4
$\begingroup$

In many cases, a large cardinal property for $\kappa$ says that there's a nontrivial elementary embedding $j: V \rightarrow M$ with $\mathsf{crit}(j) = \kappa$ and appropriate other requirements on $M$ and $j$.

What happens if we require elementarity not for $\mathcal{L}_{\omega\omega}$ but for a larger infinitary language? Clearly, there can't be a nontrivial $\mathcal{L}_{\kappa^+\kappa^+}$-elementary $j: V \rightarrow M$, because we could just write a sentence $\phi$ with $\kappa$-many quantifiers specifying the cardinality of $\kappa$, and $\phi^M$ wouldn't hold. But can we have $\mathcal{L}_{\kappa\kappa}$-elementarity? If so, does it get us any additional strength?

Improve this question
$\endgroup$
1
  • $\begingroup$ There are subtle foundational issues regarding what it means to be $\mathcal{L}_{\kappa,\kappa}$-elementary. For example, even first-order elementarity is not directly expressible. Instead, one uses a substitute: cofinal + $\Sigma_1$-elementary. This gives you elementarity with respect to any given standard formula, by induction in the meta-theory. But it isn't the same as being elementarity with respect to an internal truth predicate, if there is one. With $\mathcal{L}_{\kappa,\kappa}$, we only have the internal account, and so one needs to take special care with what is meant. $\endgroup$
    – Joel David Hamkins
    Commented Jan 4, 2018 at 0:41

1 Answer 1

Reset to default
2
$\begingroup$

This is extremely sketchy since I have to run, but I'll fill in the details later:

I think if $j$ comes from a measurable - more generally, if $M^\kappa\subseteq M$ - then we already have $\mathcal{L}_{\kappa\kappa}$-elementarity. The key is that the truth of an infinitary sentence is witnessed in a reasonably constructive way, and as long as the rank of the sentence in question is below the critical point we can show that "witnesshood is preserved."

While truth is of course not definable even for first-order sentences, there is a "classish" notion of truth here: an $\mathcal{L}_{\kappa\kappa}$-sentence $\varphi\in V$ with parameters in $V$ is true in $V$ if there is a "witness" (think family of Skolem class functions, but appropriately "$\kappa$-ary") for it. This witness is a proper class, but there is a sense in which $j$ acts on proper classes as well as sets - send $A$ to $j``A=\{j(a): a\in A\}$ - which will help us. And, although we don't actually need it here, it's worth noting that the property "$B$ is a witness to $\psi$" is expressible in the language of set theory augmented with a new predicate symbol for $B$.

Moreover, we can also make sense of a witness for a sentence relativized to a subclass $M$. We have to be careful here, though, since there are two senses in which we might think about witnessing that a sentence is true in $M$: "internally to $M$," and "correctly." I mean the latter, that is, a witness for "$\varphi^M$." (Note that if $M$ is definable in $V$, as it will be if $j:V\rightarrow M$ comes from a measurable, this is perfectly simple. Working more generally gets tedious, but doesn't change the key idea.)

Now it's not hard to show by induction on rank that - since $M^{\kappa}\subseteq M$ - we have $A$ witnesses $\varphi(\overline{c})$ iff $j``A$ witnesses $\varphi(j(\overline{c}))^M$. The key observations are that $\varphi$ itself isn't moved by $j$, the tuple $\overline{c}$ doesn't get new elements, quantifying over $\kappa$-tuples of elements of $M$ in $M$ is the same as quantifying over $\kappa$-tuples of elements of $M$ in $V$, and that $\kappa$-completeness makes the $\kappa$-ary Boolean combinations behave appropriately.

And, of course, this all breaks down once we look at $\kappa^+$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .