Stationary preserving permutations are the identity on a club

This is not something particularly interesting, I think. But it’s a nice exercise in Fodor’s lemma.

Theorem. Suppose that $\kappa$ is regular and uncountable, and $\pi\colon\kappa\to\kappa$ is a bijection mapping stationary sets to stationary sets. Then there is a club $C\subseteq\kappa$ such that $\pi\restriction C=\operatorname{id}$.

Proof. Note that the set $\{\alpha\mid\pi(\alpha)<\alpha\}$ is non-stationary, since otherwise by Fodor's lemma there will be a stationary subset on which $\pi$ is constant and not a bijection. This means that $\{\alpha\mid\alpha\geq\pi(\alpha)\}$ contains a club. The same arguments shows that $\pi^{-1}$ is non-decreasing on a club. But then the intersection of the two clubs is a club on which $\pi$ is the identity. $\square$

This is just something I was thinking about intermittently for the past few years, but now I finally spent enough energy to figure it out. And it’s cute. (Soon I will post more substantial posts, on far more exciting topics! Don’t worry!)

5 thoughts on “Stationary preserving permutations are the identity on a club”

  1. One can formulate this theorem in a category theoretic context. The category of filters in the category whose objects are pairs $(X,\mathcal{F})$ where $X$ is a set and $\mathcal{F}$ is a filter on $X$. If $\mathcal{F}$ is a filter on $X$ and $\mathcal{G}$ is a filter on $Y$, then let $\mathcal{A}_{(X,\mathcal{F}),(Y,\mathcal{G})}$ be the set of all functions $f:X\rightarrow Y$ where $f^{-1}[R]\in\mathcal{F}$ whenever $R\in\mathcal{G}$. Let $\simeq$ be the equivalence relation on $\mathcal{A}_{(X,\mathcal{F}),(Y,\mathcal{G})}$ where $f\simeq g$ iff $\{x\in X|f(x)=g(x)\in\mathcal{F}\}$. Then $\mathcal{A}_{(X,\mathcal{F}),(Y,\mathcal{G})}/\simeq$ is the set of all morphisms from $(X,\mathcal{F})$ to $(Y,\mathcal{G})$ in the category of filters. This result that you have proven therefore states that if $\kappa$ is a regular cardinal and $\mathcal{F}$ is the club filter, then the object $(\kappa,\mathcal{F})$ has no non-trivial automorphisms. As all set theorists are aware, the pairs $(X,\mathcal{U})$ where $\mathcal{U}$ is an ultrafilter do not contain any non-trivial endomorphisms. It seems like this property of no non-trivial endomorphisms should be quite prevalent and that there should be many different kinds of normal filters on regular cardinals with no non-trivial automorphisms. Have you found any other examples of filters with no non-trivial automorphisms?

    1. That’s interesting. I haven’t thought about other examples, though. The motivation for this, actually, comes from symmetric extensions. If one adds a subset, say of $\omega_1$, and you want to use permutations of $\omega_1$ to induce automorphisms of the forcing, what this right here shows, is that if your conditions are stationary sets, then your approach leads to some sort of pseudo-rigidity. (There might be automorphisms, but not ones induced by permutations of $\omega_1$.)

      I guess we can formulate this notion as the filter being rigid. Then one of the first obvious question would be, are there rigid filters on $\omega$ which are not ultrafilters?

  2. This has an interesting “application” due to Komjath: Suppose X is a set of reals such that the ideal of meager subsets of X is isomorphic to the non stationary ideal on omega_1 (Komjath showed that such X can consistently exist). Then X^2 is a non meager subset of plane each of whose non meager subsets contains three collinear points. Shelah and I proved that a similar result holds for the null ideal.

  3. Showing that every subset Y of X^2 that does not contain three collinear points is meager boils down to showing that Y can be covered by the diagonal and two other meager sets and this uses the rigidity of the non-stationary ideal.

    This is not really an application but just that this fact was exploited here.

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