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In July, Asaf Karagila asked three questions about elementary substructures of the universe of sets. The latter two were answered, the upshot being that the hypothesis $V_\kappa \prec V$ doesn't alone endow $\kappa$ with any large cardinal properties. However, Asaf's first question went unanswered:

For given $n$, is there a large cardinal property $\Phi_n$ such that, if $\kappa$ is a $\Phi_n$ cardinal, then $V_\kappa\ \prec_n\ V\ $?

As he noted, for $n = 1, 2, 3$ we can take $\Phi_n$ to be inaccessibility, supercompactness, and extendibility, respectively. (See Kanamori, pp. 299 and 319.) What of larger $n$?

I'm curious about this in connection with Zermelo's "Grenzzahlen" picture, which suggests a view of set existence on which there is no fixed universe of sets, but rather an indefinitely extensible sequence of natural universes, each contained as a set in all its successors. To salvage the separate claim that each set theoretic statement has a determinate truth value, someone with this "no fixed universe" view might argue that, for each $n$, there are arbitrarily large $\Phi_n$ cardinals; faced with a $\Sigma_n$ sentence $\varphi$, she could then declare that $\varphi$ is true simpliciter iff $V_\kappa \vDash \varphi$ for some (or each) $\Phi_n$ cardinal $\kappa$.

Of course, someone with a "no fixed universe" view wouldn't formulate things as I did in the displayed line above, since on her view there is no such object as $V$. Rather, she might wonder:

For given $n$, is there a large cardinal property $\Psi_n$ that "transcends" $\Psi_m$ for all $m < n$ and is such that, if $\kappa < \lambda$ are $\Psi_n$ cardinals, then $V_\kappa\ \prec_n\ V_\lambda\ $?

Given a positive answer, she could define truth simpliciter much as above, so that the truths are those propositions that hold in all sufficiently rich natural models.

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The $\Sigma_n$ reflecting cardinals, of course, have precisely the property that you mention, by definition. A cardinal $\kappa$ is $\Sigma_n$-reflecting if $V_\kappa\prec_{\Sigma_n} V$. One sometimes sees this term defined in such a way to insist also that $\kappa$ is inaccessible, but to my way of thinking, these should simply be called the inaccessible $\Sigma_n$-reflecting cardinals, as the concept is coherent and useful without the extra inaccessibility hypothesis.

One can prove in ZFC, for any particular $n$, that there is a closed unbounded class of $\Sigma_n$-reflecting cardinals, and this is essentially the content of the Reflection Theorem. If one adds the hypothesis that $\kappa$ is inaccessible, then for $n\geq 2$ it follows that there is a proper class of inaccessible cardinals and more, but the strength is bounded below a Mahlo cardinal (that is, very low in the large cardinal hierarchy), because if $\delta$ is Mahlo, then $V_\delta$ has a club in $\delta$ of $\kappa$ with $V_\kappa\prec V_\delta$, and this club will then contain many inaccessible cardinals.

One may easily combine the reflecting idea with other large cardinal concepts to obtain new stronger large cardinal concepts that also exhibit your desired property. For example, rather than considering an inaccessible $\Sigma_n$-reflecting cardinal, one may consider a measurable $\Sigma_n$ reflecting cardinal, or a supercompact $\Sigma_n$-reflecting cardinal, and so on. These large cardinal concepts will again exhibit the property you request, but be much stronger in large cardinal consistency strength.

Perhaps the right perspective is that the $\Sigma_n$-reflecting idea is a way of adding "epsilon" to a large cardinal concept. The $\Sigma_5$-reflecting measurable cardinal hypothesis, for example, is definitely stronger than mere measurability, but it is not as strong as a cardinal $\delta$ that is stationary for measurability---meaning that $\delta$ is regular and the measurable cardinals below $\delta$ are a stationary subset of $\delta$---since if the set of measurable cardinals $\kappa$ below $\delta$ is stationary in $\delta$, then since there is a club subset of $\delta$ of $\alpha$ with $V_\alpha\prec V_\delta$, there will be many measurable $\kappa$ that are fully reflecting in $V_\delta$.

In particular, if $\kappa$ has nontrivial Mitchell rank, then the set of measurable cardinals below $\kappa$ is stationary, and this hypothesis is satisfied. So for example, if $\kappa$ is only $(\kappa+2)$-strong (also known as the $P^2(\kappa)$-hypermeasurable cardinals), then this is plenty strong enough.

The concept of a large cardinal having property $X$ and also being $\Sigma_n$-reflecting is weaker in consistency strength than a cardinal such that the $X$-cardinals below are stationary, which is another common way of adding "epsilon", in other words, of making a mild increase in consistency strength over a given large cardinal concept.

For any fixed large cardinal concept $X$, the notion $\Phi_n(\kappa)=\kappa$ has property $X$ and is $\Sigma_n$-reflecting has the properties that you request in the second part of your question, while also being stronger than $X$. In most cases, however, this property is at the same time weaker than the typical next cardinal $Y$ above $X$, since as I mentioned, insisting on this level of reflection on top of $X$ is only adding a little, and is usually swamped by the next larger large cardinal concept. Thus, the following sequence provides a way of strengthening (in consistency strength) any given large cardinal concept:

  • $\kappa$ has large cardinal property $X$. (e.g. inaccessibility, measurability, supercompactness, etc.)

  • There are a proper class of cardinals with property $X$.

  • There is a $\Sigma_2$-reflecting cardinal with property $X$.

  • There is a $\Sigma_3$-reflecting cardinal with property $X$, or $\Sigma_n$-reflecting, etc.

  • There is a stationary class of cardinals with property $X$.

  • There is a regular cardinal $\delta$ with the $X$-cardinals below $\delta$ being stationary.

Each step increases strictly in consistency strength (for the reflecting case, you have to get above the complexity of the $X$ notion to get strict increases). So this shows that asking for reflection along with a given large cardinal notion is something in between asking for a proper class of those cardinals and asking for a stationary proper class of those cardinals. This is the same as the move from inaccessible cardinals to Mahlo cardinals, and the move from Mahlo to $1$-Mahlo, or hyper-Mahlo, and so on. But ultimately, that move is subsumed by the moves to higher levels of the large cardinal hierarchy.

It is traditional in set theory to consider individual large cardinal properties, rather than hypotheses implying a proper class or a stationary class of cardinals with the property, but these extra hypotheses are merely strenthenings of a given notion that typically have strength less than the next higher large cardinal notion.

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  • $\begingroup$ Thanks, Joel. Do you know of a reference for $\Sigma_n$-reflecting cardinals? As far as I can tell, they aren't mentioned in Jech or Kanamori. $\endgroup$ – Cole Leahy Sep 21 '11 at 2:58
  • $\begingroup$ Incidentally, we now call these the $\Sigma_n$-correct cardinals, with $\Sigma_n$-reflecting used only when the cardinal is also inaccessible. So some of the terminology in this answer does not match current usage. $\endgroup$ – Joel David Hamkins Apr 26 '14 at 21:15
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I would like to present a type of a formula $\psi_n(\lambda)$ such that $\psi_n(\lambda)$ implies $V_\lambda\prec_{\Sigma_{n+5}} V$. Call a cardinal $\kappa$ $n-$tremendous with target $\lambda$ if and only if there is a non-trivial elementary embedding $j:V\rightarrow M$ with critical point $\kappa$ such that $M$ is a transitive inner model such that $j(\kappa)>\lambda$, and $V_\kappa\prec V_{j(\kappa)}$, and $M\prec_{\Sigma_n} V$. Then define that $\kappa$ is $n-$tremendous if and only if it is $n-$tremendou with target 0, and $\psi_n(\lambda)\leftrightarrow \forall\delta(\lambda$ is $n$-tremendous with target $\delta$).

It is clear that every $n-$tremendous cardinal is $n-$huge, and moreover $\psi_n(\lambda)$ implies $\lambda$ is $n-$superhuge. I claim every $\psi_n(\lambda)$ cardinal is $\Sigma_{n+5}-$reflecting. It is clear every $n-$tremendous cardinal is $\Sigma_1-$reflecting, and as every $\psi_n(\lambda)$ cardinal is superhuge it is $\Sigma_5-$reflecting. Suppose $\psi_{n+1}(\lambda)$ and $\psi_n(\lambda)$ implies that $\lambda$ is $\Sigma_{n+5}-$reflecting. Suppose we have $\exists x(\phi(x,x_0...x_n))$, where $\phi(x,x_0...x_n)$ is $\Pi_{n+5}$. Let $z$ be a witness to that, and let $\lambda$ be $\Pi_{n+5}-$reflecting, such that $z\in V_\lambda$. Then, we can find some $j: V\rightarrow M$, we have, such that $M\vDash(\Pi_{n+5}$ formulas are\,downward absolute in $V_{j(\kappa)})$ and $j(\kappa)>\lambda$. Then $z\in V_{j(\kappa)}$, and so $M\vDash(V_{j(\kappa)}\vDash\exists x(\phi(x,x_0...x_n)))$, and so $V_{j(\kappa)}\vDash\exists x(\phi(x,x_0...x_n))$. As $V_\kappa\prec V_{j(\kappa)}$, we have $V_\kappa\vDash\exists x(\phi(x,x_0...x_n))$ and so $V_\kappa\prec_{\Sigma_{n+6}} V$. By a similar argument $\psi_n(\kappa)$ implies there is a normal measure $D$ on $\kappa$ such that $U\in D$, where $U=\{\lambda<\kappa|\lambda$ is $\Sigma_{n+5}$-reflecting$\}$.

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  • $\begingroup$ I've fixed some formatting (and added a paragraph break). In general it's better to use more "\$"s than to put a lot of "\,"s into a single mathmode block (also keep in mind the command "\mbox"). $\endgroup$ – Noah Schweber Jun 8 '19 at 4:01

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