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I am studying with Jech's book. He claims that

The existence of inaccessible cardinals is not provable in $\mathsf{ZFC}$. Moreover, it cannot be shown that the existence of inaccessible cardinals is consistent with $\mathsf{ZFC}$.

And he gives the proof for the second part:

To prove the second part, assume that it can be shown that the existence of inaccessible cardinals is consistent with $\mathsf{ZFC}$; in other words, we assume if $\mathsf{ZFC}$ is consistent, then so is $\mathsf{ZFC} + \mathsf{I}$ where $\mathsf{I}$ is the statement “there is an inaccessible cardinal.” We naturally assume that $\mathsf{ZFC}$ is consistent. Since $\mathsf{I}$ is consistent with $\mathsf{ZFC}$, we conclude that $\mathsf{ZFC} + \mathsf{I}$ is consistent. It is provable in $\mathsf{ZFC} + \mathsf{I}$ that there is a model of $\mathsf{ZFC}$. Thus the sentence “$\mathsf{ZFC}$ is consistent” is provable in $\mathsf{ZFC} + \mathsf{I}$. However, we have assumed that “$\mathsf{I}$ is consistent with $\mathsf{ZFC}$” is provable, and so “$\mathsf{ZFC} + \mathsf{I}$ is consistent” is provable in $\mathsf{ZFC} + \mathsf{I}$. This contradicts Gödel’s Second Incompleteness Theorem.

And he adds

“it cannot be shown” means: It cannot be shown by methods formalizable in $\mathsf{ZFC}$.

So his proof claims that if we assume that $\mathsf{ZFC}$ is consistent, $\mathsf{ZFC} \nvdash \mathrm{Con}(\mathsf{ZFC}) \to \mathrm{Con}(\mathsf{ZFC} + \mathsf{I})$, right? I am confused here. Can we claim that $\mathsf{ZFC} + \mathsf{I}$ is consistent in the metatheoretical sense provided that a consistent theory($\mathsf{ZFC}$) proves that if itself is consistent, then $\mathsf{ZFC} + \mathsf{I}$ is consistent? If $\mathsf{ZFC} \vdash \neg\mathrm{Con}(\mathsf{ZFC})$ (then $\mathsf{ZFC}$ is not 1-consistent) still $\mathsf{ZFC} \vdash \mathrm{Con}(\mathsf{ZFC}) \to \mathrm{Con}(\mathsf{ZFC} + \mathsf{I})$ holds, but then since $\mathsf{ZFC} + \mathsf{I} \vdash \mathrm{Con}(\mathsf{ZFC})$, $\mathsf{ZFC} + \mathsf{I}$ is inconsistent.

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  • $\begingroup$ You raise a good point. If ZFC proves itself inconsistent, then it does prove the relative consistency of ZFC and ZFC + I... namely, it proves they're both inconsistent. (But then again, we can't trust it on these matters, since it's arithmetically unsound, if not inconsistent.) But I'm not sure I understand the question you're asking, or what point you're trying to make by showing (perfectly correctly) that if ZFC proves itself inconsistent, then ZFC+I is inconsistent. $\endgroup$ Apr 15, 2020 at 14:47
  • $\begingroup$ @spaceisdarkgreen I am confused while reading his proof. To use the Gödel’s Second Incompleteness Theorem to make a contradiction, we should know that $\mathsf{ZFC} + \mathsf{I}$ is consistent metatheoretically. $\endgroup$
    – Ris
    Apr 15, 2020 at 14:58
  • $\begingroup$ Of course we don't know that, though. But what if it is inconsistent? What can we conclude then? $\endgroup$ Apr 15, 2020 at 15:03
  • $\begingroup$ @spaceisdarkgreen If $\mathsf{ZFC} + \mathsf{I}$ is inconsistent, then $\mathsf{ZFC} + \mathsf{I} \vdash \mathrm{Con}(\mathsf{ZFC} + \mathsf{I})$ does not lead to any contradiction, so I think his proof is wrong. $\endgroup$
    – Ris
    Apr 15, 2020 at 15:04
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    $\begingroup$ No, but only in the corner case you mentioned where ZFC proves itself inconsistent. If ZFC+I is inconsistent, then ZFC can prove that, so the only way it can prove Con(ZFC)-> Con(ZFC+I) is to prove not Con(ZFC). $\endgroup$ Apr 15, 2020 at 15:28

1 Answer 1

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After writing this answer, I realized that spaceisdarkgreen already explained this in the comment thread above; if they leave an answer, I'll delete this one.


Yes, there's an issue here. What we really have is the following:

"In $\mathsf{ZFC}$ (or indeed much less$^1$), we can prove that the following are equivalent:

  1. $\mathsf{ZFC}\not\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$.

  2. $\mathsf{ZFC}\not\vdash \neg Con(\mathsf{ZFC})$.

Note that the latter is intermediate between $Con(\mathsf{ZFC})$ and $\Sigma_1$-$Sound(\mathsf{ZFC})$ (the latter of which in turn is a very weak fragment of arithmetical soundness).


The $\neg 2\rightarrow \neg 1$ direction is exactly what you've observed: if $\mathsf{ZFC}\vdash \neg Con(\mathsf{ZFC})$, then $\mathsf{ZFC}\vdash Con(\mathsf{ZFC})\rightarrow\varphi$ for every sentence $\varphi$.

Now we want to show $\neg1\rightarrow\neg 2$. This basically parallels Jech's argument. There are three steps, each of which is provable in $\mathsf{ZFC}$ (or indeed much less):

  • Monotonicity. Suppose $\mathsf{ZFC}\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$. Then a fortiori we have $\mathsf{ZFC+I}\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$, and so $\mathsf{ZFC+I}\vdash Con(\mathsf{ZFC+I})$.

  • Godel's second incompleteness theorem. From this and the previous bulletpoint we get $\neg Con(\mathsf{ZFC+I})$.

    • Note - addressing one of your comments - that no additional assumption here is necessary: "if $\mathsf{ZFC+I}$ is consistent then GSIT applies and so $\mathsf{ZFC+I}$ is inconsistent" is already a deduction of $\neg Con(\mathsf{ZFC+I})$.
  • $\Sigma_1$-completeness. The previous bulletpoint implies $\mathsf{ZFC}\vdash\neg Con(\mathsf{ZFC+I})$. But now combining this with our original hypothesis $\neg 1$, we get $$\mathsf{ZFC}\vdash \neg Con(\mathsf{ZFC+I})\wedge[Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})],$$ which in turn yields $$\mathsf{ZFC}\vdash\neg Con(\mathsf{ZFC})$$ as desired.


$^1$Mathematical limbo - how low can we go?

As the argument above shows, we really just need our metatheory to prove three things:

  • Monotonicity of $\vdash$.

  • Godel's second incompleteness theorem.

  • The $\Sigma_1$-completeness of $\mathsf{ZFC}$.

The first is basically trivial (e.g. even Robinson arithmetic does that), while this fascinating paper of Visser mentions $\mathsf{EA}$ as an upper bound for the third ($\mathsf{EA}$ is incredibly weak, as that same paper demonstrates). Meanwhile, I believe - but don't have a source for the claim - that $\mathsf{EA}$ also proves GSIT, which would make $\mathsf{EA}$ in fact a sufficient metatheory!

However, going all the way down to $\mathsf{EA}$ - if we even can - is really just showing off. For almost all purposes it's enough to observe that $I\Sigma_1$ (a weak fragment of $\mathsf{PA}$) is enough. $I\Sigma_1$ has a number of nice properties which in my opinion do make it a better stopping point than the more-famous $\mathsf{PA}$: basically, it's the weakest "natural" theory capable of "naturally" developing basic computability theory (for example, the provably total functions of $I\Sigma_1$ are exactly the primitive recursive functions). It's also finitely axiomatizable, which is sometimes quite useful. And finally, it's the first-order part of $\mathsf{RCA_0}$, meaning that a reduction to $I\Sigma_1$ fits quite nicely in the program of reverse mathematics.

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