A FIVE ELEMENT BASIS FOR THE UNCOUNTABLE LINEAR ORDERS

JUSTIN TATCH MOORE

Abstract. In this paper I will show that it is relatively consistent with the usual axioms of mathematics (ZFC) together with a strong form of the axiom of infinity (the existence of a supercompact cardinal) that the class of uncountable linear orders has a five element basis. In fact such a basis follows from the Proper Forcing Axiom, a strong form of the Baire Category Theorem. The elements are X, ω_{1}, ω_{1}^{∗}, C, C^{∗} where X is any suborder of the reals of cardinality ℵ_{1} and C is any Countryman line. This confirms a longstanding conjecture of Shelah.

1. Introduction

Our focus in this paper will be to show that the Proper Forcing Axiom (PFA) implies that any uncountable linear order must contain an isomorphic copy of one of the following five orders: X, ω_{1}, ω_{1}^{∗}, C, and C^{∗}. Here X is any fixed set of reals of cardinality ℵ_{1} and C is any fixed Countryman line. Such a list is called a basis.

The simplest example of an uncountable linear order is ℝ, the real line. This object serves as the prototype for the class of linear orders and as the canonical example of an uncountable set. Early on in modern set theory, Baumgartner proved the following deep result which suggested that it might be possible to prove more general classification results for uncountable linear orders.

Theorem 1.1. [3] (PFA) If two sets of reals are ℵ_{1}-dense,^{1} then they are isomorphic. In particular if X is a set of reals of cardinality ℵ_{1},

2000 Mathematics Subject Classification. Primary: 03E35, 03E75, 06A05; Secondary: 03E02 .

Key words and phrases. Aronszajn, basis, Countryman, forcing axiom, linear order, MRP, PFA, Shelah’s conjecture.

Revisions and updates to the paper were supported by NSF grant DMS–0401893. Travel support to present these results in Kobe, Japan was provided by Grant-in-aid for Scientific Research (C)(2)15540120, Japanese Society for the Promotion of Science.

1I.e. every interval meets them at a set of cardinality ℵ_{1}.

then X serves as a single element basis for the class of uncountable separable linear orders.

PFA is a strengthening of the Baire Category Theorem and is independent of the usual axioms of set theory. Frequently — as in Baumgartner’s result above — this axiom can be used to find morphisms between certain structures or to make other combinatorial reductions (see [1], [3], [7], [24], [25], [27]). Some additional assumption is necessary in Baumgartner’s result because of the following classical construction of Sierpiński.

Theorem 1.2. [19] There is a set of reals X of cardinality continuum such that if f ⊆ X^{2} is a continuous injective function, then f differs from the identity function on a set of cardinality less than continuum.

From this it is routine to prove that under the Continuum Hypothesis there is no basis for the uncountable separable linear orders of cardinality less than |𝒫(ℝ)|. This gives a complete contrast to the conclusion of Baumgartner’s result.

The simplest example of a linear order which is separable only in the trivial instances is a well order. The uncountable well orders have a canonical minimal representative, the ordinal ω_{1}.^{2} Similarly, the converse ω_{1}^{∗} of ω_{1} obtained by reversing the order relation forms a single element basis for all of the uncountable converse well orders.

Those uncountable linear orders which do not contain uncountable separable suborders or copies of ω_{1} or ω_{1}^{∗} are called Aronszajn lines.^{3} They are classical objects considered long ago by Aronszajn and Kurepa who first proved their existence. Some time later Countryman made a brief but important contribution to the subject by asking whether there is an uncountable linear order C whose square is the union of countably many chains.^{4} Such an order is necessarily Aronszajn. Furthermore, it is easily seen that no uncountable linear order can embed into both a Countryman line and its converse. Shelah proved that such orders exist in ZFC [16] and made the following conjecture:

Shelah’s Conjecture. [16] (PFA) Every Aronszajn line contains a Countryman suborder.

This soon developed into the following equivalent basis conjecture — see [22].

2The canonical representation of well orders mentioned here is due to von Neumann.

3Or Specker types.

4Here chain refers to the coordinate-wise partial order on C^{2}.

Conjecture. [4] (PFA) The orders X, ω_{1}, ω_{1}^{∗}, C and C^{∗} form a five element basis for the uncountable linear orders any time X is a set of reals of cardinality ℵ_{1} and C is a Countryman line.

Notice that by our observations such a basis is necessarily minimal.

This problem was exposited, along with some other basis problems for uncountable structures, in Todorčević’s [21]. It also appears as Question 5.1 in Shelah’s problem list [18]. A related and inspirational analysis of Aronszajn trees was also carried out in [22].

In this paper I will prove Shelah’s conjecture. In doing so, I will introduce some new methods for applying PFA which may be relevant to solving other problems. I would like to thank Ilijas Farah, Jean Larson, Paul Larson, Bill Mitchell, and Boban Veličković for carefully reading the paper and offering their suggestions and comments. I would also like to thank Jörg Brendle for supporting my visit to Japan where I presented the results of this paper in a series of lectures at Kobe University in December 2003.

This work is dedicated to Fennel Moore.

2. Background

This paper should be readily accessible to anyone who is well versed in set theory and the major developments in the field in the 70s and 80s. The reader is assumed to have proficiency in the areas of Aronszajn tree combinatorics, forcing axioms, the combinatorics of [X]^{ℵ_{0}}, and Skolem hull arguments. Jech’s [12] and Kunen’s [14] serve as good references on general set theory. They both contain some basic information on Aronszajn trees; further reading on Aronszajn trees can be found in [5], [20], and [26]. The reader is referred to [6], [16], [22], [23], or [26] for information on Countryman lines. It should be noted, however, that knowledge of the method of minimal walks will not be required.

The set theoretic assumption we will be working with is the Proper Forcing Axiom. We will be heavily utilizing Todorcevic’s method of building proper forcings using models as side conditions. Both [25] and the section on PFA in [24] serve as good concise references on the subject. See [15] for information on the Mapping Reflection Principle. For basic forcing technology, the reader is referred to [11] and [14]. Part III of Jech’s [11] gives a good exposition on the combinatorics of [X]^{ℵ_{0}}, the corresponding closed unbounded (or club) filter, and related topics.

The notation in this paper is mostly standard. If X is an uncountable set, then [X]^{ℵ_{0}} will be used to denote the collection of all countable subsets of X. All ordinals are von Neumann ordinals — they are the set of their predecessors under the ∈ relation. The collections H(θ) for regular cardinals θ consist of those sets of hereditary cardinality less than θ. Hence H(2^{θ}^{+}) contains H(θ^{+}) as an element and 𝒫(H(θ^{+})) as a subset. Often when I refer to H(θ) in this paper I will really be referring to the structure (H(θ), ∈, ◁) where ◁ is some fixed well ordering of H(θ) which can be used to generate the Skolem functions.

3. The axioms

The working assumption in this paper will be the Proper Forcing Axiom introduced by Shelah and proved relatively consistent from a supercompact cardinal. We will often appeal to the bounded form of this axiom isolated by Goldstern and Shelah [9]. We will use an equivalent formulation due to Bagaria [2]:

BPFA: If ϕ is a formula in language of H(ℵ_{1}^{+}) with only bounded quantifiers and there is a proper partial order which forces ∃Xϕ(X), then H(ℵ_{1}^{+}) already satisfies ∃Xϕ(X).

At a crucial point in the proof we will also employ the Mapping Reflection Principle introduced recently in [15]. In order to state it we will need the following definitions.

Definition 3.1. If X is an uncountable set, then there is a natural topology — the Ellentuck topology — on [X]^{ℵ_{0}} defined by declaring

[x, N] = {Y ∈ [X]^{ℵ_{0}} : x ⊆ Y ⊆ N}

to be open whenever N is in [X]^{ℵ_{0}} and x is a finite subset of N.

This topology is regular and 0-dimensional. Moreover, the closed and cofinal sets generate the club filter on [X]^{ℵ_{0}}.

Definition 3.2. If M is an elementary submodel of some H(θ) and X is in M, then we say a subset Σ ⊆ [X]^{ℵ_{0}} is M-stationary if whenever E ⊆ [X]^{ℵ_{0}} is a club in M, the intersection Σ ∩ E ∩ M is non-empty.

Definition 3.3. If Σ is a set mapping defined on a set of countable elementary submodels of some H(θ) and there is an X such that Σ(M) ⊆ [X]^{ℵ_{0}} is open and M-stationary for all M, then we say Σ is an open stationary set mapping.

The Mapping Reflection Principle is the following statement:

MRP: If Σ is an open stationary set mapping defined on a club of models, then there is a continuous ∈-chain ⟨N_{ξ} : ξ < ω_{1}⟩ in the domain of Σ such that for every ν > 0 there is a ν_{0} < ν such that N_{ξ} ∩ X is in Σ(N_{ν}) whenever ν_{0} < ξ < ν.

The sequence ⟨N_{ξ} : ξ < ω_{1}⟩ postulated by this axiom will be called a reflecting sequence for the set mapping Σ.

4. A combinatorial reduction

Rather than prove Shelah’s basis conjecture directly, I will appeal to the following reduction.

Theorem 4.1. (BPFA) The following are equivalent:

(1) The uncountable linear orders have a five element basis.

(2) There is an Aronszajn tree T such that for every K ⊆ T there is an uncountable antichain X ⊆ T such that ∧(X)^{5} is either contained in or disjoint from K.

Remark. This result seems to essentially be folklore; reader interested in the historical aspects of this are referred to p 79 of [1], [4], [16]. A detailed proof of this theorem can be found in the last section of [22]. I will sketch the proof for completeness.

The implication (1) implies (2) does not require BPFA and in fact (1) implies that the conclusion of (2) holds for an arbitrary Aronszajn tree T. To see why it is true, suppose that (T, ≤) is an Aronszajn tree equipped with a lexicographical order and suppose that K ⊆ T witnesses a failure of (2). If (T, ≤) doesn’t contains a Countryman suborder, then (1) must fail. So without loss of generality, we may assume that (T, ≤) is Countryman.

Define s ≤′ t iff s ∧ t is in K and s ≤ t or s ∧ t is not in K and t ≤ s. It is sufficient to check that neither (T, ≤) nor its converse (T, ≥) embeds an uncountable suborder of (T, ≤′). This is accomplished with two observations. First, since (T, ≤) and its converse are Countryman, any such embedding can be assumed to be the identity map. Second, if ≤ and ≤′ agree on X ⊆ T, then ∧(X) ⊆ K; disagreement on X results in ∧(X) ∩ K = ∅.

For the implication (2) implies (1) we first observe that, by Baumgartner’s result mentioned above, it suffices to show that the Aronszajn lines have a two element basis. Fix a Countryman line C which is a lexicographical order ≤ on an Aronszajn tree T. The club isomorphism of Aronszajn trees under BPFA [1] together with some further appeal to MA_{ℵ_{1}} implies that any Aronszajn line contains a suborder isomorphic to some (X, ≤′) where X ⊆ T is uncountable and binary and ≤′ is a — possibly different — lexicographical order on T. Statement (2) is used to compare ≤ and ≤′ and find an uncountable Y ⊆ X on which these orders always agree or always disagree. Applying MA_{ℵ_{1}}, C embeds into all its uncountable suborders, thus finishing the proof.

5This will be defined momentarily.

5. The proof of the main result

In this section we will prove the basis conjecture of Shelah by proving the following result and appealing to Theorem 4.1.

Theorem 5.1. (PFA) There is an Aronszajn tree T such that if K ⊆ T, then there is an uncountable antichain X ⊆ T such that ∧(X) is either contained in or disjoint from K.

The proof will be given as a series of lemmas. In each case, I will state any set theoretic hypothesis needed to prove a lemma. This is not so much to split hairs but because I feel that it will help the reader better understand the proof.

For the duration of the proof, we will let T be a fixed Aronszajn tree which is contained in the complete binary tree, coherent, closed under finite changes, and special.^{6} It will be convenient to first make some definitions and fix some notation.

Definition 5.2. If s and t are two elements of T, then diff(s, t) is the set of all ξ such that s(ξ) and t(ξ) are defined and not equal. If F ⊆ T, then diff(F) is the union of all diff(s, t) such that s and t are in F. The coherence of T is the assertion that diff(s, t) is a finite set for all s, t in T.

Definition 5.3. If X is a subset of T and δ < ω_{1}, then X ↾ δ is the set of all t ↾ δ such that t is in X. Here t ↾ δ is just functional restriction.

Definition 5.4. If s and t are in T, then Δ(s, t) is the least element of diff(s, t). If s and t are comparable, we leave Δ(s, t) undefined.^{7} If Z ⊆ T and t is in T, then Δ(Z, t) = {Δ(s, t) : s ∈ Z}.

Definition 5.5. If X is a finite subset of T, then X(j) will denote the j^{th} least element of X in the lexicographical order inherited from T.

Definition 5.6. If s, t are incomparable in T, then the meet of s and t — denoted s ∧ t — is the restriction s ↾ Δ(s, t) = t ↾ Δ(s, t). If X is a subset of T, then ∧(X) = {s ∧ t : s, t ∈ X}.^{8}

The following definition provides a useful means of measuring subsets of an elementary submodel’s intersection with ω_{1}.

6The tree T (ρ_{3}) of [23] is such an example. Coherence is defined momentarily.

7This is somewhat non-standard but it will simplify the notation at some points.

For example, in the definition of Δ(Z, t) we only collect those values where Δ is defined.

8The domain of ∧ is the same as the domain of Δ; the set of all incomparable pairs of elements of T.

Definition 5.7. If P is a countable elementary submodel of H(ℵ_{1}^{+}) containing T as an element, define ℐ_{P} (T) to be the collection of all I ⊆ ω_{1} such that for some uncountable Z ⊆ T in P and some t of height P ∩ ω_{1} which is in the downward closure of Z, the set Δ(Z, t) is disjoint from I.

The following propositions are routine to verify using the coherence of T and its closure under finite changes (compare to the proof that 𝒰 (T) is a filter in [22] or [26]).

Proposition 5.8. If I is in ℐ_{P} (T) and t is in T with height P ∩ ω_{1}, then there is a Z ⊆ T in P such that t is in the downward closure^{9} of Z and Δ(Z, t) is disjoint from I.

Proposition 5.9. If I is in ℐ_{P} (T), Z_{0} is a subset of T in P and t is an element of the downward closure of Z_{0} of height P ∩ ω_{1}, then there is a Z ⊆ Z_{0} in P which also contains t in its downward closure and satisfies Δ(Z, t) ∩ I is empty.

Proposition 5.10. ℐ_{P} (T) is a proper ideal on ω_{1} which contains I ⊆ ω_{1} whenever I ∩ P is bounded in ω_{1} ∩ P.

Proposition 5.11. Suppose P is a countable elementary submodel of H(ℵ_{1}^{+}) such that Z ⊆ T is an element of P, and there is a t ∈ T of height P ∩ ω_{1} in the downward closure of Z. Then Z is uncountable.

Let K ⊆ T be given. The following definitions will be central to the proof. The first is the naïve approach to forcing an uncountable X such that ∧(X) is contained in K.

Definition 5.12. ℋ (K) is the collection of all finite X ⊆ T such that ∧(X) is contained in K.^{10}

It is worth noting that ℋ (K) is the correct forcing to work with if K is union of levels of T; this is demonstrated in [22]. This and other successes in [22] emboldened me to attempt the more general case in which K is an arbitrary subset of T.

The second is the notion of rejection which will be central in the analysis of ℋ (K). For convenience we will let ℰ denote the collection of all clubs E ⊆ [H(ℵ_{1}^{+})]^{ℵ_{0}} which consist of elementary submodels which contain T and K as elements. Let E_{0} denote the element of ℰ which consists of all such submodels.

9The downward closure of Z is the collection of all s such that s ≤ s^{∗} for some s^{∗} in Z

10A collection of finite sets such as this becomes a forcing notion when given the order of reverse inclusion (q ≤ p means that q is stronger than p). A collection of ordered pairs of finite sets becomes a forcing by coordinate-wise reverse inclusion.

Definition 5.13. If X is a finite subset of T, then let K(X) denote the set of all γ < ω_{1} such that for all t in X, if γ is less than the height of t, then t ↾ γ is in K.

Definition 5.14. If P is in E_{0} and X is a finite subset of T, then we say that P rejects X if K(X) is in ℐ_{P} (T).

The following trivial observations about P in E_{0} and finite X ⊆ T are useful and will be used tacitly at times in the proofs which follow.

Proposition 5.15. If P does not reject X, then it does not reject any of its restrictions X ↾ γ.

Proposition 5.16. P rejects X iff it rejects X ↾ (P ∩ ω_{1}) iff it rejects X \ P.

Proposition 5.17. If X is in P, then P does not reject X.

The forcing notion ∂(K) which we are about to define seeks to add a subset of T in which rejection is rarely encountered.^{11}

Definition 5.18. ∂(K) consists of all pairs p = (X_{p}, 𝒩_{p}) such that:

(1) 𝒩_{p} is a finite ∈-chain such that if N is in 𝒩 , then T and K are in N and N is the intersection of a countable elementary submodel of H(2^{2^{ℵ_{1}}}^{+}) with H(2^{ℵ_{1}}^{+}).

(2) X_{p} ⊆ T is a finite set and if N is in 𝒩_{p}, then there is an E in

ℰ ∩ N such that X_{p} is not rejected by any element of E ∩ N. We will also be interested in the suborder

∂ℋ (K) = {p ∈ ∂(K) : X_{p} ∈ ℋ (K)}

which seems to be the correct modification of ℋ (K) from the point of view of forcing the conclusion of the main theorem.

In order to aid in the presentation of the lemmas, I will make the following definition.

Definition 5.19. ∂(K) is canonically proper if whenever M is a countable elementary submodel of H(|2^{∂(K)}|^{+}) and ∂(K) is in M, any condition p which satisfies M ∩ H(2^{ℵ_{1}}^{+}) is in 𝒩_{p} is (M, ∂(K))-generic. An analogous definition is made for ∂ℋ (K).

We will eventually prove that, assuming the Proper Forcing Axiom, ∂ℋ (K) is canonically proper. The following lemma shows that this is sufficient to finish the argument.

11The symbol ∂ is being used here because there is a connection to the notion of a Cantor-Bendixon derivative. In a certain sense we are removing the parts of the partial order ℋ (K) which are causing it to be improper.

Lemma 5.20. (BPFA) If ∂ℋ (K) is canonically proper, then there is an uncountable X ⊆ T such that ∧(X) is either contained in K or disjoint from K.

Remark. This conclusion is sufficient since the properties of T imply that X contains an uncountable antichain.

Proof. Let M be a countable elementary submodel of H(|2^{∂ℋ (K)} |^{+}) containing ∂ℋ (K) as an element. Let t be an element of T of height M ∩ ω_{1}. If

p = ({t}, {M ∩ H(2^{ℵ_{1}}^{+})})

is a condition in ∂ℋ (K), then it is (M, ∂ℋ (K))-generic by assumption. Consequently p forces that the interpretation of

Ẋ = {s ∈ T : ∃q ∈ Ġ(s ∈ X_{q})}

is uncountable. Since Ẋ will then be forced to have the property that ∧(Ẋ) ⊆ Ǩ, we can apply BPFA to find such an X in V.

Now suppose that p is not a condition. It follows that there is a countable elementary submodel P of H(ℵ_{1}^{+}) in M such that T is in P and K({t}) is in ℐ_{P} (T). Therefore there is a Z ⊆ T in P such that t ↾ (P ∩ ω_{1}) is in the downward closure of Z and for all s in Z, s ∧ t is not in K. Let Y consist of all those w in ∧(Z) such that if u, v are incomparable elements of Z and u ∧ v ≤ w, then u ∧ v is not in K. Notice that Y is an element of P. Y is uncountable since it contains s ∧ t for every s in P ∩ Z which is incomparable with t and the heights of elements of this set are easily seen to be unbounded in P ∩ ω_{1}. We are therefore finished once we see that ∧(Y) is disjoint from K. To this end, suppose that w_{0} and w_{1} are incomparable elements of Y. Let u_{0}, u_{1}, v_{0}, v_{1} be elements of Z such that u_{i} and v_{i} are incomparable and w_{i} = u_{i} ∧ v_{i}. Since w_{0} and w_{1} are incomparable,

u_{0}(Δ(w_{0}, w_{1})) = w_{0}(Δ(w_{0}, w_{1})) ≠ w_{1}(Δ(w_{0}, w_{1})) = v_{1}(Δ(w_{0}, w_{1})).

It follows that u_{0} ∧ v_{1} = w_{0} ∧ w_{1}. Since w_{0} extends u_{0} ∧ v_{1} and is in Y, it must be that u_{0} ∧ v_{1} is not in K. Hence w_{0} ∧ w_{1} is not in K. This completes the proof that ∧(Y) is disjoint from K. □

The following lemma is the reason for our definition of rejection. It will be used at crucial points in the argument.

Lemma 5.21. Suppose that E is in ℰ and ⟨X_{ξ} : ξ < ω_{1}⟩ is a sequence of disjoint n-element subsets of T so that no element of E rejects any X_{ξ} for ξ < ω_{1}. Then there are ξ ≠ η < ω_{1} such that X_{ξ}(j) ∧ X_{η}(j) is in K for all j < n.

Proof. By the pressing down lemma we can find a ζ < ω_{1} and a stationary set Ξ ⊆ ω_{1} such that:

(1) For all ξ in Ξ, X_{ξ} contains only elements of height at least ξ.

(2) X_{ξ}(j) ↾ ζ = X_{η}(j) ↾ ζ for all j < n and ξ, η ∈ Ξ.

(3) For all ξ in Ξ the set diff(X_{ξ} ↾ ξ) is contained in ζ. Now let P be an element of E which contains ⟨X_{ξ} : ξ ∈ Ξ⟩. Let η be an element of Ξ outside of P and pick a ξ in Ξ ∩ P such that X_{ξ}(0) ↾ ξ and X_{η}(0) ↾ η are incomparable and for all j < n

X_{η}(j) ↾ Δ(X_{η}(0), X_{ξ}(0))

is in K. This is possible since otherwise Z = {X_{ξ}(0) ↾ ξ : ξ ∈ Ξ} and t = X_{η}(0) ↾ (P ∩ ω_{1}) would witness K(X_{η}) is in ℐ_{P} (T) and therefore that P rejects X_{η}.

Notice that if j < n, then

Δ(X_{η}(j), X_{ξ}(j)) = Δ(X_{η}(0), X_{ξ}(0))

since

diff(X_{ξ} ↾ ξ) ∪ diff(X_{η} ↾ η) ⊆ ζ,
X_{η}(j) ↾ ζ = X_{ξ}(j) ↾ ζ.

Hence the meets

X_{ξ}(j) ∧ X_{η}(j) = X_{η}(j) ↾ Δ(X_{ξ}(0), X_{η}(0))

are in K for all j < n. □

The next lemma draws the connection between ∂ℋ (K) and the forcing ∂(K). We will then spend the remainder of the paper analyzing ∂(K).

Lemma 5.22. (BPFA) If ∂(K) is canonically proper, so is ∂ℋ (K).

Proof. We will show that otherwise the forcing ∂(K) introduces a counterexample to Lemma 5.21 which would then exist in V by an application of BPFA. Let M be a countable elementary submodel of H(|2^{∂(K)}|^{+}) which contains K as an element and let r ∈ ∂ℋ (K) be such that M ∩ H(2^{ℵ_{1}}^{+}) is in 𝒩_{r} and yet r is not (M, ∂ℋ (K))-generic. By extending r if necessary, we may assume that there is a dense open set 𝒟 ⊆ ∂ℋ (K) in M which contains r such that if q is in 𝒟 ∩ M, then q is ∂ℋ (K)-incompatible with r.

Let E ∈ ℰ ∩ M be such that no element of E ∩ M rejects X_{r} and let E′ be the elements of E which are the union of their intersection with E. Put Y_{r} = (X_{r} \ M) ↾ (M ∩ ω_{1}).

Claim 5.23. No element of E′ rejects Y_{r}.

Proof. First observe that no element of E′ ∩ M rejects Y_{r}; the point is to generalize this to arbitrary elements of E′. Let P be an element of E′. We need to verify that K(Y_{r}) is not in ℐ_{P} (T). If P ∩ ω_{1} is greater than M ∩ ω_{1}, then Y_{r} ⊆ P and this is trivial. Now suppose that Z ⊆ T is in P and t is an element of T of height P ∩ ω_{1} which is in the downward closure of Z. Let P_{0} be an element of E ∩ P which contains Z as a member. Such a P_{0} will satisfy

P_{0} ∩ ω_{1} < P ∩ ω_{1} ≤ M ∩ ω_{1}.

Let ν = P_{0} ∩ ω_{1}. If Δ(Z, t ↾ (P ∩ ω_{1})) is disjoint from K(Y_{r}), then it witnesses K(Y_{r} ↾ ν) is in ℐ_{P_{0}}(T). But then we could use the elementarity of M to find such a P_{0} in M ∩ E, which is contrary to our choice of E. Hence no element of E′ rejects Y_{r}. □

Let ζ ∈ M ∩ ω_{1} be an upper bound for diff(Y_{r}) and let n = |Y_{r}|. If j < n, let A_{j} ⊆ T be an antichain in M which contains Y_{r}(j).^{12} Put 𝒟_{∗} to be the collection of all q in 𝒟 such that:

(4) X_{r} ∩ M = X_{q} ∩ N(q) where N(q) is the least element of 𝒩_{q} which is not in 𝒩_{r} ∩ M.

(5) Y_{q} ↾ ζ = Y_{r} ↾ ζ where Y_{q} = (X_{q} \ N(q)) ↾ (N(q) ∩ ω_{1}).

(6) No element of E′ rejects Y_{q}.

(7) Y_{q}(j) is in A_{j} whenever j < n. Note that 𝒟_{∗} is in M.

Let G be a ∂(K)-generic filter which contains r. Notice that r is (M, ∂(K))-generic. Working in V [G], let ℱ be the collection of all Y_{q} where q is in 𝒟_{∗} ∩ G. Now M[G ∩ M] is an elementary submodel of H(|2^{∂(K)}|^{+})[G]^{13} which contains ℱ as an element but not as a subset (since Y_{r} is in ℱ). Therefore ℱ is uncountable. Notice that every element of ℱ has the property that it is in ℋ (K) but that for every countable ℱ_{0} ⊆ ℱ there is a Y_{q} in ℱ \ ℱ_{0} such that Y_{q} ∪ Y_{q_{0}} is not in ℋ (K) for any Y_{q_{0}} in ℱ_{0}. This follows from the elementarity of M[G ∩ M] and from the fact that Y_{r} ∪ Y_{q} is not in ℋ (K) for any Y_{q} in ℱ ∩ M[G ∩ M]. Now it is possible to build an uncountable sequence ⟨X_{ξ} : ξ ∈ Ξ⟩ of elements of ℱ such that:

(8) X_{ξ} has size n for all ξ ∈ Ξ and is a subset of the ξ^{th} level of T.

(9) X_{ξ} ∪ X_{η} is not in ℋ (K) whenever ξ ≠ η are in Ξ.

(10) There is a ζ < ω_{1} such that X_{ξ} ↾ ζ = X_{η} ↾ ζ has size n for all ξ, η < ω_{1}.

12Here we are using that T is special.

13By Theorem 2.11 of [17].

It follows from item 9 that if ξ < η < ω_{1}, then there are j, j′ < n such that X_{ξ}(j) ∧ X_{η}(j′) is not in K. By item 10, it must be the case that j = j′ since this condition ensures that

X_{ξ}(j) ∧ X_{η}(j′) = X_{ξ}(j) ∧ X_{ξ}(j′)

whenever j ≠ j′ < n and hence this meet would be in K by virtue of X_{ξ} being in ℋ (K). Applying BPFA we get a sequence of sets satisfying 1–3 in V and therefore a contradiction to Lemma 5.21 since no elements of ℱ are rejected by any member of E′. Hence ∂ℋ (K) must also be canonically proper. □

Next we have a typical “models as side conditions” lemma.

Lemma 5.24. If ∂(K) is not canonically proper, then there are disjoint sets 𝒜, ℬ and a function Y : 𝒜 ∪ ℬ → [T]^{<ℵ_{0}} such that:

(1) 𝒜 ⊆ [H(2^{ℵ_{1}}^{+})]^{ℵ_{0}} and {N ∩ H(ℵ_{1}^{+}) : N ∈ 𝒜 } is stationary.

(2) ℬ ⊆ [H(2^{2^{ℵ_{1}}}^{+})]^{ℵ_{0}} is stationary.

(3) If M is in 𝒜 ∪ ℬ, then (Y (M), {M ∩ H(2^{ℵ_{1}}^{+})}) is in ∂(K).

(4) For every M in ℬ and N in 𝒜 ∩ M, (Y (N) ∪ Y (M), {N}) is not a condition in ∂(K).

Proof. Let M be a countable elementary submodel of H(|2^{∂(K)}|^{+}) and r in ∂(K) be a condition which is not (M, ∂(K))-generic such that M ∩ H(2^{ℵ_{1}}^{+}) is in 𝒩_{r}. By extending r if necessary, we can find dense open 𝒟 ⊆ ∂(K) in M which contains r such that no element of 𝒟 ∩ M is compatible with r. Furthermore we may assume that if q is in 𝒟, N is in 𝒩_{q}, and t is in X_{q}, then t ↾ (N ∩ ω_{1}) is also in X_{q}.

Define r_{0} = (X_{r} ∩ M, 𝒩_{r} ∩ M). If q is in 𝒟, let N(q) be the ∈-least element of 𝒩_{q} \ 𝒩_{r_{0}}. Let k = |𝒩_{r} \ 𝒩_{r_{0}}| and ζ be the maximum of all ordinals of the forms ht(s) + 1 for s ∈ X_{r_{0}} and N ∩ ω_{1} for N ∈ 𝒩_{r_{0}}. Let 𝒯_{k} be the set of all q ≤ r_{0} in 𝒟 such that:

(11) X_{q} ∩ N(q) = X_{r_{0}} and 𝒩_{q} ∩ N(q) = 𝒩_{r_{0}}.

(12) X_{q} ↾ ζ = X_{r} ↾ ζ.

(13) |X_{q}| = |X_{r}| = m and |𝒩_{q} \ 𝒩_{r_{0}}| = k. Let N_{i}(q) denote the i^{th} ∈-least element of 𝒩_{q} \ N(q) and define 𝒯_{i} recursively for i ≤ k. Given 𝒯_{i+1}, define 𝒯_{i} to be the collection of all q such that

{N_{i+1}(q^{∗}) ∩ H(ℵ_{1}^{+}) : q^{∗} ∈ 𝒯_{i+1} and q = q^{∗} ↾ N_{i+1}(q^{∗})}

is stationary where

q^{∗} ↾ N_{i+1}(q^{∗}) = (X_{q^{∗}} ∩ N_{i+1}(q^{∗}), 𝒩_{q^{∗}} ∩ N_{i+1}(q^{∗})).

Let 𝒯 be the collection of all q in ⋃_{i≤k} 𝒯_{i} such that if q is in 𝒯_{i}, then

q ↾ N_{i′+1}(q) is in 𝒯_{i′} for all i′ < i.

Claim 5.25. r is in 𝒯 .

Proof. If q is in ∂(K), define

q̃ = (X_{q}, {N ∩ H(ℵ_{1}^{+}) : N ∈ 𝒩_{q}}).

While elements of 𝒩_{r} \M need not contain 𝒯_{i} as an element for a given i ≤ k, they do contain 𝒯̃_{i} = {q̃ ∈ 𝒯_{i}} as an element for each i ≤ k. Define r_{k} = r and r_{i} = r_{i+1} ↾ N_{i+1}(r). Suppose that r_{i+1} is in 𝒯_{i+1}. Since 𝒯̃_{i+1} and r_{i} are in N_{i+1}(r) = N_{i+1}(r_{i+1}) and since N_{i+1}(r)∩H(ℵ_{1}^{+}) is in every club in ℰ ∩ N_{i+1}(r), it follows by elementarity of N_{i+1}(r) that the set

{N_{i+1}(q^{∗}) ∩ H(ℵ_{1}^{+}) : q^{∗} ∈ 𝒯_{i+1} and r_{i} = q^{∗} ↾ N_{i+1}(q^{∗})} =
{N_{i+1}(q̃^{∗}) : q̃^{∗} ∈ 𝒯̃_{i+1} and r̃_{i} = q̃^{∗} ↾ N_{i+1}(q̃^{∗})}

is stationary. Hence r_{i} is in 𝒯_{i}. □

Notice that 𝒯 is in M. 𝒯 has a natural tree order associated with it induced by restriction. Since no element of 𝒯_{k} ∩ M is compatible with r and since r_{0} is in 𝒯 ∩ M, there is a q in 𝒯 ∩ M which is maximal in the tree order such that q is compatible with r but such that none of q’s immediate successors in 𝒯 ∩M are compatible with r. Let l denote the height of q in 𝒯 and put 𝒜 to be equal to the set of all N_{l+1}(q^{∗}) such that q^{∗} is an immediate successor of q in 𝒯 . Notice that if q^{∗} is in 𝒯_{l+1} and q is a restriction of q^{∗}, then q^{∗} is in 𝒯 . Hence we have arranged that {N ∩ H(ℵ_{1}^{+}) : N ∈ 𝒜 } is stationary. For each N in 𝒜, select a fixed q^{∗} which is an immediate successor of q in 𝒯 such that N_{l+1}(q^{∗}) = N and put

Y (N) = X_{q^{∗}} \ X_{q}.

Claim 5.26. For all N in 𝒜 ∩ M the pair (X_{r} ∪ Y (N), {N}) is not a condition in ∂(K).

Proof. Let N be in 𝒜 ∩ M and fix an immediate successor q^{∗} of q in 𝒯 such that N_{l+1}(q^{∗}) = N and Y (N) = X_{q^{∗}} \ X_{q}. Observe that

(X_{r} ∪ X_{q^{∗}}, 𝒩_{q^{∗}} ∪ 𝒩_{r})

is not a condition in ∂(K) but that

(X_{r} ∪ X_{q}, 𝒩_{q} ∪ 𝒩_{r})

is a condition. Furthermore, (X_{r} ∪X_{q^{∗}}, 𝒩_{q^{∗}} ∪𝒩_{r}) fails to be a condition only because it violates item 2 in the definition of ∂(K). Observe that 𝒩_{q^{∗}} \ 𝒩_{q} = {N}. If N′ is an element of 𝒩_{r} ∪ 𝒩_{q}, then the sets of restrictions

{t ↾ (N′ ∩ ω_{1}) : t ∈ X_{r} ∪ X_{q^{∗}}},
{t ↾ (N′ ∩ ω_{1}) : t ∈ X_{r} ∪ X_{q}}

are equal by definitions of 𝒯_{l} and q and by our initial assumptions about the closure of X_{q} for q in 𝒟 under taking certain restrictions. Since (X_{r} ∪X_{q}, 𝒩_{r} ∪𝒩_{q}) is a condition, such an N′ cannot witness the failure of 2. Therefore it must be the case that the reason (X_{r} ∪X_{q^{∗}}, 𝒩_{q^{∗}} ∪𝒩_{r}) is not in ∂(K) is that N witnesses a failure of item 2. Now, the elements of X_{q^{∗}} which have height at least N ∩ ω_{1} are exactly those in Y (N) = X_{q^{∗}} \ X_{q}. This finishes the claim. □

Notice that by elementarity of M, Y ↾ 𝒜 can be chosen to be in M. Now M satisfies “There is a stationary set of countable elementary submodels M_{∗} of H(2^{2^{ℵ_{1}}}^{+}) such that for some Y (M_{∗}) with (Y (M_{∗}), {M_{∗} ∩H(2^{ℵ_{1}}^{+})}) in ∂(K) we have that for every N in 𝒜 ∩M_{∗} the pair (Y (N) ∪ Y (M_{∗}), {N}) is not a condition in ∂(K).” By elementarity of M, we are finished. □

The following definition will be useful.

Definition 5.27. A function h is a level map if its domain is a subset of ω_{1} and h(δ) is a finite subset of the δ^{th} level of T whenever it is defined.

The next proposition is useful and follows easily from the fact that all levels of T are countable.

Proposition 5.28. If N ∩ H(ℵ_{1}^{+}) is in E_{0}, δ = N ∩ ω_{1}, and X is a finite subset of the δ^{th} level of T, then there is a level map h in N such that h(δ) = X.

The next lemma will represent the only use of MRP in the proof.

Lemma 5.29. (MRP) Suppose that M is a countable elementary submodel of H(2^{2^{ℵ_{1}}}^{+}) which contains T and K as members. If X is a finite subset of T, then there is an E in ℰ ∩ M such that either every element of E ∩ M rejects X or no element of E ∩ M rejects X.

Remark. Notice that the latter conclusion is just a reformulation of the statement that (X, {M ∩ H(2^{ℵ_{1}}^{+})}) is a condition in ∂(K).

Proof. Let δ = M ∩ω_{1}. Without loss of generality, we may assume that X = X ↾ δ. Applying Proposition 5.28, select a level map g in M such that g(δ) = X. If N is a countable elementary submodel of H(2^{ℵ_{1}}^{+}) with T and K as members, define Σ(N) as follows. If the set of all P in E_{0} which reject g(N ∩ ω_{1}) is N-stationary, then put Σ(N) to be equal to this set unioned with the complement of E_{0}. If Σ(N) is defined in this way, it will be said to be defined non-trivially. Otherwise put Σ(N) to be the interval [∅, N ∩ H(ℵ_{1}^{+})].

Observe that Σ is an open stationary set mapping which is moreover an element of M. Applying MRP and the elementarity of M, it is possible to find a reflecting sequence ⟨N_{ξ} : ξ < ω_{1}⟩ for Σ which is an element of M. Let E be the collection of all \overline{P} in E_{0} which contain

(14) the sequence ⟨N_{ξ} ∩ H(ℵ_{1}^{+}) : ξ < ω_{1}⟩ and

(15) some δ_{0} < N ∩ ω_{1} such that N_{ξ} ∩ H(ℵ_{1}^{+}) is in Σ(N_{δ}) whenever

ξ is in (δ_{0}, δ). Notice that E is in M ∩ ℰ.

To finish the proof, suppose that the set of all P in M ∩ E_{0} which reject X is M-stationary (i.e. the second conclusion does not hold).

Claim 5.30. Σ(N_{δ}) is defined non-trivially.

Proof. Suppose that E′ ⊆ E_{0} is a club in N_{δ}. Since the reflecting sequence is continuous, N_{δ} is a subset of M and therefore E′ is also in M. By assumption, there is a P in E′ ∩ M such that P rejects X. Let ν = P ∩ ω_{1} and note that ν < δ. Applying elementarity of N_{δ}, Proposition 5.16, and the fact that X ↾ ν is in N_{δ}, it is possible to find such a P in E′ ∩ N_{δ} which rejects X ↾ ν — and hence X. It follows that Σ(N_{δ}) is defined non-trivially. □

Now suppose that \overline{P} is in E ∩ M. We are finished once we see that \overline{P} rejects X. Let ν = \overline{P} ∩ ω_{1}. Since δ_{0} < ν < δ, P_{ν} = N_{ν} ∩ H(ℵ_{1}^{+}) is in Σ(N_{δ}). So P_{ν} rejects X or — equivalently — K(X) is in ℐ_{P_{ν}}(T). Observe that P_{ν} ∩ ω_{1} = \overline{P} ∩ ω_{1} and P_{ν} ⊆ \overline{P} by continuity of the reflecting sequence. Hence ℐ_{P} (T) ⊆ ℐ\overline{P} (T). It follows that \overline{P} rejects X. □

The next lemma finishes the proof of the main theorem.

Lemma 5.31. (MRP + MA_{ℵ_{1}}) There are no 𝒜, ℬ, and Y which satisfy the conclusion of Lemma 5.24. In particular, ∂(K) is canonically proper.

Proof. We will assume that there are such 𝒜, ℬ, and Y and derive a contradiction by violating Lemma 5.21. Without loss of generality we may suppose that elements of ℬ contain 𝒜 as a member. By modifying Y we may assume that all elements of Y (M) have height M ∩ ω_{1} whenever M is in 𝒜 ∪ ℬ. Further, we may assume that Y (M) has the same fixed size n for all M in ℬ and that there is a ζ_{0} and E_{∗} ∈ ℰ such that:

(16) If M is in ℬ, then diff(Y (M)) ⊆ ζ_{0}.

(17) If M, M′ are in ℬ, then Y (M) ↾ ζ_{0} = Y (M′) ↾ ζ_{0}.

(18) If M is in ℬ, then E_{∗} is in M and no element of E_{∗} rejects

Y (M). This is achieved by the pressing down lemma and the proof of Claim 5.23.^{14} Let ℱ be the collection of all finite X ⊆ T such that all elements of X have the same height γ > ζ_{0} and the set

{M ∈ ℬ : Y (M) ↾ γ = X}

is stationary. Notice that, for a fixed γ, we can define ℬ to be a union over the finite subsets X of T_{γ} of the collection

ℬ[X] = {M ∈ ℬ : Y (M) ↾ γ = X}

and hence at least one such ℬ[X] must be stationary. Consequently ℱ must be uncountable. Also, no element of E_{∗} rejects any element of ℱ. Now define 𝒬 to be the collection of all finite F ⊆ ℱ such that if X ≠ X′ are in F, then the heights of elements of X and X′ are different and there is a j < n such that X(j) ∧ X′(j) is not in K.

Claim 5.32. (MRP) 𝒬 satisfies the countable chain condition.

Proof. Suppose that ⟨F_{ξ} : ξ < ω_{1}⟩ is a sequence of distinct elements of 𝒬. We will show that {F_{ξ} : ξ < ω_{1}} is not an antichain in 𝒬. By a Δ-system argument, we may assume that the sequence consists of disjoint sets of the same cardinality m. Let F_{ξ}(i) denote the i^{th}-least element of F_{ξ} in the order induced by T ’s height function. If j < n, let F_{ξ}(i, j) denote the j^{th} element of F_{ξ}(i) in the lexicographical order on F_{ξ}(i) (i.e. F_{ξ}(i, j) = F_{ξ}(i)(j)).

Let N be an element of 𝒜 which contains ζ_{0} and ⟨F_{ξ} : ξ < ω_{1}⟩ as members. Put δ = N ∩ ω_{1} and fix a β in ω_{1} \ N. Let E be a club in N such that Y (N) is not rejected by any element of E ∩ N.

For each i < m, pick an M_{i} in ℬ such that N ∈ M_{i} and F_{β}(i) is a restriction of Y (M_{i}). Let M be a countable elementary submodel of H(2^{2^{ℵ_{1}}}^{+}) such that N = M ∩ H(2^{ℵ_{1}}^{+}). Applying Lemma 5.29 to M for each i < m and intersecting clubs,^{15} it is possible to find a P in E ∩ N such that

14To get the last item, find an E_{1} in ℰ such that no element of E_{1} ∩ M rejects Y (M) for stationary many M in ℬ, put E_{∗} to be the elements P of E_{1} which are equal to the union of their intersection with E_{1}.

15This is the only place where Lemma 5.29 and hence MRP is applied.

I = ⋃_{i<m} K(Y (N) ∪ Y (M_{i})) ∈ ℐ_{P} (T).

Put ν = P ∩ ω_{1}. Pick a ζ < ν such that diff(∪F_{β} ↾ ν) is contained in ζ and if u, v are distinct elements of ∪F_{β} ↾ ν, then u ↾ ζ and v ↾ ζ are distinct.

Subclaim 5.33. There is a sequence ⟨α_{ξ} : ξ < ω_{1}⟩ in P such that for each ξ < ω_{1} we have the following conditions:

(1) ξ ≤ α_{ξ},

(2) ∪F_{α_{ξ}} ↾ ζ = ∪F_{β} ↾ ζ,

(3) diff(∪F_{α_{ξ}} ↾ ξ) = diff(∪F_{β} ↾ ν), and

(4) ∪F_{α_{ν}} ↾ ν = ∪F_{β} ↾ ν.

Proof. The only part which is non-trivial is to get the sequence to be a member of P and to satisfy item 4. By Proposition 5.28, there is a level map g in P such that g(ν) = ∪F_{β} ↾ ν. Now working in P, we can define α_{ξ} to be an ordinal such that ∪F_{α_{ξ}} ↾ ξ = g(ξ) if g(ξ) is defined, is a restriction of this form, and satisfies diff(g(ξ)) = diff(∪F_{β} ↾ ν) and g(ξ) ↾ ζ = ∪F_{β} ↾ ζ. If α_{ξ} is left undefined, then simply select a α_{ξ} with the necessary properties. Notice that α_{ν} is defined using g. □

Subclaim 5.34. There is an uncountable Ξ ⊆ ω_{1} in P such that if

Z = {F_{α_{ξ}}(0, 0) ↾ ξ : ξ ∈ Ξ}

and t = F_{β}(0, 0) ↾ ν, then t is in the downwards closure of Z and Δ(Z, t) is disjoint from

I = ⋃_{i<m} K(Y (N) ∪ Y (M_{i})).

Proof. By Proposition 5.9 there is a Ξ_{0} ⊆ ω_{1} such that for some t_{0} in T of height ν in the downward closure of Z_{0} = {F_{α_{ξ}}(0, 0) ↾ ξ : ξ ∈ Ξ_{0}} the set Δ(Z_{0}, t_{0}) is disjoint from I. Let Z_{1} be all elements s in T obtained from some F_{α_{ξ}}(0, 0) ↾ ξ by changing its values on the finite set

diff (F_{α_{ν}}(0, 0) ↾ ν, t_{0}) ∩ ξ.

Let Ξ be the collection of all ξ such that F_{α_{ξ}}(0, 0) ↾ ξ is an initial part of some element of Z_{1}. Notice that Ξ is in P and is uncountable since it contains ν. Furthermore, if t = F_{α_{ν}}(0, 0) ↾ ν = F_{β}(0, 0) ↾ ν, then Δ(Z, t) is contained in

Δ(Z_{1}, t) = Δ(Z_{0}, t_{0})

and hence is disjoint from I. □

The key observation — and the reason why the main theorem goes through — is the following. Since P is in E ∩ N, it does not reject Y (N) and therefore it is the case that there is a ξ in Ξ ∩ P such that for all j < |Y (N)| the restriction

Y (N)(j) ↾ Δ(F_{α_{ξ}}(0, 0), t)

is in K where t = F_{β}(0, 0) ↾ ν. By the choice of Ξ this means that for all i < m there is a j < n such that

F_{β}(i, j) ↾ Δ(F_{α_{ξ}}(0, 0), t) = F_{β}(i, j) ∧ F_{α_{ξ}}(i, j)

is not in K. Let α = α_{ξ}.

Now we claim that F_{α}∪F_{β} is in 𝒬. To see this, suppose that i, i′ < m. If F_{β}(i) ↾ ν ≠ F_{β}(i′) ↾ ν, then pick a j < n such that F_{α}(i, j) ∧ F_{α}(i′, j) is not in K. Since

Δ(F_{α}(i′, j), F_{β}(i′, j)) ≥ ζ > Δ(F_{α}(i, j), F_{α}(i′, j))

it must be the case that

Δ(F_{α}(i, j), F_{β}(i′, j)) = Δ(F_{α}(i, j), F_{α}(i′, j))

and so

F_{α}(i, j) ∧ F_{β}(i′, j) = F_{α}(i, j) ∧ F_{α}(i′, j)

is not in K.

If F_{β}(i) ↾ ν = F_{β}(i′) ↾ ν, then we have that for all j < n that

Δ(F_{α}(i, j), F_{β}(i′, j)) = Δ(F_{α}(i, j), F_{β}(i, j)) = Δ(F_{α}(0, 0), F_{β}(0, 0)).

By arrangement there is a j such that

F_{α}(i, j) ∧ F_{β}(i, j) = F_{β}(i, j) ↾ Δ(F_{α}(0, 0), F_{β}(0, 0))

is not in K. Hence for all i, i′ < m there is a j < n such that

F_{α}(i, j) ∧ F_{β}(i′, j)

is not in K and therefore we have that F_{α} ∪ F_{β} is in 𝒬. □

Applying MA_{ℵ_{1}} to the forcing 𝒬 it is possible to find an uncountable ℱ_{0} ⊆ ℱ such that whenever X ≠ X′ are in ℱ_{0}, there is a j < n such that X(j) ∧ X′(j) is not in K. This contradicts Lemma 5.21 since no element of E_{∗} rejects any element of ℱ. □

6. Closing remarks

The use of MRP in the argument above is restricted to proving Lemma 5.29. Working from a stronger assumption, the following abstract form of the lemma can be deduced. The interested reader is encouraged to supply a proof and see why a stronger assumption is apparently needed for the abstract statement while MRP suffices in the proof of Lemma 5.29.

0-1 law for open set mappings. (SMRP^{16}) Suppose that Σ is an open set mapping defined on a club and that Σ has the following properties:

(1) If N is in the domain of Σ, then Σ(N) is closed under end extensions.^{17}

(2) If N and \overline{N} are in the domain of Σ and \overline{N} is an end extension

of N, then Σ(N) = Σ(\overline{N}) ∩ N. Then for a closed unbounded set of N in the domain of Σ, there is a club E ⊆ [X_{Σ}]^{ℵ_{0}} in N such that E ∩N is either contained in or disjoint from Σ(N).

It seems quite possible that this 0-1 law will be useful in analyzing related problems such as Fremlin’s problem on perfectly normal compacta (see [10], [21]).

The conventional wisdom had been that if it were possible to prove the consistent existence of a five element basis for the uncountable linear orders, then such a basis would follow from BPFA. MRP has considerable consistency strength [15], while BPFA can be forced if there is a reflecting cardinal [9]. The following is left open.

Question 6.1. Does BPFA imply Shelah’s conjecture?

Recently König, Larson, Veličković, and I have shown that a certain saturation property of Aronszajn trees taken together with BPFA implies Shelah’s conjecture [13]. This saturation property can be forced if there is a Mahlo cardinal. This considerably reduces the upper bound on the consistency strength of Shelah’s conjecture to that of a reflecting Mahlo cardinal. It is possible, however, that Shelah’s conjecture cannot follow from BPFA simply on grounds of its consistency strength. It should be remarked though that Shelah’s conjecture is not known to have any large cardinal strength.

16SMRP is the Strong Mapping Reflection Principle obtained by replacing “club”

in the statement of MRP with “projective stationary” (see [8]). This axiom follows from Martin’s Maximum via the same proof that MRP follows from PFA (see [15]).

17Here we define \overline{N} end extends N as meaning that N ∩ω_{1} = \overline{N} ∩ω_{1} and N ⊆ \overline{N}.

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Department of Mathematics, Boise State University, Boise, Idaho 83725–1555

  E-mail address: justin@math.boisestate.edu