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Let a cardinal $\kappa$ be $n$-shadow iff $\kappa$ does not have cofinality $\omega$ and for any $n$-th order sentence $\varphi$ in the language $\mathcal{L}_\in$, $\varphi\Leftrightarrow V_\kappa\models\varphi$. If this is unclear, for a standard $M\models\text{ZFC}$ and some $x\in M$ which is a cardinal in $M$, $x$ is $n$-shadow in $M$ iff $V_{\kappa+n}\cap M$ is elementarily equivalent to $\mathcal{P}^n (M)$.

The property of being $0$-shadow is (much) weaker than being Mahlo; in fact, much weaker than $\text{Ord is Mahlo}$.

The property of being $n$-shadow is also (much, much, much) weaker than being $n$-extendible. Furthermore, if there is an $n$-extendible cardinal, then $\text{Con}(\text{ZFC}^2+n\text{-shadow})$.

The reason I bring this up is because of the interesting properties of $1$-shadow cardinals when given an inner model $M$ of ZFC.

Assuming $M$ is an inner model and there is a $1$-shadow cardinal in $M$, then there is no nontrivial elementary embedding from $M$ into itself (in most cases this is equivalent to it's sharp not existing).

So, if $0^{\#}$ exists then there are no $1$-shadow cardinals in $L$. If $0^{\dagger}$ exists then there are no $1$-shadow cardinals in $L[U]$ for the standard $U$.

Specifically, if there is a nontrivial elementary embedding from $K$ into itself, then no cardinal is $1$-shadow in $K$, even though it should be that every uncountable cardinal has most large cardinal properties in $K$ if such an embedding exists, because of the tendencies of the core models.

Questions: What is the consistency strength of $n$-shadow cardinals? What properties result from these properties of inner models?

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  • $\begingroup$ Can you explain a little more about the background theory here? Your definition involves asserting $\phi$, an arbitrary $n$-order sentence, but there are various distinct ways to understand what this means. For example, the property of being a 1-shadow cardinal is expressible in GBC, but is this what you mean? Or do you want to use the full second-order semantics? And similarly with the higher order case. $\endgroup$
    – Joel David Hamkins
    Commented Dec 3, 2017 at 15:40
  • $\begingroup$ I notice you refer to $\text{ZFC}^2$, but if you mean the full second-order semantics, then there are no inner models of this theory, since they have to have all subsets of their elements. So I'm unsure what you mean. $\endgroup$
    – Joel David Hamkins
    Commented Dec 3, 2017 at 16:23
  • $\begingroup$ I didn't claim any inner models that satisfied $\text{ZFC}^2$ also satisfied the existence of a $1$-shadow cardinal. $\endgroup$
    – Zetapology
    Commented Dec 4, 2017 at 1:58
  • $\begingroup$ @JoelDavidHamkins Sorry for the late reply. I have now edited the question in order to better express my definition. $\endgroup$
    – Zetapology
    Commented Dec 11, 2017 at 7:30
  • $\begingroup$ You give two definitions of $n$-shadow cardinal in the first paragraph, but they are not equivalent. In the first definition, you say $\varphi\iff V_\kappa\models\varphi$, which suggests that we are using the full second-order semantics for the higher-order quantifiers, but then you say $V_{\kappa+n}\cap M\prec P^n(M)$, where in the former part we take only the higher-order semantics provided by $M$, which might be fewer subsets than provided by $V$. In my answer, I had used the second definition. $\endgroup$
    – Joel David Hamkins
    Commented Dec 11, 2017 at 13:25

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You consider the strength of having a $1$-shadow cardinal in an inner model. But I claim that no proper inner model $M$ has any $1$-shadow cardinals at all.

Assume $M\subsetneq V$ is a proper inner model, by which I mean that $M$ is a transitive class model of ZF, containing all ordinals, and $M\neq V$. Suppose that $\kappa$ is a $1$-shadow cardinal, which according to your definition means that $V_{\kappa+1}\cap M\prec P(M)$. Since $M\neq V$, there is a set $A\in M$ for which there is a subset $B\subset A$ with $B\in P(M)$ but $B\notin M$. This property is expressible in $P(M)$, whether we use the full second-order semantics or the GBC semantics, where the classes of $V$ are given as a collection. But notice that $V_{\kappa+1}\cap M$ has no such sets $A$ and $B$, since every element of $V_\kappa\cap M$ has all its subsets that are in $V_{\kappa+1}\cap M$, since they are all in $M$. $\Box$

This seems to refute your claim that $n$-shadow is necessarily weaker than $n$-extendible, since an inner model can have extendible cardinals. Perhaps the resolution of this issue will be that the definition of the shadow cardinals is not yet expressing exactly what you want to express?

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  • $\begingroup$ It is only weaker in consistency strength, not direct implication. $\endgroup$
    – Zetapology
    Commented Dec 11, 2017 at 15:27
  • $\begingroup$ What does it mean for you when you refer to the "consistency strength" of a second-order or higher-order assertion? $\endgroup$
    – Joel David Hamkins
    Commented Dec 11, 2017 at 15:38
  • $\begingroup$ The same way correct cardinals are equiconsistent with ZFC. If there is an extendible cardinal, there is a (transitive set) model of $\text{ZFC}^2+n\text{-shadow}$. $\endgroup$
    – Zetapology
    Commented Dec 11, 2017 at 15:39
  • $\begingroup$ You're right that my definition wasn't quite what I wanted to express. $\endgroup$
    – Zetapology
    Commented Dec 11, 2017 at 15:40
  • $\begingroup$ Well, since you are using the full second-order semantics, you can't seem to go from the assumption that there is a model of an extendible cardinal to the existence of a set model of an $n$-shadow cardinal. So Con(ZFC extendible) will not imply the existence of a model with an $n$-shadow cardinal. For example, the consistency of first-order theories is the same in V as in L, but this is not true for second-order or higher-order theories, if you is using the full second-order semantics. I think all these issues are much more delicate and problematic than your post suggests. $\endgroup$
    – Joel David Hamkins
    Commented Dec 11, 2017 at 15:44

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