In the first-order context, "reflection" of a formula $\varphi(x)$ below $\kappa$ refers to the the following situation:

There are many ordinals $\alpha<\kappa$ such that for all $a \in V_\alpha$, $V_\alpha \models \varphi(a)$ iff $V_\kappa \models \varphi(a)$.

If $\kappa$ is inaccessible, then this holds for any expansion of $V_\kappa$ in a countable language. "Many" can be taken to mean on a club set.

When we move to higher-order logic, talk of reflection usually shifts to talk of indescribability. A cardinal is $\Pi^m_n$ indescribable if for any $A \subseteq V_\kappa$ and any $\Pi_n$ sentence $\sigma$ in $(m+1)$-order logic with a predicate for $A$, if $(V_\kappa, \in, A) \models \sigma$, then there is $\alpha<\kappa$ such that $(V_\alpha,\in,A\cap V_\alpha) \models \sigma$. It is a standard fact that if $\kappa$ is measurable, then there is a measure-one set of $\alpha< \kappa$ that are $\Pi^m_n$-indescribable for every $m,n$. One can also show something stronger: If $\kappa$ is measurable, there is a measure-one set of $\alpha < \kappa$ such that if $A \subseteq V_\alpha$, then there is $\beta < \alpha$ such that $(V_\alpha,\in,A)$ and $(V_\beta,\in,A\cap V_\beta)$ have the same $\omega$-order theory.

Now this is not completely analogous to reflection because we're no longer talking about elementary substructures, but just elementarily equivalent structures, albeit with a common interpretation of a particular predicate. So my question is, what kind of large cardinal $\kappa$ is needed to get the following statement?

For any $A \subseteq V_\kappa$ and any $n \in \omega$, there are many ordinals $\alpha < \kappa$ such that $(V_\alpha,\in,A \cap V_\alpha) \prec^n (V_\kappa,\in,A)$, where $\prec^n$ means elementary in $(n+1)$-order logic.

It happens at an $\omega$-strong cardinal, but this is clearly not optimal.

  • 3
    $\begingroup$ How exactly are you defining $\prec^n$? If it only allows first-order parameters, then by using a truth predicate for the language with A and only first-order parameters it should be straightforward to show that these are just the totally indescribable cardinals. If we allow second-order parameters, then the claim is inconsistent - just consider "$\forall x\in X$" where $X = V_\alpha$. $\endgroup$ Sep 17, 2014 at 7:49
  • $\begingroup$ @SamRoberts, why not post that as an answer? Part of the point as I see it is that the remark about "no longer talking about elementary substructures" is not really accurate, since any desired set parameters can be coded into $A$, and so $\sigma$ may in effect refer to them. $\endgroup$ Sep 17, 2014 at 11:27

1 Answer 1


It depends on how we define elementarity for the $(n+1)^{th}$-order language of set theory, $\mathcal L^{n}_\in$. For $X\subseteq V_\kappa$, two salient definitions are:

$V_\alpha \prec^n_X V_\kappa$ iff $\forall \vec{x}\in V_\alpha(V_\alpha\vDash \phi(X\cap V_\alpha,\vec{x}) \leftrightarrow V_\kappa\vDash \phi(X,\vec{x}))$, for all $\phi\in \mathcal L^{n}_\in$ with parameters among $X, \vec{x}$.

$V_\alpha \prec^{*n}_X V_\kappa$ iff $\forall \vec{x}\in V_\alpha\forall \vec{Y}\subseteq V_\alpha(V_\alpha\vDash \phi(X\cap V_\alpha, \vec{Y}, \vec{x}) \leftrightarrow V_\kappa\vDash \phi(X,\vec{Y},\vec{x}))$, for all $\phi\in \mathcal L^{n}_\in$ with parameters among $X,\vec{Y}, \vec{x}$.

${\bf Theorem}$ ${\bf 1:}$ There are no $\alpha<\kappa$ such that $V_\alpha\prec^{*n} V_\kappa$ for $n\geq 1$.

${\it Proof.}$ If $Y = V_\alpha$, $V_\alpha\vDash \forall x(x\in Y)$ but not $V_\kappa\vDash \forall x(x\in Y)$.

${\bf Theorem}$ ${\bf 2:}$ If $\kappa$ is $\Pi^{n}_1$-indescribable for $n\geq 1$, then for any $X\subseteq V_\kappa$ there are unbounded $\alpha<\kappa$ such that $V_\alpha\prec^{n-1}_X V_\kappa$.

${\it Proof.}$ In $n+1^{th}$-order Zermelo set theory we can define a $\Delta^{n}_1$ satisfaction predicate for $\mathcal L^{n-1}_\in$. Denote it $Sat$. Then we can define $Y\subseteq V_\kappa$ such that $V_\kappa$ satisfies:

(*) $\langle \vec{x},\phi\rangle\in Y \leftrightarrow Sat(\phi, X, \vec{x})$, for all $\phi\in \mathcal L^{n-1}_\in$ with parameters among $X, \vec{x}$.

Clearly, (*) is $\Delta^n_1$ and since $\kappa$ is $\Pi^{n}_1$-indescribable, it follows that there are unbounded $\alpha<\kappa$ which satisfy:

(**) $\langle \vec{x},\phi\rangle\in Y\cap V_\alpha \leftrightarrow Sat(\phi, X\cap V_\alpha, \vec{x})$, for all $\phi\in \mathcal L^{n-1}_\in$ with parameters among $X, \vec{x}$.

For any such $\alpha$ it is easy to see that $V_\alpha\prec^{n-1}_X V_\kappa$, as required.

(Note that we can easily strengthen "unbounded" to "stationary" but not "club" - any $\alpha$ for which $V_\alpha\prec^1 V_\kappa$ will be inaccessible.)

${\bf Corollary:}$ $\kappa$ is totally indescribable just in case for any $X\subseteq V_\kappa$ and $n<\omega$ there are unbounded $\alpha$ such that $V_\alpha\prec^n_X V_\kappa$.

  • $\begingroup$ Isn't your corollary in fact an equivalence? That is, doesn't your argument show that the desired property is equivalent to total indescribability? $\endgroup$ Sep 17, 2014 at 13:25
  • $\begingroup$ @JoelDavidHamkins That's right (and what I claimed in my comment!). Will modify it. Thanks! $\endgroup$ Sep 17, 2014 at 13:28
  • $\begingroup$ Thanks, I did mean the first definition. Notation question: When you write $\mathcal{L}^n_\in$, you mean the $n^{th}$ order formulas? (i.e. We have a 15th order predicate for 14th order satisfaction.) Wouldn't it be better for that to denote $(n+1)$-order formulas, keeping with the $\Pi^n_1$ notation? $\endgroup$ Sep 17, 2014 at 14:27
  • $\begingroup$ That's right. I was thinking i should perhaps bring that notation in line with the $\Pi^m_n$ notation. I will go through and change it when I get a chance. $\endgroup$ Sep 17, 2014 at 15:46

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