Indescribable cardinal

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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 $\Pi_{m}$ sentence $\phi$:

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

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

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.

By consistency strength, quite a few equivalences are known about these cardinals. For example:

  • $\Pi_1^1$-Indescribability is equivalent to weak compactness.
  • $\Pi_0^2$-Indescribability is equivalent to $n$-Indescribability, Strong Inaccessibility, $\Sigma_1^1$-Indescribablity, $\Pi_0^n$-Indescribability given any $n>2$, and $\Pi_0^1$-Indescribability.
  • $\Pi_n^1$-Indescribability is equivalent to $\Sigma_{n+1}^1$-Indescribability.
  • If $m>1$, $\Pi_{n+1}^m$-Indescribability is stronger (consistency-wise) than $\Sigma_{n+1}^m$ 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 $m>1$.
  • $\Pi_n^m$-Indescribablity is equivalent to $m$-$\Pi_n$-Shrewdness.

Totally indescribable

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

$\alpha$-indescribable

$\alpha $-indescribable cardinals are $\Pi_{\omega}^{\alpha}$-indescribable when $\alpha$ is finite.

More information required on $\alpha$-indescribable