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?



Set theory bootstrapping
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.


