OPEN PROBLEMS IN TOPOLOGY

18 Steprans / Steprans’ Problems [ch. 2In his proof that T = A Shelah introduced the ideal of sets on which anautohomeomorphism is trivial.3.1. Definition. If Φ ∈ A define J (Φ) = { X ⊂ ω :(∃f: X ↦→ ω) f isone-to-one and Φ|P(X) is induced by f }Hence Φ is trivial precisely if J (Φ) is improper—that is, contains ω. Itwas shown in Shelah’s argument that, under certain circumstances, if J (Φ)is merely sufficiently large then Φ is trivial. This is of course not true in generalbecause if there is a P -point of character ℵ 1 then there is an autohomeomorphismof βN \ N which is trivial on precisely this P -point. It might betempting to conjecture however, that if J (Φ) is either, improper or a primeideal for every autohomeomorphism Φ, then this implies that all such autohomeomorphismsare trivial. This is true but only for the reason that thehypothesis is far too strong—after all if Φ i : P(A i ) ↦→ P(A i ) is an autohomeomorphismfor i ∈ k and the sets A i are pairwise disjoint, then it is easy tosee how to define⊕{Φ i ; i ∈ k}: ∪{P(A i ); i ∈ k} ↦→ ∪{P(A i ); i ∈ k}in such a way that J (⊕{Φ i ; i ∈ k}) =⊕{J (Φ i ); i ∈ k} Notice that this impliesthat {J (Φ); Φ ∈ A} is closed under finite direct sums; but not much else isknown. In particular, it is not known what restrictions on {J (Φ); Φ ∈ A}imply that every member of A is trivial.? 29.Question 3.2. Suppose that for every Φ, J (Φ) is either improper or thedirect sum of prime ideals. Does this imply that every automorphism istrivial?Even the much weaker hypothesis has not yet been ruled out.? 30.Question 3.3. If J (Φ) ≠ ∅ for each Φ ∈ A does this imply that each Φ ∈ Ais trivial?Rudin’s proof of the existence of non-trivial autohomeomorphisms showseven more than has been stated. He showed in fact that, assuming CH, forany two P -points there is an autohomeomorphism of βN \ N which takes oneto the other.3.2. Definition. R H (κ) is defined to be the statement that, given two setsof P -points, A and B, bothofsizeκ, thereisΦ∈ A such that Φ(A) =Φ(B).Define R T (κ) to mean that, given two sequences of P -points of length κ, aand b, thereisΦ∈ A such that Φ(a(α)) = Φ(b(α)) for each α ∈ κ.In this notation, Rudin’s result says that CH implies that R T (1)holds. Itiseasy to see that, in general, R T (1) implies R T (n) for each integer n. Observe