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Let $I$ be the statement "there is an inaccessible cardinal".

I'm aware of two proofs of "If $ZFC\vdash I$ then $ZFC$ is inconsistent".

One proof uses the Second Incompleteness Theorem, which I understand.

The other proof goes something like:

1) Suppose $ZFC\vdash I$.

2) Then we have $ZFC\vdash ^{\backprime\backprime} V_\kappa\models ZFC ^{\prime\prime}$ where $\kappa$ is the least inaccessible cardinal.

3) Then by absoluteness results, we have $ZFC\vdash ^{\backprime\backprime}V_\kappa\models\neg I ^{\prime\prime}$.

4) Thus we have a model of $ZFC+\neg I$

And this is where this version of the proof ends in all sources I've come across.

I was wondering how the rest of the proof should go to get that $ZFC$ really is inconsistent.

My idea was it is essential to use the fact:

(*) If $ZFC\vdash\varphi$ then $ZFC\vdash ^{\backprime\backprime}ZFC\vdash\varphi ^{\prime\prime}$ for any sentence $\varphi$.

Then to finish:

5) Since we've assumed $ZFC\vdash I$, then we have $ZFC\vdash ^{\backprime\backprime}ZFC\vdash I ^{\prime\prime}$ by (*).

6) And since $ZFC\vdash ^{\backprime\backprime}V_\kappa\models ZFC^{\prime\prime}$, then we have $ZFC\vdash ^{\backprime\backprime}V_\kappa\models I ^{\prime\prime}$ by using 5).

7) By 3), we have $ZFC\vdash ^{\backprime\backprime}V_\kappa\nvDash I ^{\prime\prime}$.

8) So since $ZFC\vdash ^{\backprime\backprime}V_\kappa\models I ^{\prime\prime}$ and $ZFC\vdash ^{\backprime\backprime}V_\kappa\nvDash I ^{\prime\prime}$, then $ZFC$ is inconsistent.

Is this the way the proof is typically finished or is it somehow not necessary to appeal to (*)?

Any help is appreciated, thanks.

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    $\begingroup$ I'm not a set theorist, so I could be missing a lot of detial, but I think 1) and 4) contradict one another. To wit if $ZFC\vdash I$ then by the completeness theorem $M \vDash I$ for every model $M$ of ZFC. Hence there is no model of $ZFC + \neg I$, a contradiction? $\endgroup$
    – James
    Sep 1, 2014 at 22:30
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    $\begingroup$ @James Even your method does not avoid the basic idea of what the OP wrote. 4 is not written properly. "Thus we have a model of $ZFC + \neg I$" is not accurate since it loses track of where this proof is going on. Really it should be written $\mathsf{ZFC} \vdash "\mathsf{ZFC} + \neg I$ has a model". Afterward, by noting that $ZFC \vdash$ "Godel Completeness Theorem" and applying the Godel completeness theorem inside $\mathsf{ZFC}$, then you would the contradict you described. This is essentially the OP proof. Regardless everything is happening syntaxically in $\mathsf{ZFC}$. $\endgroup$
    – William
    Sep 2, 2014 at 2:14
  • $\begingroup$ @ William, I know, 4) is a bit ambiguous, because it is really a theorem of $ZFC$ that that there is a model of $ZFC+\neg I$. The reason I left it that way is because that is how some sources word it (for instance Jech). I'm not sure what the motivation is behind the ambiguity. Also, @James, the authors restrict the theory to finitistic reasoning, so we must restrict our use of such model theoretic arguments in establishing the inconsistency of $ZFC$. $\endgroup$
    – user52534
    Sep 2, 2014 at 5:17
  • $\begingroup$ @William Are you also using (*) to obtain your contradiction in $ZFC$? $\endgroup$
    – user52534
    Sep 2, 2014 at 5:23
  • $\begingroup$ Essentially yes. What (*) really mean is that $\mathsf{ZFC}$ can formalize it own version of $\mathsf{ZFC}$. Then you can talk about models of $\mathsf{ZFC}$ (like $V_\kappa$) inside $\mathsf{ZFC}$. $\endgroup$
    – William
    Sep 2, 2014 at 5:27

1 Answer 1

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You could run the argument schematically. For instance, using induction in the metatheory you could show that (1) when $\phi$ is an axiom of ZFC, ZFC proves "if $V_\kappa$ is an inaccessible rank, then $V_\kappa\vDash \phi$"; and (2) when $\psi$ is a consequence of $\phi$, ZFC proves "if $M\vDash \phi$, then $M\vDash \psi$".

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  • $\begingroup$ Can you prove that $\sf ZFC$ proves that the object it internalizes as $\sf ZFC$ is the same object as the meta-theory's $\sf ZFC$ (in semantical terms, can you always prove that $\sf ZFC$ as perceived by the meta-universe and the universe are the same object)? Because if not, then just showing that $V_\kappa$ satisfies all the axioms we agree make "$\sf ZFC$" might not be sufficient. You need the fact that $(V_\kappa,\in)$ has a truth predicate. $\endgroup$
    – Asaf Karagila
    Sep 2, 2014 at 13:45
  • $\begingroup$ I'm not sure I follow, @AsafKaragila. Are you denying that (1) or (2) are provable in the metatheory, or that the argument to "ZFC is inconsistent" can't be carried out using them? $\endgroup$
    – user104955
    Sep 2, 2014 at 13:55
  • $\begingroup$ No, I stipulate that in the case that $M$ is a model of $\sf ZFC+I$ which has non-standard integers, $V_\kappa^M$ will satisfy more than just $\sf ZFC$, because $(\sf ZFC)^\it M\neq\sf ZFC$. $\endgroup$
    – Asaf Karagila
    Sep 2, 2014 at 13:57
  • $\begingroup$ Sorry for being slow! Here's the argument I'm proposing, using (1) and (2). Suppose ZFC proves In. Then In is a consequence of some conjunction $\phi$ of finitely many axioms of ZFC. By (1) ZFC will prove that if $V_\kappa$ is the least inaccessible rank, $V_\kappa\vDash \phi$. So by (2) ZFC will prove that if $V_\kappa$ is the least inaccessible rank, $V_\kappa\vDash In$. But ZFC proves that least inaccessible rank doesn't satisfy In. So ZFC is inconsistent. Where does the argument go wrong? Thanks in advance! $\endgroup$
    – user104955
    Sep 2, 2014 at 14:07
  • $\begingroup$ That argument is fine, but it's not quite the argument given in your answer. Your answer suggests that (1) We prove that $V_\kappa$ satisfy each individual axiom of $\sf ZFC$. In the comment you suggest to only prove that for a finite fragment sufficient for proving an inaccessible cardinal exist, and that's different. $\endgroup$
    – Asaf Karagila
    Sep 2, 2014 at 14:15

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