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In the mathoverflow question , "Godel's Constructible Universe in Infinitary Logics (A Possible Solution to $HOD$ Problem), Prof Hamkins answered user46667's question 2

What is $\mathrm L_{\infty}$? ($\mathrm L_{\infty}$ is Goedel's constructible universe over the infinitary language $\mathcal L_{\infty, \infty}$, where $\mathcal L_{\infty, \infty}$ allows conjunctions and disjunctions of arbitrary (infinite) length, and arbitrary (infinite) quantifications

as follows:

Thm. (Hamkins): $\mathrm L_{\infty}$ is the entire set-theoretic universe $V$. (What does "entire" mean in this case?)

His proof is based on the validity of the following claim:

What I claim specifically is that for every set $a$ there is a $\mathcal L_{\infty,\infty}$ formula $\psi_a(x)$ such that in any transitive set $M$ with $a$$\subset$$M$ we have $a$={$x$|$M$$\vDash$ $\psi_a(x)$} [via $\in$-induction--my comment]

My questions revolve around which sets can be defined by $\mathcal L_{\infty,\infty}$-formulas:

i) Can choice functions be defined by $\mathcal L_{\infty,\infty}$-formulas (my guess: certainly yes) so that $AC$ holds for $V$?

ii) Can amorphous sets be defined by $\mathcal L_{\infty,\infty}$-formulas (my guess: possibly?) so that $AC$ fails for $V$?

iii) Can Cohen and Random reals (and sets thereof) be defined by $\mathcal L_{\infty,\infty}$-formulas? (I have a paper in my collection titled "Forcing with Propositional Lindenbaum Algebras" by Alexandar Preovic, so my guess is that this question can be answered--yes, if the results in that paper aren't false.)

In general, I would like any references of any research on treating a set-theoretic universe $V$ as $\mathrm L_{\infty}$.

Thank in advance.

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Let me first address the issue of what $L_\infty$ really is: given a model $M$ of ZFC, there is a class $\mathcal{L}_{\infty,\infty}^M$ - the class of infinitary formulas belonging to $M$. Within $M$, we can then define $L_\infty^M$ - a subclass of $M$. This is exactly analogous to $L$: each model of ZFC has its own constructible universe, which may or may not be the whole model. In the case of $L_\infty$, we can in fact prove that $L_\infty^M=M$. This is what is meant by "$L_\infty$ is the entire set-theoretic universe $V$".

Note that the infinitary formulas used in construting $L_\infty$ inside some model $M$, are exactly those formulas in $M$. We're not using any "external" infinitary formulas; this is an internal construction.

Note, though, that this begins by assuming $M$ is a model of ZFC! So asking whether we can use this to conclude "AC holds for $V$" is circular.

Is choice necessary, though? Good question. The issue is how exactly one defines "$\mathcal{L}_{\infty,\infty}$" in the absence of choice. If we use well-ordered infinitary Boolean operations, then the answer is no; otherwise, yes, for the same reasons as in the ZFC case.

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  • $\begingroup$ Very helpful-- Thanks! Could you give me an example of a 'non-well-ordered but well-founded' infinitary Boolean operation? $\endgroup$ Commented Jul 8, 2016 at 13:56
  • $\begingroup$ Also, does it matter whether $M$ is a set model or class model of $ZFC$? $\endgroup$ Commented Jul 8, 2016 at 14:10
  • $\begingroup$ @ThomasBenjamin Re: your first comment, consider "$\bigvee_{a\in A} x=a$" for $A$ a non-well-orderable set. Re: your second comment, no - I just prefer set models because then I don't have to do (as much) class theory. $\endgroup$ Commented Jul 8, 2016 at 14:38
  • $\begingroup$ @ThomasBenjamin I don't understand your question. It doesn't make sense to talk about a "model of $L_\infty$" - theories have models, not (possibly-class-sized) structures. More importantly, the whole point of this answer is that $L_\infty^M$ is literally just $M$. $\endgroup$ Commented Aug 18, 2023 at 21:08
  • $\begingroup$ Can one have the following situation: $L_{\infty}$ is countable (and if so then since $L^{M}_{\infty}$ = $M$ if one has that $M$ is a ctm of $ZF$, can one have generic extensions of $L^{M}_{\infty}$)? $\endgroup$ Commented Oct 27, 2023 at 4:13

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