Asaf Karagila
I don't have much choice...

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.

By
(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?

By Asaf Karagila
(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?

By Ashutosh
(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.

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

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

By Ashutosh
(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.

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