Let $\mathcal{L}$ be the first order language of ZFC set theory, and let $\mathcal{L}_{\infty,\infty}$ be the usual infinitary extension of the language allowing arbitrary long disjunctions/conjunctions and quantifications.

What happens if one replaces the usual axiom schema of replacement for ZFC by a new schema over (the class of) formulas from $\mathcal{L}_{\infty,\infty}$ instead of from the first-order theory? Does one get a "standard" model of set theory?

Do set theorists believe that this improved version of replacement is true? If so, why still work over the less expressive language $\mathcal{L}$? If not, why not?

Does the connection between replacement and transfinite induction completely disappear under this extension?

Motivation: I was thinking about Skolem's paradox, and thought that working in $\mathcal{L}_{\infty,\infty}$ might resolve part of the problem---namely that first order language just wasn't expressive enough to talk about important aspects of big sets. This seemed to be backed up by other things I read. But then, it struck me that if we are working with a more expressive language, why limit our axioms to the simpler language?

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    You get a language which is not quite first order, is more expressive, and may get you into trouble. It may be possible to do something like Russell's paradox, where for each sub formula in a conjunct you get another formula in a different conjunct which contradicts the sub formula except for one small piece, and you end up with a mess in which each initial segment is consistent and has a model, but the sentence in its full glory does not. My guess is that Replacement on this scale is scary if not downright contradictory. Gerhard "How Would A Proof Look?" Paseman, 2018.07.30. – Gerhard Paseman Jul 30 at 7:07
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    "Why still work over the less expressive language $\mathcal L$?" Because we have a complete axiomatization of the logic of $\mathcal L$. We can dream in more expressive systems, and I think we can "see" that replacement should still be true in that context, but we can't prove things without knowing what the correct logical axioms are (nor could we actually write down infinitary statements). – Andreas Blass Jul 30 at 12:57
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    @AndreasBlass I'm not sure I understand your comment. Why would I need a complete axiomatization of the logic? Wouldn't a partial axiomatization (which would, naturally, include those methods from first order logic as applied to first order statements) suffice to get me started? In fact, this seems to match what I actually do in mathematics, as a ring theorist. I prove first order statements all the time, but I also prove all sorts of things which cannot be stated in the first order language of rings. – Pace Nielsen Jul 31 at 2:39
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    I think one can show that for every cardinal $\kappa$, $M$ is a model of $\mathcal L_{\kappa,\kappa}$-ZFC iff $M \cong V_\kappa$ and $\kappa$ is inaccessible. – Monroe Eskew Aug 2 at 18:07
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    What I said isn’t right— we can only say that an initial segment of $M$ is isomorphic to $V_\kappa$. – Monroe Eskew Aug 4 at 12:18
up vote 12 down vote accepted
+100

Let me describe how I understand the question. You want to consider assertions in the infinitary language $\newcommand\L{\mathcal{L}}\L_{\infty,\infty}$, and assert instances of replacement for formulas in this language.

Thus, our meta-theoretic context should have considerable set-theoretic resources, in order to handle the manipulations of these infinitary assertions, even if we interpret those assertions only in the object theory.

So basically, we have the realm of objects, a universe $V$ of the sets that the theory is about, and then we have a meta-theoretic or background set-theoretic context, let me call it $V^*$, which is a set-theoretic world of its own in which the formulas live and in which we undertake the truth analysis of those formulas. Let me assume ZFC in the meta-theory.

You ask:

Does one get a 'standard' model of set theory?

The answer is yes. I claim that $V$ must be well-founded with respect to $V^*$, and the reason is that with your replacement axiom, we can easily derive the $\L_{\infty,\infty}$-separation axiom, and with this, we can proceed to define the class of all sets $a$ that have no actual $\in$-descending $\omega$-sequence below them. This is expressible by an $\L_{\infty,\infty}$ formula of $V^*$, using the standard $\omega$ as $V^*$ sees it, and it would define the well-founded part of $V$, as $V^*$ sees it. In particular, by separation, using this definable property, the collection of nonstandard sets inside any given set would itself be a set. But this cannot happen unless every set is standard, for otherwise by the foundation axiom there would have to be an $\in$-minimal nonstandard set, which is impossible. Therefore, all the $V$-sets are standard with respect to $V^*$.

It follows now that every element $a\in V$ is definable an $\L_{\infty,\infty}$ formula. Namely, $a$ will be the unique satisfying instance of the formula $\phi_a$, defined as follows: $$\phi_a(x)=\forall y\left[ y\in x\leftrightarrow\bigvee_{b\in a} \phi_b(y)\right].$$ That is, $a$ is the unique set $x$ whose elements are the objects that satisfy the definitions of the various elements of $a$. This idea is used all over admissible set theory.

From this, it follows from your theory that $V$ must have the actual power set operation, since for any set $A\in V$, and any subset $B\subseteq A$ in $V^*$, we can write down the formula $\phi(x)=\bigvee_{b\in B}\phi_b(x)$, which asserts that $x$ is one of the objects in $B$. By separation, we will have deduced that $B$ is in $V$.

It follows in turn that $V=V^*_\kappa$ for some cardinal. One can show furthermore that $\kappa$ must be a strong limit, as well as regular. So $\kappa$ is an inaccessible cardinal, and these models $V_\kappa$ are also known as Zermelo-Grothendieck universes.

Conversely, I claim that if $\kappa$ is inaccessible, then $V_\kappa$ satisfies your theory. The reason is that $V_\kappa$ satisfies every instance of replacement using any means whatsoever in $V^*$, not just $\L_{\infty,\infty}$ assertions, because every set $A\in V_\kappa$ has size less than $\kappa$, and every subset of $V_\kappa$ of size less than $\kappa$ is bounded in some $V_\alpha$ for $\alpha<\kappa$, by the inaccessibility of $\kappa$. From this, it follows that we can define exactly that range using only a $\L_{\kappa,\kappa}$ formula.

So the models of your theory are exactly the $V_\kappa$ for $\kappa$ inaccessible, in the meta-theory.

All the preceding analysis is essentially similar to the result of Zermelo proved a century ago, when he proved that second-order ZFC set theory is true exactly in the $V_\kappa$ when $\kappa$ is inaccessible.

What I take the analysis to show, is that questions of these infinitary assertions amount to first-order assertions of set theory in the meta-theory. The subject becomes in a sense the same as the model theory of set theory undertaken in the meta-theory. In this sense, the infinitary theory amounts to a change in the level of analysis, moving from infinitary assertions in the object theory to first-order assertions about formulas in the meta-theory.

  • Is this exactly right? Suppose $\kappa$ is inaccessible-- can't we define $\kappa$ using an $\mathcal L_{\kappa^+,\kappa^+}$ formula? Furthermore, if $M \subseteq V$ is $<\kappa$-closed, doesn't $M$ satisfy $\mathcal L_{\kappa,\kappa}$-ZFC? – Monroe Eskew Aug 6 at 16:59
  • I'm not sure what part of my answer you are objecting to. (Your $M$ is my $V$ and your $V$ is my $V^*$.) My argument is that if $\kappa$ is inaccessible, then $V_\kappa$ satisfies all instances of replacement using any $\mathcal{L}_{\infty,\infty}$ formula, even far beyond $\kappa$, because $\kappa$ is really regular: witnesses for any property over a set in $V_\kappa$ will still be bounded in $V_\kappa$ and hence an element. It doesn't matter that $\kappa$ itself is definable in the meta-theory, you cannot use that to make a failing instance of replacement in $V_\kappa$. – Joel David Hamkins Aug 6 at 17:13
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    Oh yes, if you cut off the logic at $\mathcal{L}_{\kappa,\kappa}$, then you can get more models that reach beyond $\kappa$, I agree. – Joel David Hamkins Aug 6 at 17:38
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    @JoelDavidHamkins Perhaps this will make my question clearer. Run through your argument when $V^{\ast}$ has no inaccessibles and do not necessarily assume that the model $V$ must be set sized according to $V^{\ast}$. Don't we get $V=V^{\ast}$? Or am I missing something more? – Pace Nielsen Aug 7 at 2:48
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    Yes, I agree with that. But another way to describe that case is that $V^*$ thinks that there are no set models of the theory. $V^*$ isn't necessarily able to evaluate the truth of the formulas on itself, but only on a set model, unless you assume a stronger theory in $V^*$ than I have. – Joel David Hamkins Aug 7 at 10:34

The idea of studying ZF and its subsystems formulated in infinitary languages, to my knowledge, seems to have begun and ended with the work of Klaus Gloede, in the 1970s.

The following paper of Gloede provides a useful synopsis of his work, which began with his doctoral dissertation (it is available here behind a paywall, the first two pages are freely visible, they consist of the table of contents and the first page of the paper).

K. Gloede, Set theory in infinitary languages. ⊨ISILC Logic Conference (Proc. Internat. Summer Inst. and Logic Colloq., Kiel, 1974), pp. 311–362. Lecture Notes in Math., Vol. 499, Springer, Berlin, 1975.

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