# Stationary preserving permutations are the identity on a club

There are 5 comments on this post.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 \(\{\alpha\mid\alpha\leq\pi(\alpha)\}\) 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!)

## There are 5 comments on this post.

(Apr 10 2017, 02:13)

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?

(Apr 10 2017, 08:35 In reply to Joseph Van Name)

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?

(Apr 16 2017, 00:00)

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.

(Apr 16 2017, 00:27 In reply to Ashutosh)

How do you mean that this is an application of this?

(Apr 16 2017, 01:23)

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.