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The update gives a cardinal $\bar\kappa$ that is $\alpha$-extendible for $\alpha$ much larger than many measurable cardinals above $\bar\kappa$.

The update gives a cardinal $\bar\kappa$ that is $\alpha$-extendible for $\alpha$ much larger than many measurable cardinals above $\bar\kappa$. In particular, $\bar\kappa$ is highly partially supercompact.

Proof. Suppose that $\kappa$ has the stated reflection property. Consider some large ordinal $\theta$ and let $B=\langle V_{\theta+1},{\in}\rangle$. By the reflection property, there is some structure $A$ with lots of elementary embeddings $j:A\to B$ hitting any desired target. We may assume $A$ is a transitive set. Let $\bar\kappa$ be the smallest critical point of such an embedding. For any $X\subseteq\bar\kappa$, it is in $B$ and so there is some $x\in A$ and $j:A\to B$ with $j(x)=X$. Since $x$ and $j(x)=X$ must agree up to $\bar\kappa$, this implies $X\in A$. So $P(\bar\kappa)\subseteq A$. This implies that $\bar\kappa$ is measurable, since we can define the induced normal measure $X\in\mu\iff \bar\kappa\in j(X)$ for such a $j$ with critical point $\bar\kappa$.

The main point is that the embeddings $j:A\to B$ with critical point $\bar\kappa$ now are witnesses of $\bar\kappa_1$-extendibility, which is already a high degree of extendibility. And we can continue with $\kappa_\alpha$ etc. getting very high degrees of extendibility of $\bar\kappa$ this way.

The update gives a cardinal $\bar\kappa$ that is $\alpha$-extendible for $\alpha$ much larger than many measurable cardinals above $\bar\kappa$.

Proof. Suppose that $\kappa$ has the stated reflection property. Consider some large ordinal $\theta$ and let $B=\langle V_{\theta+1},{\in}\rangle$. By the reflection property, there is some structure $A$ with lots of elementary embeddings $j:A\to B$ hitting any desired target. We may assume $A$ is a transitive set. Let $\bar\kappa$ be the smallest critical point of such an embedding. For any $X\subseteq\bar\kappa$, it is in $B$ and so there is some $x\in A$ and $j:A\to B$ with $j(x)=X$. Since $x$ and $j(x)=X$ must agree up to $\bar\kappa$, this implies $X\in A$. So $P(\bar\kappa)\subseteq A$. This implies that $\bar\kappa$ is measurable, since we can define the induced normal measure $X\in\mu\iff \bar\kappa\in j(X)$ for such a $j$ with critical point $\bar\kappa$.

The main point is that the embeddings $j:A\to B$ with critical point $\bar\kappa$ now are witnesses of $\bar\kappa_1$-extendibility, which is already a high degree of extendibility. And we can continue with $\kappa_\alpha$ etc. getting very high degrees of extendibility of $\bar\kappa$ this way. It seems we get that $\bar\kappa$ is $\alpha$-extendible, where $\alpha$ is the $\bar\kappa$th measurable cardinal above $\bar\kappa$. And much more.

Let $\bar\kappa_0=\bar\kappa$ be the least critical point that arises with $j:A\to B$, and let $\kappa_1$ be the next critical point. Such a critical point arise because there must be $j:A\to B$ with $\bar\kappa$ in the range of $j$. What I claim is that $P(\bar\kappa_1)\subseteq A$. To see this, consider any $X\subseteq \bar\kappa_1$, and then find a $j:A\to B$ with $\{\bar\kappa,X\}\in\text{ran}(j)$. It follows that $\bar\kappa\in\text{ran}(j)$ and so the critical point of $j$ is at least $\bar\kappa_1$, and so $X\in A$ by the same reasoning as before.

Let $\bar\kappa_0=\bar\kappa$ be the least critical point that arises with $j:A\to B$, and let $\kappa_1$ be the next critical point. Such a critical point must arise because there must be $j:A\to B$ with $\bar\kappa$ in the range of $j$, and such an embedding must have critical point above $\bar\kappa$. What I claim is that $P(\bar\kappa_1)\subseteq A$. To see this, consider any $X\subseteq \bar\kappa_1$, and then find a $j:A\to B$ with $\{\bar\kappa,X\}\in\text{ran}(j)$. It follows that $\bar\kappa\in\text{ran}(j)$ and so the critical point of $j$ is at least $\bar\kappa_1$, and so $X\in A$ by the same reasoning as before.