# Sets of Fixed Points in an Ultrapower Embedding

Let $$\kappa$$ be a measurable cardinal and $$j: V \rightarrow M$$ be the elementary embedding with critical point $$\kappa$$. Let $$X$$ be a set such that for all $$x\in X$$, $$j(x) = x$$. My question is: when is $$X$$ a member of $$M$$?

We know that $$X \in M$$ when $$|X| \leq \kappa$$. But given everything in $$X$$ is unmoved by $$j$$, can $$X$$ be larger, or even arbitrarily large? In particular, is there such an $$X$$ of cardinality $$\kappa^+$$?

• For every cardinal $\kappa$, the cardinal $\sup_{n\in\mathbb{N}}j^n(\kappa)$ is fixed by $j$. So just take a really complicated set of such cardinals and you get a set all of whose elements are in $M$ and fixed by $j$ which is not itself in $M$. Commented Sep 18, 2020 at 22:18
• @NoahSchweber So for example, if A is a set of $\kappa^+$-many such cardinals, do we know that A is not in $M$? Commented Sep 18, 2020 at 22:35

In general, we can certainly have such a set in $$M$$, since if $$j:V\to M$$ is a $$\lambda$$-supercompactness embedding then in fact $$M^\lambda\subseteq M$$, so if $$\kappa$$ is supercompact then we can get such embeddings and sets $$X$$ with $$X$$ as large as desired.

You said "We know $$X\in M$$ when $$|X|\leq\kappa$$". But this isn't true in general. Maybe you are assuming that $$M=\mathrm{Ult}(V,U)$$ for some nonprincipal $$\kappa$$-complete ultrafilter $$U$$ over $$\kappa$$? Under this assumption, there can be no such $$X\in M$$ of cardinality $$>\kappa$$:

Theorem. Let $$M=\mathrm{Ult}(V,U)$$ where $$U$$ is a nonprincipal $$\kappa$$-complete ultrafilter over $$\kappa$$. Let $$j:V\to M$$ be the ultrapower embedding. Let $$X\in M$$ be a set such that $$j(x)=x$$ for all $$x\in X$$. Then $$V\models$$"$$|X|\leq\kappa$$" and hence $$M\models$$"$$|X|\leq\kappa$$" (since $$M^\kappa\subseteq M$$).

(The proof below uses an idea from the proof of Lemma 2.2/Theorem 2.3 of Reinhardt cardinals and iterates of V.)

Proof. Suppose otherwise. Then by considering a wellorder of $$X$$ in $$M$$ and cutting $$X$$ down to an initial segment of $$X$$ under this wellorder, we can assume $$V\models$$"$$|X|=\kappa^+$$". Now fix a bijection $$\pi:\kappa^+\to X$$ in $$V$$. Then $$M\models$$"$$j(\pi):j(\kappa^+)\to j(X)$$ is a bijection".

Claim: $$j(\pi)^{-1}X=j\kappa^+$$.

Proof of Claim: Equivalently, since $$j(\pi)$$ is a bijection, we must show $$j(\pi)\circ j\kappa^+=X$$. For this, first observe that by the elementarity of $$j$$, we have $$j(\pi(\alpha))=j(\pi)(j(\alpha))$$ for all $$\alpha<\kappa^+$$, and therefore $$j\circ\pi=j(\pi)\circ j\upharpoonright\kappa^+.$$ But $$j\circ\pi=\pi$$, since $$j\upharpoonright X=\mathrm{id}$$, so $$\pi=j(\pi)\circ j\upharpoonright\kappa^+,$$ so $$X=\pi\kappa^+=j(\pi)\circ j\kappa^+,$$ as desired.

Now by the claim, and since $$j(\pi),X\in M$$, we get $$j\kappa^+\in M$$, but this is false, because $$j\kappa^+$$ is cofinal in $$j(\kappa^+)$$, which is a regular cardinal in $$M$$, and $$j\kappa^+$$ has ordertype $$\kappa^+$$, and $$\kappa^+, so $$j\kappa^+$$ singularizes $$j(\kappa^+)$$ in $$M$$, a contradiction.

(Remark: Actually if $$M$$ contains the element $$jX$$ where $$X$$ has cardinality $$\lambda$$ in $$V$$, then $$j\lambda\in M$$, and so $$j\upharpoonright\lambda\in M$$; this is by a slight variant of the argument above; note we don't need that $$X$$ is fixed pointwise for this.)

Suppose that $$j$$ is an ultrapower embedding given by $$U$$. We know that: (1) $$U$$ has size $$2^\kappa$$, and (2) $$U\notin M$$.

Consider the set $$T$$ of the first $$2^\kappa$$ limit cardinals which are also fixed points of $$j$$, and let $$\lambda_\xi$$ be the $$\xi$$th member of $$T$$. Next fix a set of ordinals $$U'$$ such that the transitive collapse of $$U'$$ is the transitive closure of $$\{U\}$$ (which happens to be $$\{U\}\cup U\cup\kappa$$).

Finally, define $$X=T\cup\{\lambda_\xi^+\mid \xi\in U'\}$$. First note that if $$\lambda$$ is a fixed point, then $$\lambda^+$$ is a fixed point as well, so for all $$x\in X$$, $$j(x)=x$$. But we can read $$U'$$ from $$X$$, since we can enumerate all the limit cardinals in $$X$$, and $$U'$$ will be the set of the indices of those that also have their successor in $$X$$.

But since from $$U'$$ we can read $$U$$, and $$U\notin M$$, we have that $$X\notin M$$ either.

This is robust, in the sense that even if $$j$$ is not an ultrapower embedding, we can choose any $$U\in V\setminus M$$ and code it in a similar way.

(There are probably simpler examples, though.)

• Thanks so much. This helps a lot. I was still wondering if we can find such $X \notin M$ with smaller cardinality (in particular, $\kappa ^ +$). So take the set of first $\kappa ^+$ limit cardinals fixed by $j$. Do we know if this set is in $M$? Commented Sep 19, 2020 at 18:04