Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If $\kappa$ is an inaccessible cardinal then $V_\kappa$ is a model of $\sf ZFC_2$ ($\sf ZFC$ with a second-order replacement axiom).

If there are many inaccessible cardinals then there are many models of $\sf ZFC_2$, but one can add all sort of $\varphi$ which describe $V_\kappa$ completely. For example if $\varphi$ is "there is no inaccessible cardinal" then $\sf ZFC_2+\varphi$ is only satisfied by $V_\kappa$ if $\kappa$ is the least inaccessible cardinal. We can continue and state that there is exactly one, or two, or $\alpha$ inaccessible cardinals for every "small enough" $\alpha$. We can continue by adding more and more information (e.g. there is no inaccessible which is the limit of inaccessibles; there is only one inaccessible which has a stationary set of inaccessibles; and so on).

Let us say that $\varphi$ in $n$-order logic is a categorizer for $\kappa$ if $V_\kappa$ is the unique model of $\sf ZFC_2+\varphi$. We say that $\kappa$ is $\Pi^n_m$-categorical (or $\Sigma^n_m$-categorical) if there is a $\Pi^n_m$ ($\Sigma^n_m$) sentence $\varphi$ which is a categorizer for $\kappa$.

On the other hand, we say that $\kappa$ is $\Pi^n_m$-indescribable if for every $R\subseteq V_\kappa$ and a $\Pi^n_m$ sentence $\psi$ such that $\langle V_\kappa,\in,R\rangle\models\psi$ there is some $\alpha<\kappa$ such that $\langle V_\alpha,\in,R\cap V_\alpha\rangle\models\psi$. (Similarly, of course, we define $\Sigma^n_m$-indescribable cardinals.)

Note that inaccessible cardinals are $\Pi^1_0$-indescribable, but the least inaccessible is $\Pi^0_n$-categorical for some $n$ (because the statement "there are no inaccessible cardinals is a first-order statement).

Question: Is there a [deep?] connection between the two notions?

share|improve this question
    
Also, $\Pi^n_m$-categorical is just a name I came up with. Is this a known large cardinal definition, and does it have a proper name? –  Asaf Karagila Jul 7 '13 at 11:58
2  
I'm intrigued by the downvote. –  Asaf Karagila Jul 7 '13 at 12:16
7  
Probably a finitist, Asaf... –  Todd Eisworth Jul 7 '13 at 14:13
add comment

1 Answer

up vote 6 down vote accepted

Categoricity should perhaps be conceived of not as a large cardinal notion, but rather as an anti-large cardinal notion, since most large cardinal concepts express some degree of reflection, which is the opposite of categoricity.

For example, if $\kappa$ is $\Pi^n_m$-categorical, then clearly it is not $\Pi^n_m$-indescribable. This is one connection between your concepts.

But let me argue that there cannot be an equivalence here between levels of categoricity and failure of indescribability. First, let's observe the following.

Lemma. Every $\Pi^n_m$-categorical cardinal is $\Delta_2$-definable in set theory.

Proof. If $\kappa$ is $\Pi^n_m$-categorical, using formula $\varphi$, then $\kappa$ is the unique cardinal such that $V_\kappa$ satisfies $\varphi$, and this is a first-order expressible fact $\varphi^\ast$ about $V_{\kappa+n}$ (since the second order quantifiers of $\varphi$ are interpreted as intended as first-order quantifiers inside $V_{\kappa+n}$. So $\kappa$ is the unique ordinal such that $V_{\kappa+n}\models\varphi^\ast$. This is $\Pi_2$ expressible as follows: a given ordinal $\xi$ is the $\kappa$ we are talking about if and only if $\forall Z\ (Z=V_{\xi+n}\to Z\models\varphi^\ast)$. This is a $\Pi_2$ assertion, since saying that $Z=V_{\xi+n}$ has complexity $\Pi_1$---one must say that $Z$ computes the power sets of its members correctly---and the latter part has all quantifiers bounded by $Z$. Similarly, $\xi=\kappa$ if and only if $\exists Z\ Z=V_{\xi+n}$ and $Z\models\varphi^\ast$, which is a $\Sigma_2$ formulation. QED

Now, recall that a cardinal $\delta$ is $\Sigma_2$-correct, if $V_\delta\prec_{\Sigma_2} V$. The reflection theorem shows that there is a closed unbounded class of $\Sigma_2$-correct cardinals. A $\Sigma_2$-reflecting cardinal is a regular $\Sigma_2$-correct cardinal, and these have the consistency strength strictly weaker than a Mahlo cardinal. Meanwhile, every strong cardinal, every strongly unfoldable cardinal (a transfinite continuation of the totally indescribable cardinals), every supercompact cardinal is $\Sigma_2$-reflecting.

Theorem. Every $\Pi^n_m$-categorical cardinal is smaller than every $\Sigma_2$-correct ordinal.

Proof. If $\kappa$ is $\Pi^n_m$-categorical, then it is $\Sigma_2$-definable by some formula $\psi$. In particular, the assertion "$\exists\kappa\ \psi(\kappa)$" is true in $V$. Since this is a $\Sigma_2$-assertion, it follows that if $\delta$ is $\Sigma_2$-correct, then $V_\delta$ will agree that there is such a $\kappa$. Since $\kappa$ is unique with $\psi(\kappa)$, it follows that $\kappa\lt\delta$. QED

So from a large cardinal perspective, the categorical cardinals must lie low in the hierarchy of ordinals.

Notice that since there are only countably many formulas, and in set theory we have a uniformly expressible account of whether a cardinal is $\Pi^n_m$-categorical by a given formula, it follows that there will be only countably many $\Pi^n_m$-categorical cardinals. True reflection does not even begin until you are above them all. But meanwhile, the indescribability and non-indescribability phenomena are unbounded in the ordinals.

Lastly, let me mention that the strongly unfoldable cardinals are best understood as a transfinite extension of the indescribability notions of indescribable cardinals. That is, they in effect replace $\Pi^n_m$ with $\Pi^\alpha_m$ for transfinite $\alpha$. But at this level (and even at the finite levels in my opinion), it is better to characterize the property in terms of embeddings than higher-order logic.

share|improve this answer
    
Joel, I don't understand something. If the least inaccessible is $\Pi^0_n$ categorical (where $n=5$ is probably enough), how can it be $\Pi^1_0$ indescribable? Also, we can describe the first inaccessible larger than the first measurable cardinal, "there is a maximal inaccessible cardinal and it is the only measurable", and that too is a $\Pi^0_n$ statement (because measurable are first-order definable). So either $\Sigma_2$-correct ordinals are very very very large (regardless to the shape of the universe), or I'm missing a point. –  Asaf Karagila Jul 7 '13 at 15:16
    
First, I think the least inaccessible is $\Pi^0_2$-categorical, that is, $n=2$ is enough, since the least inaccessible $\kappa$ is unique such that $V_\kappa\models\text{ZFC}_2+$"there is no inaccessible cardinal", and this latter assertion is $\Pi_2$ expressible in set theory, and hence $\Pi^0_2$ expressible as an assertion about $V_\kappa$. –  Joel David Hamkins Jul 7 '13 at 15:34
    
Second, yes, the first inaccessible above the least measurable are also below the first $\Sigma_2$-correct ordinal, since they are $\Sigma_2$-definable. These are rather small in terms of the much larger large cardinals. Consistency-wise, $\Sigma_2$-reflecting ordinals are very weak, but their size in a given universe is governed in effect by what cardinals exist there. –  Joel David Hamkins Jul 7 '13 at 15:36
    
As for your first question, yes, the first inaccessible is $\Pi^0_2$-categorical. And it seems to be $\Pi^1_0$-reflecting, since the subscript $0$ means that there actually are no second-order quantifiers in the statement $\psi$ to be reflected, and so we may just apply Lowenheim-Skolem to find $\alpha\lt\kappa$ with $\langle V_\alpha,R\cap V_\alpha\rangle\prec\langle V_\kappa,R\rangle$. –  Joel David Hamkins Jul 7 '13 at 15:41
    
I see! Thank you very much. –  Asaf Karagila Jul 7 '13 at 16:37
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.