It is possible I am missing something exceedingly obvious. We know that $V_{\kappa} \vDash ZFC$ when $\kappa$ is strongly inaccessible.

Pretty much every proof of $ZFC \not \vdash "\exists$ inaccessible cardinals" goes like:

Suppose $ZFC \vdash "\exists$ strong inaccessibles" (I'll shorten this from now one as $\exists I$). Then pick the least such one, $\kappa$. $V_{\kappa} \vDash "\exists I"$. By standard absoluteness, what $V_{\kappa}$ thinks are strongly inaccessible will indeed be strongly inaccessibles, contradicting the choice of $\kappa$ as the smallest such cardinal.

And this argument is fine to me. But I don't see why we have to go through the trouble of estabilishing absoluteness, picking the smallest one and everything. Why can't we simply do:

Suppose $ZFC \vdash "\exists I"$. Since we may formalize model theory, the truth predicate for set models, consistency statements, etc, we have that $ZFC \vdash "\exists I" \rightarrow Con(ZFC)$ (by exhibiting a set model of $ZFC$). Then we have by basic modus ponens that $ZFC \vdash Con(ZFC)$. This contradicts Godel's incompleteness theorem and then we're done.

This argument seems fine to me, and more natural too. However, most arguments I can see seem to follow the first argument instead. And the first one isn't hard per se, but it makes me wonder why we're doing all these extra steps.

Am I missing something? Does the second argument not work?


1 Answer 1


They both work. However, I don't actually think that the second one is more natural: it involves first proving the second incompleteness theorem, which is nontrivial and not obviously relevant to the specific problem at hand.

By contrast, in the first approach we give a method for building models of $\mathsf{ZFC}$ + "There are no inaccessibles" from models of $\mathsf{ZFC}$. Moreover, this method "preserves niceness" in many senses, e.g. if the starting $\mathsf{ZFC}$-model is well-founded then the $\mathsf{ZFC\neg I}$-model is well-founded as well. Note that this is not something incompleteness can do for us alone - e.g. there is no nice way to take a well-founded model of $\mathsf{ZFC}$ and produce a well-founded model of $\mathsf{ZFC+\neg Con(ZFC)}$ since there are no well-founded models of $\mathsf{ZFC+\neg Con(ZFC)}$ in the first place.

So I think the first approach actually is the better one.

  • $\begingroup$ Thanks for the clarification! I was just worried something was wrong with the argument so I wanted to make sure. I suppose you're right about which argument is nicer. I guess I just disliked the first one because I'm not fully comfortable with absoluteness yet. $\endgroup$ Jan 19, 2021 at 2:31
  • $\begingroup$ Why is there no well founded model of $ZFC + \neg Con(ZFC)$? From what little I know, the only thing I can tell about a model of $ZFC + \neg Con(ZFC)$ is that it has non standard natural numbers $\endgroup$ Jan 19, 2021 at 2:32
  • 1
    $\begingroup$ @WhyIsSetTheorySoHard If it's got nonstandard natural numbers, it's not well-founded! (The interesting situation is the converse: ill-founded $\omega$-models of $\mathsf{ZFC}$ exist and are quite fun beasties, but that's highly nontrivial.) $\endgroup$ Jan 19, 2021 at 2:32
  • $\begingroup$ I don't see how that follows :( I suspect I'm missing something stupid again. As I understand it, there will just be a bunch of $\mathbb{Z}$ chains after $\mathbb{N}$ which code "contradictions". $\endgroup$ Jan 19, 2021 at 2:34
  • $\begingroup$ By the same token, this line of thinking establishes that inaccessibles are strictly stronger than Con(ZFC), which is good to know. $\endgroup$ Jan 19, 2021 at 2:36

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