con$(ZFC)$, so that we have that $ZFC+\neg$Con$(ZFC)$ is consistent, so there is a set model $V$ s.t. $V\models ZFC+\neg$Con$(ZFC)$. Is this correct? Also, how are you producing the countable set model $M$ of $ZFC_n$? -Thanks – user52534 Aug 11 '13 at 0:26
• @user52534: Sorry, I was a little sloppy. I expanded my answer in response to your questions. As the completeness theorem holds in $V$, and $V$ knows $\mathrm{ZFC}_n$ is consistent, the model $V$ is able to produce a (countable) model of $\mathrm{ZFC}_n$. If anything else is unclear, feel free to request for further clarifications. – Lawrence Wong Aug 11 '13 at 12:42
• Thanks again for your patience, I think I've almost got it. My understanding is that you've proved the sentence "$ZFC\vdash\exists M[M$is countable$\wedge M\models ZFC]$" by using a model theoretic argument formalized inside of $ZFC$. In other words, if we let $\psi$ be the aforementioned sentence, then you've shown that $ZFC\vdash\psi$ by first showing that $ZFC+\neg$Con$(ZFC)\vdash\psi$ and then showing that $ZFC+$Con$(ZFC)\vdash\psi$. Is this correct? – user52534 Aug 13 '13 at 18:46
• If what I said above is not correct, then my guess would be that if we let $\theta$ be the sentence $\exists M[M$is countable$\wedge ZFC\vdash [M\models ZFC]]$, then you've shown that $ZFC\vdash\theta$. Thanks in advance. – user52534 Aug 13 '13 at 18:47