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From wikipedia:

Shoenfield's absoluteness theorem shows that $\Pi^1_2$ and $\Sigma^1_2$ sentences in the analytical hierarchy are absolute between a model $V$ of ZF and the constructible universe $L$ of the model, when interpreted as statements about the natural numbers in each model.

Question. Is there an "ordinal" version of the arithmetical and/or analytic hierarchies such that an analogue of Shoenfield absoluteness can be proven for that hierarchy? (With $\omega$ replaced by $\mathrm{On}$). My gut feeling is that we have to choose our language very carefully, in order to block $\aleph_1$ and/or $\beth_1$ from being definable by formulae somewhere in the hierarchy.

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  • $\begingroup$ Just to clarify: a $\Pi^1_2$ sentence is of the form $(\forall X \in 2^\omega) (\exists Y \in 2^\omega)\ \varphi(x,y)$ where $\varphi$ is a quantifier-free formula in the language of arithmetic. I assume you are suggesting looking at sentences of the form $(\forall X \in 2^\mathrm{On}) (\exists Y \in 2^\mathrm{On})\ \varphi(x,y)$ where $\varphi$ is a quantifier-free formula in the "language of ordinal arithmetic" suitably defined. Is this correct? $\endgroup$
    – Jason Rute
    Commented May 24, 2014 at 18:20
  • $\begingroup$ To continue my last comment, this may be a silly question, but is there even a good way to express the quantifier $(\exists Y \in 2^\mathrm{On})$? It seems to me that $2^\mathrm{On}$ is too big to be a proper class. (For example, replacement is an axiom schema since one can't quantify over all class functions in a single sentence.) $\endgroup$
    – Jason Rute
    Commented May 24, 2014 at 18:23
  • $\begingroup$ @JasonRute, yes. I don't know how to express such a quantifier within the language of ZFC, but one option would be to work with set-sized transitive models of ZFC. Another would be to using second-order ZFC with Henkin semantics. A third option would be to just dilute the theorem we're trying to prove, and stick to first-order stuff. $\endgroup$
    – goblin GONE
    Commented May 25, 2014 at 3:06
  • $\begingroup$ Perhaps a more modest question is: Is there an analog of Shoenfield's (or even Mostowski's) absoluteness theorem for $2^{\kappa}$ where $\omega<\kappa$ (perhaps under additional assumptions)? I know that Sy Friedman and others have been working recently on generalized descriptive set theory, but I don't think this question was addressed yet (it sure seems like an interesting one to me). $\endgroup$
    – Haim
    Commented May 25, 2014 at 3:54
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    $\begingroup$ Asaf, this is how I think about this problem (at least for the $\Sigma_1^1$ case): You may regard generalized analytic sets as projections of closed sets. Now the question is whether we can find a reasonable tree representation for such sets as in the case of $\kappa=\omega$, and then we would like to have some absoluteness theorems about the existence of $\kappa$-branches (obviously this is not as simple as the original case). Perhaps we need some general absoluteness theorems for infinitary logics, I don't know. $\endgroup$
    – Haim
    Commented May 25, 2014 at 13:06

1 Answer 1

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Perhaps the solution lies in the Levy absoluteness theorem.

We define the Levy hierarchy of formulas as follows.

  1. $\Sigma_0,\Pi_0,\Delta_0$ are formulas in the language of set theory which only have bounded quantifiers.

  2. $\Sigma_{n+1}$ formulas equivalent to formulas of the form $\exists x\varphi$, where $\varphi$ is $\Pi_n$; and similarly $\Pi_{n+1}$ are formulas equivalent to $\forall x\varphi$ where $\varphi$ is $\Sigma_n$.

  3. $\Delta_n$ are formulas which are both $\Sigma_n$ and $\Pi_n$.

Now we have that if $M$ is a transitive class (set or otherwise), then $\Delta_0$ formulas are absolute between $M$ and $V$ (as long as $M$ includes all the parameters). Next we have that $\Sigma_1$ formulas are upwards absolute and $\Pi_1$ formulas are downwards absolute, so $\Delta_1$ formulas are absolute.

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    $\begingroup$ The last sentence is just wrong. For instance, CH is $\Delta_2$, but not upwards absolute. $\endgroup$
    – Emil Jeřábek
    Commented May 24, 2014 at 20:14
  • $\begingroup$ Err, yes. You're right. I have no idea why that sentence entered the answer! :| $\endgroup$
    – Asaf Karagila ♦
    Commented May 24, 2014 at 20:23
  • $\begingroup$ It appears though that Boolean combinations of $\Sigma_1$ sentences (but not formulas with arbitrary parameters) are absolute, the reason being that a $\Sigma_1$ sentence is equivalent to the existence of its countable transitive model, and that’s a $\Sigma^1_2$ statement. $\endgroup$
    – Emil Jeřábek
    Commented May 24, 2014 at 20:54
  • $\begingroup$ @Emil: Yes, I wanted to add that at first, but for some reason (which I cannot recall through these two hours) decided against this. Note that this absoluteness requires $\sf DC$ to hold in $V$, since it is needed for the existence of the countable transitive model (although there might be a way to go around that, or weaken the general requirement of $\sf DC$). $\endgroup$
    – Asaf Karagila ♦
    Commented May 24, 2014 at 21:01
  • $\begingroup$ @Emil: By the way, as phrased above, $\Sigma_1$ statements are closed under conjunction and disjunction. $\endgroup$
    – Asaf Karagila ♦
    Commented May 24, 2014 at 23:36

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