# Indescribable cardinal

(Redirected from Totally indescribable)
The Structure of Indescribability in Consistency Strength

A cardinal $\kappa$ is indescribable if it holds the reflection theorem up to a certain point. This is important to mathematics because of the concern for the reflection theorem. In more detail, a cardinal $\kappa$ is $\Pi_{m}^n$-indescribable if and only if for every $\Pi_{m}$ first-order sentence $\phi$:

$$\forall S\subseteq V_{\kappa}(\langle V_{\kappa+n};\in,S\rangle\models\phi\rightarrow\exists\alpha<\kappa(\langle V_{\alpha+n};\in,S\cap V_{\alpha}\rangle\models\phi))$$

Likewise for $\Sigma_{m}^n$-indescribable cardinals.

Here are some other equivalent definitions:

• A cardinal $\kappa$ is $\Pi_m^n$-indescribable for $n>0$ iff for every $\Pi_m$ first-order unary formula $\phi$:

$$\forall S\subseteq V_\kappa(V_{\kappa+n}\models\phi(S)\rightarrow\exists\alpha<\kappa(V_{\alpha+n}\models\phi(S\cap V_\alpha)))$$

• A cardinal $\kappa$ is $\Pi_m^n$-indescribable iff for every $\Pi_m$ $n+1$-th-order sentence $\phi$:

$$\forall S\subseteq V_\kappa(\langle V_\kappa;\in,S\rangle\models\phi\rightarrow\exists\alpha<\kappa(\langle V_\alpha;\in,S\cap V_\alpha\rangle\models\phi))$$

In other words, if a cardinal is $\Pi_{m}^n$-indescribable, then every $n+1$-th order logic statement that is $\Pi_m$ expresses the reflection of $V_{\kappa}$ onto $V_{\alpha}$. This exercises the fact that these cardinals are so large they almost resemble the order of $V$ itself. This definition is similar to that of shrewd cardinals, an extension of indescribable cardinals.

## Variants

Totally indescribable cardinals are $\Pi_m^n$-indescribable for every natural $m$ and $n$ (equivalently $\Sigma_m^n$-indescribable for every natural m and n, equivalently $\Delta_m^n$-indescribable for every natural $m$ and $n$). This means that every (finitary) formula made from quantifiers, $\in$ and a subset of $V_{\kappa}$ reflects from $V_{\kappa}$ onto a smaller rank.

$Q$-indescribable cardinals are those which have the property that for every $Q$-sentence $\phi$:

$$\forall S\subseteq V_\kappa(\langle V_\kappa;\in,S\rangle\models\phi\rightarrow\exists\alpha<\kappa(\langle V_\alpha;\in,S\cap V_\alpha\rangle\models\phi))$$

$\beta$-indescribable cardinals are those which have the property that for every first order sentence $\phi$:

$$\forall S\subseteq V_\kappa(\langle V_{\kappa+\beta};\in,S\rangle\models\phi\rightarrow\exists\alpha<\kappa(\langle V_{\alpha+\beta};\in,S\cap V_\alpha\rangle\models\phi))$$

There is no $\kappa$ which is $\kappa$-indescribable. A cardinal is $\Pi_{<\omega}^m$-indescribable iff it is $m$-indescribable for finite $m$. Every $\omega$-indescribable cardinal is totally indescribable.

## Facts

Here are some known facts about indescribability:

• $\Pi_1^1$-indescribability is equivalent to weak compactness. [1]
• $\Pi_2^0$-indescribability is equivalent to $n$-indescribability, strong inaccessibility, $\Sigma_1^1$-indescribablity, $\Pi_n^0$-indescribability given any $n>2$, and $\Pi_0^1$-indescribability. [1], [2]
• $\Pi_n^1$-indescribability is equivalent to $\Sigma_{n+1}^1$-Indescribability. [2]
• If $m>1$, $\Pi_{n+1}^m$-indescribability is stronger (consistency-wise) than $\Sigma_n^m$ and $\Pi_n^m$-indescribability; every $\Pi_{n+1}^m$-indescribable cardinal is also both $\Sigma_n^m$ and $\Pi_n^m$-indescribable and a stationary limit of such for $m>1$. [2]
• If $m>1$, the least $\Pi_n^m$-indescribable cardinal is less than the least $\Sigma_n^m$-indescribable cardinal, which is in turn less than the least $\Pi_{n+1}^m$-indescribable cardinal. [2]
• $\Pi_n^m$-indescribablity is equivalent to $m$-$\Pi_n$-shrewdness (similarly with $\Sigma_n^m$). [3]
• Every measurable cardinal is $\Pi_1^2$-indescribable. Although, the least measurable is $\Sigma_1^2$-describable. [1]
• Every critical point of a nontrivial elementary embedding $j:M\rightarrow M$ for some transitive inner model $M$ of ZFC is totally indescribable in $M$. (For example, rank-into-rank cardinals, $0^{\#}$ cardinals, and $0^{\dagger}$ cardinals). [1]
• If $2^\kappa\neq\kappa^+$ for some $\Pi_1^2$-indescribable cardinal, then there is a smaller $\lambda$ such that $2^\lambda\neq\lambda^+$. However, assuming the consistency of the existence of a $\Pi_n^1$-indescribable cardinal $\kappa$, it is consistent for $\kappa$ to be the least cardinal such that $2^\kappa\neq\kappa^+$. [4]

## References

1. Jech, Thomas J. Set Theory. Third, Springer-Verlag, Berlin, 2003. (The third millennium edition, revised and expanded) www   bibtex
2. Kanamori, Akihiro. The higher infinite. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www   bibtex
3. Rathjen, Michael. The art of ordinal analysis. , 2006. www   bibtex
4. Hauser, Kai. Indescribable Cardinals and Elementary Embeddings. 56(2):439 - 457, 1991. www   DOI   bibtex
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