To motivate things, let me start with a special case of the question I'm interested in. Let $\mathsf{In}(x)\equiv$ "$x$ is an inaccessible cardinal."



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> **Question 1**: Is it consistent with the theory $$\mbox{$\mathsf{ZFC}$ + "There is a transitive model of $\mathsf{ZFC}$ + 'There are $2$ inaccessibles'"}$$ that there is a transitive set $A\models\mathsf{ZFC}$ and an $\alpha\in\mathsf{In}^A$ such that no transitive set $B\supseteq A$ has $B\models\mathsf{ZFC}+\mathsf{In(\alpha)}+ \exists \beta>\alpha[\mathsf{In}(\beta)]$?
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(Of course to make this answerable in principle I'm assuming the consistency of $\mathsf{ZFC}$ + "There is a transitive model of $\mathsf{ZFC}$ + 'There are $2$ inaccessibles'" in the first place.)


Intuitively, I'm imagining building a model of $\mathsf{ZFC}$ stage-by-stage and changing the large cardinal requirements as we go ("actually, the client wants *two* inaccessibles"); this question is my attempt to make precise the naive question of whether we can be forced to 'undo previous progress' for non-consistency-strength-flavored reasons.




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More generally, suppose we have two formulas $P(x),Q(y)$ in the language of set theory. Say that $(P,Q)$ is a **fragile pair** iff the following hold:


* $\mathsf{ZFC}$ proves that both $P$ and $Q$ only ever (if at all) hold of ordinals.
* The conjunction of the following axioms is consistent:


	+ each axiom of $\mathsf{ZFC}$,
	+ there is a transitive model of $\mathsf{ZFC+\exists \alpha<\beta[P(\alpha)\wedge Q(\beta)]}$, and
	+ there is a transitive $A$ and an $\alpha\in A$ such that $$A\models\mathsf{ZFC+P(\alpha)}$$ but there is no transitive $B\supseteq A$ such that $$B\models \mathsf{ZFC} + P(\alpha) + \exists \beta>\alpha[ Q(\beta)].$$


In this language, and letting $\mathsf{In}(x)\equiv$ "$x$ is an inaccessible cardinal," Question 1 above is asking whether $(\mathsf{In,In})$ is a fragile pair.



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> **Question 2**: Heuristically speaking, when should we expect a pair $(P,Q)$ of naturally-occurring large cardinal properties to be (non-)fragile?
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This question is so vague that it deserves some unpacking. In my experience, questions like Q1 above tend to have very "coarse" answers in the sense that they are applicable to a broad class of properties in place of the specific ones considered. So, in fact, I expect that any answer to Q1 will already make progress on Q2; I'm asking Q2 separately to open the door to partial results which don't directly bear on Q1 or its more direct analogues.


I'm especially interested, re: Q2, in the case where $P$ and $Q$ are compatible with $\mathsf{V=L}$. To see why this is (potentially) special, consider the following simple proof that $(\mathsf{In},\mathsf{Meas})$ is fragile where $\mathsf{Meas}(x)\equiv$ "$x$ is a measurable cardinal:" if $A\models$ "$\mathsf{V=L}$ and $\alpha$ is the least inaccessible" then there is no end extension of $A$ with a measurable in which $\alpha$ stays inaccessible. More generally, $(P,\mathsf{Meas})$ is fragile whenever $P$ is compatible with $\mathsf{V=L}$, via the same argument. But nothing like this argument seems to work for weaker large cardinal properties.



Fragility of large cardinals with respect to transitive end extensions
Question 1: Yes, in fact if $M\models$ZFC+"There is a transitive model of $T$" where $T$ is the theory ZFC + "There are two distinct inaccessibles", then there is such an $(A,\alpha)$ - just let $A$ be the least model of ZFC + "There is an inaccessible". Then $A=L\_\gamma$ for some ordinal $\gamma$, and there is a unique $\alpha\in A$ such that $A\models$"$\alpha$ is inaccessible". Let $\alpha$ be this. Then $(A,\alpha)$ has the property, since $A$ is pointwise definable, and note that if $A\subseteq B$ and $B$ is transisitve and $B\models$ZFC+"$\alpha$ is inaccessible", then $\mathrm{Ord}^B=\gamma=\mathrm{Ord}^A$, and $B\models$"$\alpha$ is the unique inaccessible", since otherwise letting $\beta\in B$ be such that $\alpha\neq\beta$ and $B\models$"$\beta$ is inaccessible", then actually $\alpha<\beta$ and $L\_\beta\models$ZFC+"$\alpha$ is inaccessible", contradicting the minimality of $A$.


Here is a more general argument that covers the previous case and various other ones. Assume all the hypotheses of fragility for $(P,Q)$ excluding the last clause, i.e. regarding the transitive set $A$. Assume also that
ZFC + "$P(x)\wedge Q(y)\wedge x<y$" proves (i) "$x$ is uncountable" and (ii) "$V\_y\models$ ZFC + $P(x)$". Then $(P,Q)$ is fragile.


For just let $(\alpha,\beta)$ be the lexicographically least pair of ordinals such that $\alpha<\beta$ and there is a transitive set $M$ with $\alpha,\beta\in M$ and such that $M\models$ ZFC + "$P(\alpha)\wedge Q(\beta)$".
Then by assumption (ii), $V\_\beta^M\models$ ZFC + $P(\alpha)$.
But then working in $M$, we can find a countable elementary $X\preccurlyeq V\_\beta^M$ with $\alpha\in X$. Let $A$ be the transitive collapse of $X$
and $\pi:A\to V\_\beta^M$ the uncollapse. Let $\bar{\alpha}=\pi^{-1}(\alpha)$. By assumption (i), $\alpha$ is uncountable in $M$, so $\bar{\alpha}<\alpha$.


Now $A$ is transitive and $A\models$ ZFC + $P(\bar{\alpha})$. But by the minimality of $(\alpha,\beta)$, there is no $\bar{\beta}$ and transitive $B$ such that $B\models$ ZFC + $P(\bar{\alpha})\wedge Q(\bar{\beta})$
(and in particular no such $B$ with $A\subseteq B$).


So e.g. if $P(x)$ says "$x$ is Woodin" and $Q(y)$ says "$y$ is measurable" then $(P,Q)$ is fragile.


