1
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Very helpful-- Thanks! Could you give me an example of a 'non-well-ordered but well-founded' infinitary Boolean operation? $\endgroup$ Jul 8, 2016 at 13:56
  • $\begingroup$ Also, does it matter whether $M$ is a set model or class model of $ZFC$? $\endgroup$ 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. $\endgroup$ 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 $M$. $\endgroup$ 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 $L^{M}_{\infty}$)? $\endgroup$ Oct 27, 2023 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.