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Motivated by the answer Hanul Jeon kindly provided to my previous question, I have another question.

Suppose $\kappa$ is inaccessible, then by using elementary methods, we can show that for each axiom $\varphi$ of ZFC, $V_\kappa \models \varphi$.

In the books and references I have seen, such as Jech's book, they immediately derive Con(ZFC). Now in my mind, it seems we are using a crucial hypothesis, namely the $\omega$-consistency of ZFC.

What I mean by this is that saying ZFC + "$\kappa$ is an inaccessible cardinal" $\vdash (V_\kappa \models \ulcorner \text{ZFC}\urcorner)$, is ultimately about natural numbers, and we have shown $V_\kappa \models \varphi$ for $\varphi$, which are coded by standard natural numbers.

So my question boils down to: Is ZFC $\omega$-consistent? Or we can prove ZFC + "$\kappa$ is an inaccessible cardinal" $\vdash (V_\kappa \models \ulcorner \text{ZFC}\urcorner)$, with other methods?

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    $\begingroup$ When you prove "for each axiom $\varphi$ of ZFC, $V_\kappa \models \varphi$", you are proving that within ZFC, not in the metatheory. $\endgroup$
    – Eric Wofsey
    Apr 19, 2020 at 21:20
  • $\begingroup$ @EricWofsey, Yes, if we prove that statement, it's okay I guess. But what ends up happening in Jech, for example, he proves Extentionality, Pairing, etc... hold in $V_\kappa$, which is not the same as proving "for each axiom $\varphi$ of ZFC, $V_\kappa \models \varphi$", no? $\endgroup$
    – Shervin Sorouri
    Apr 19, 2020 at 21:32
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    $\begingroup$ @ShervinSorouri Except that's not what happens in Jech: Jech gives a ZFC proof of the internal statement "$V_\kappa\models ZFC$." This really boils down to just the Separation and Replacement schemes: what Jech's doing is producing a ZFC-proof of "For all inaccessible $\kappa$, $V_\kappa$ satisfies Union, Extensionality, Pairing, Powerset, Choice, Foundation, Infinity, and every instance of Replacement and Separation." This yields a genuine ZFC-proof. $\endgroup$
    – Noah Schweber
    Apr 19, 2020 at 21:33
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    $\begingroup$ @ShervinSorouri I highly doubt he proves each replacement axiom one by one. $\endgroup$
    – spaceisdarkgreen
    Apr 19, 2020 at 21:37
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    $\begingroup$ @ShervinSorouri No, he doesn't: he gives a ZFC-proof of the single sentence "$V_\kappa$ satisfies every instance of separation" (and similarly for replacement), for a total of nine (as opposed to infinitely many) sub-claims. That's a crucial distinction. $\endgroup$
    – Noah Schweber
    Apr 19, 2020 at 21:37

2 Answers 2

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No, there is no additional assumption needed here.

What is being proved in ZFC specifically is the following set of nine (or more importantly, finitely many) sentences:

  • If $\kappa$ is inaccessible then $V_\kappa\models$ Extensionality.

  • If $\kappa$ is inaccessible then $V_\kappa\models$ Pairing.

... boringboringboring ...

  • If $\kappa$ is inaccessible then $V_\kappa\models$ Powerset.

  • If $\kappa$ is inaccessible then $V_\kappa\models$ the whole Separation scheme.

  • If $\kappa$ is inaccessible then $V_\kappa\models$ the whole Replacement scheme.

Those last two points are the crucial ones: Jech is not giving a meta-argument that ZFC proves that $V_\kappa$ satisfies each specific scheme instance, but rather a ZFC proof of the internal statement that the whole scheme is satisfied at once. The key step here is the construction of a truth predicate over $V_\kappa$ (since the latter is only a set-sized structure); this enables us to directly talk about the truth values of arbitrary sentences in $V_\kappa$.

For example, in some detail here is the ZFC-proof of "If $\kappa$ is inaccessible then $V_\kappa$ satisfies every instance of Replacement:"

  • Let $\varphi(x,y)$ be an arbitrary formula and $u\in V_\kappa$ such that for all $a\in u$ there is a unique $b_a\in V_\kappa$ such that $V_\kappa\models\varphi(a,b_a)$.

  • In $V$, let $f=f_{\varphi,u}: a\mapsto b_a$. (Note that via that truth predicate, the definition of $f$ is uniform in $\varphi$ and $u$ - no need for nonuniformity here.)

  • By Separation in the real world we get the set $S=\{b: \exists a\in u(V_\kappa\models\varphi(a,b))\}$. Since $\kappa$ is inaccessible, we have $\alpha:=sup(ran(f))<\kappa$ and so $S\in V_\kappa$.

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  • $\begingroup$ I think, I understand what you are saying, but I am getting lost when I start thinking about talking about this formally. So, my main question is that when saying $V_\kappa \models \ulcorner\text{ZFC}\urcorner$, we generally mean that we have coded every axiom of ZFC, and every one of them holds in $V_\kappa$ with maybe some nonstandard ones. But what you are saying, is that internal to the model, we can show that all the above nine axioms hold. Can you please clarify on how we can deduce $V_\kappa \models \ulcorner\text{ZFC}\urcorner$ from your argument? $\endgroup$
    – Shervin Sorouri
    Apr 19, 2020 at 21:56
  • $\begingroup$ @ShervinSorouri ZFC is internally defined to be the conjunction of those nine axiom/axiom schemes. So once we have an internal proof of each of those nine results, in one step we combine them to get an internal proof of the whole thing. Obviously if you pick a nonstandard definition of ZFC then things get weird. E.g. if we consider the theory whose $i$th axiom is the $i$th ZFC axiom in the usual ordering if ZFC+I has no length-$<i$ contradiction and "$0=1$" otherwise, then classically this is still ZFC but ZFC+I can't prove that. $\endgroup$
    – Noah Schweber
    Apr 19, 2020 at 21:58
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    $\begingroup$ Really the right way to think about it is that we're proving in ZFC the sentence $$(*)\quad\forall \kappa\in Inacc,\varphi\in ZFC(V_\kappa\models \varphi).$$ We implicity use two important "internal definitions" here: $ZFC$, and $\models$. The proof of $(*)$ we give will follow the form of those choices of definition. $\endgroup$
    – Noah Schweber
    Apr 19, 2020 at 22:01
  • $\begingroup$ Actually, it may be helpful to ditch inaccessibility entirely here: it's probably easier to consider how we can prove in ZFC that "$\forall x,y(x=y)$" is not an axiom of ZFC. This relies exactly on looking at the way we're defining ZFC in the object language. (Note that I said "axiom of ZFC," not "theorem of ZFC.") We can indeed whip up a code for a theory ZFC' such that ZFC=ZFC' under reasonable assumptions but ZFC can't prove "$\forall x,y(x=y)$ is not a ZFC'-axiom." $\endgroup$
    – Noah Schweber
    Apr 19, 2020 at 22:03
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    $\begingroup$ @ShervinSorouri "do you know of any good references which touch on these meta-theoretical aspects?" Actually, the discussion here might be useful (both here and for your other question). "when we interpret $ZFC$ as the nine axioms you propose or other methods, the justification to do so, is philosophical(or common sense) to an extent?" Only (and exactly) to the extent that that's also the justification for, to pick an example, implementing the natural numbers as finite ordinals in set theory. (Or for working with a fixed definition of $PA$ in $PA$.) $\endgroup$
    – Noah Schweber
    Apr 19, 2020 at 22:28
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I think the confusion you're having is with a situtation that sometimes occurs where inside a (necessarily non-$\omega$-) model, there is a model of ZFC such that the model does not believe is a model of ZFC. In other words, the model satisfies all standard replacement axioms, but not the nonstandard ones of the outside model.

The difference here is that we are just working in ZFC, and showing that each replacement axiom holds in $V_\kappa$, not looking at what some particular model thinks. Thus the proof that all the replacement axioms hold in $V_\kappa$ for some arbitrary inaccessible $\kappa$ converts to truth in any model (for that model's $V_\kappa$, of course).

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