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Can $\Pi^m_n$ indescribable cardinal be the first one where $\text{GCH}$ fails? Hauser showed in

Hauser,K.: Indescribable cardinals and elementary embeddings. J. Symb. Logic 56, 439457 (1991)

that the answer is positive for $m=1$. But for $m \ge 2$, he only violated $\text{GCH}$ at indescribable while violating it on every inaccessible below.

What is known for $m \ge 2$?

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The answer is that this is impossible. The GCH cannot fail for the first time at a $\Pi^2_1$-indescribable cardinal. To see this, take any $\Pi^2_1$-indescribability embedding $j:M\to N$, meaning that $M$ is a transitive model of ZFC of size $\kappa$ with $M^{\lt\kappa}\subset M$, and $N$ is transitive, with $\text{cp}(j)=\kappa$ and $N$ is $\Sigma^2_1$-correct, meaning that $(V_{\kappa+2})^N\prec_{\Sigma_1} V_{\kappa+2}$ and $N^{|V_{\kappa}|}\subset N$. In particular, $N$ has the correct $P(\kappa)$. If the GCH holds below $\kappa$ in $V$, then it holds in $M$ up to $\kappa$, and so it holds in $N$ up to $j(\kappa)$, and so it holds at $\kappa$ in $N$, and so it holds in $V$ since $N$ has the correct $P(\kappa)$.

Concerning Mohammad's answer, the point to make is that my paper Unfoldable cardinals and the GCH shows how to make the GCH fail for the first time at an unfoldable cardinal. But this doesn't provide an example of total indescribability, since in fact "being the first failure of GCH" is a description that ruins indescribability.

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  • $\begingroup$ Very interesting. $\endgroup$ Commented Nov 11, 2013 at 2:58

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