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

The exercise reads:

"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?

Now the hint reads:

"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.

  • $\begingroup$ Thanks for the reply, but I'm still a bit confused about your explanation. $\endgroup$ – user52534 Aug 11 '13 at 0:10
  • $\begingroup$ 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 $\endgroup$ – user52534 Aug 11 '13 at 0:26
  • $\begingroup$ @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. $\endgroup$ – Lawrence Wong Aug 11 '13 at 12:42
  • $\begingroup$ 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? $\endgroup$ – user52534 Aug 13 '13 at 18:46
  • $\begingroup$ 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. $\endgroup$ – user52534 Aug 13 '13 at 18:47

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.