Rectangular axioms, perfect set properties and decomposition ^{∗}^{†}

Jörg Brendle  Paul Larson  Stevo Todorcevic

May 12, 2006

Abstract

We consider three combinatorial topics appearing in Gödel’s manuscript Some considerations leading to the probable conclusion that the true power of the continuum is ℵ_{2}. These statements concern rectangular functions, perfect set properties, and covering properties of sets of reals. We consider these statements in light of more recent work on the set theory of the reals.

1 Introduction

In 1970, Kurt Gödel circulated a manuscript in which he presented four axioms with the aim of bounding the size of the continuum. The history of this manuscript and the argument it contains have been discussed by Moore [13] and Solovay [14]. The work in this paper began by trying to understand these axioms and the corresponding argument. We isolated three statements which appear implicitly in his manuscript, and found that taken together these statements do indeed put a bound on the continuum. These statements concern dominating sequences for functions of the form f : κ^{+} → κ, perfect set axioms, and decompositions of sets of reals. In the forms we consider here, these areas are still not well understood. A special role is played by the G_{ℵ_{1}} sets, those sets which can be represented as an intersection of ℵ_{1} many open sets. In particular, perhaps the most quotable result presented here is the fact that the perfect set property for intersections of κ many open sets is equivalent to the statement that 𝔡 > κ for any cardinal κ (Theorem 5.6). Considerable attention is also given to perfect set properties that hold in the model obtained after adding ω_{2} many Sacks reals to a model of CH (see Theorem 5.11).

As for the idea of settling the continuum from reasonable hypotheses, in the thirty years since Gödel produced his argument several axioms have been studied which do imply 𝔠 ≤ ℵ_{2}. These include the Proper Forcing Axiom [8], the Mapping Reflection Principle [24], Stationary Reflection at ω_{2} [8], Rado’s Conjecture [29], Martin’s Maximum [15], ψ_{AC} [33], Bounded Martin’s Maximum [31] and others. These statements are not independent of one another, but the point is that several different proofs correspond to them. None of them is in the spirit of Gödel’s approach, however. It remains to be seen whether Gödel’s original idea, decomposing the reals into small, simple sets, can give us similar evidence as to the size of the continuum.

∗The research of the first author is partially supported by Grant–in–Aid for Scientific Research (C) (2) Nrs. 10640118 and 12640124. The second author was partially supported by NATO and JSPS.

†MSC 2000: 03E50, 03E17

2 Notation

Except where noted, the reals are considered to be the set 2^{ω}. As usual, ℝ^{+} and ℚ^{+} are the sets of positive reals and rationals respectively. Lebesgue measure is denoted by μ.

Modifying notation in [28], we let g(κ, λ) be the least η such that there is a family of functions F ⊂ λ^{κ} such that every such function is everywhere dominated by some member of F, and such that |{f↾γ | γ < κ ∧ f ∈ F }| = η.

For κ a cardinal and Γ ⊂ P(ℝ), we let PSP(κ, Γ) denote the statement that every member of Γ of cardinality κ or greater contains a perfect set. A set of reals is in G_{κ} if it is the intersection of κ many open sets, and F_{κ} if it is the union of κ many closed sets.

For Γ ⊂ P(ℝ), Cov(Γ) is the least κ such that there exists a subset of Γ of cardinality κ whose union is all the reals. Non(Γ) is the least κ such that there exists a set of reals not in Γ of cardinality κ. We let N denote the collection of subsets of the reals of measure zero, M the meager sets, and SN the sets of strong measure zero.

The cardinal invariant 𝔡 is the cardinality of the smallest set of functions from ω to ω such that every such function is dominated mod finite by a member of the set. The bounding number 𝔟 is the cardinality of the smallest set of functions from ω to ω such that no such function dominates mod finite every member of the set. The cardinality of the continuum is denoted by 𝔠.

Given a function g : ω → ℝ^{+}, a g-set is a set of the form ⋂_{i<ω} ⋃_{j>i} O_{i}, where each O_{i} is an interval of width g(i). We say that a set A ⊂ ℝ can be g-covered or is g-coverable if A is contained in a g-set. For sets A ⊂ B, g separates A from B if A can be g-covered and B cannot. This induces the partial ordering ⊲ on sets of reals A ⊂ B defined by letting A ⊲ B if there is a function separating A from B.

3 A version of Gödel’s argument

Gödel began his paper by presenting the statements G1-G4 below. Axiom G4 says that there are no (ω_{1}, ω_{1})-gaps in the scale from G3. Hausdorff showed that the existence of such a scale implies that 2^{ω} = 2^{ω_{1}}. It is still not known, however, whether the existence of such a scale is consistent with ZFC. Martin and Solovay showed [14] that that G1-G3 together do not put a bound in the size of the continuum.

G1. There exists a scale of functions ω_{n} → ω_{n} of type ω_{n+1} majorizing by end pieces every such function. It follows that there exists a set M of power ℵ_{n+1} majorizing everywhere every such function.

G2. The total number of initial segments of all the functions in this scale and in M is ℵ_{n}.

G3. There exists a complete scale in ℝ^{ω} such that all increasing or decreasing sequences in this scale have cofinality at most ω_{1}.

G4. The Hausdorff continuity axiom for this scale.

We work with the following axioms, in addition to ZFC. Each of these axioms, or something stronger, appeared explicitly or implicitly in Gödel’s argument. In particular, Gödel claimed, incorrectly (see Sections 4 and 5), to derive our Axioms 1 and 2 from G1 and G2. Axiom 3 below plays the role of G3.

Axiom 1. g(ω_{2}, ω_{1}) = ℵ_{2}.

Axiom 2. PSP(ℵ_{3}, G_{ℵ_{1}}).

Axiom 3. There exists a set H of functions from ω to ℝ^{+} such that the following hold.

(a) All sequences from H which are increasing or decreasing in the modfinite domination ordering have cofinality ω_{1} or less.

(b) For any set A ⊂ ℝ not of strong measure zero there exists a sequence

⟨(B_{α}, g_{α}) ∈ P(ℝ) × H : α < ω_{1}⟩

such that the B_{α}’s are nondecreasing under inclusion with union containing A, each B_{α} is a g_{α}-set, and each g_{α} is mod-finite less than each f ∈ H for which A is f-coverable.

Theorem 3.1 below is proved by Gödel’s argument. The main line of the argument is showing that Axioms 1 and 3 together imply that the reals are the union of ℵ_{2} many G_{ℵ_{1}} sets of strong measure zero. By Axiom 2 these sets must each have cardinality less than ℵ_{3}.

Theorem 3.1. Axioms 1-3 together imply that 𝔠 ≤ ℵ_{2}.

Proof: Let F ⊂ ω_{1}^{ω_{2}} be as given by Axiom 1, and let

F = {f↾γ : f ∈ F ∧ γ < ω_{2}}.

Fix a wellordering E : ω_{2} → F such that for all σ ⊂ τ ∈ F, E^{−1}(σ) ≤ E^{−1}(τ).

Construct a matrix

⟨(A_{α,β}, g_{α,β}) ∈ P(ℝ) × H : α < ω_{2}, β < ω_{1}⟩

and a sequence

⟨B_{α} ⊂ ℝ : α < ω_{2}⟩

of G_{ℵ_{1}} sets with the following properties.

• Each A_{α,β} is a g_{α,β}-set.

• B_{0} = ℝ, and for α ∈ ω_{2} \ {0},

B_{α} = ⋂{A_{E^{−1}(E(α)↾γ),E(α)(γ)} : γ ∈ dom(E(α))}.

• For all α < ω_{2}, ⟨A_{α,β} : β < ω_{1}⟩ is nondecreasing in the subset order and has union containing B_{α}.

• For all α < ω_{2}, if B_{α} is not of strong measure zero, then for each β < ω_{1} and γ ∈ dom(E(α)), g_{α,β} is mod-finite less than g_{E^{−1}(E(α)↾γ),E(α)(γ)}.

The construction of the matrix is straightforward, using Axiom 3 to define each column by induction.

Given such a matrix, 𝔠 ≤ ℵ_{2} as follows. For each f ∈ F, consider the sequence ⟨B_{E^{−1}(f↾γ)} : γ < ω_{2}⟩. The sets in this sequence must eventually be of strong measure zero, since otherwise one gets an ω_{2}-sequence of members of H which is decreasing in the mod-finite domination ordering. But by our perfect set property for G_{ℵ_{1}} sets, this strong measure zero set cannot be of cardinality greater than ℵ_{2}, since perfect sets cannot be of strong measure zero. Lastly, each real defines a function from ω_{2} to ω_{1} by where it first appears in each column (letting the value of the function at α be 0 if x ∉ B_{α}). This function is everywhere dominated by some f ∈ F. Since f everwhere dominates the function determined by x, x is in all the B_{α}’s corresponding to f, and so x is in the strong measure zero set corresponding to f restricted to some γ < ω_{2}. Thus ℝ is the union of ω_{2}-many sets of cardinality less than or equal to ℵ_{2}, and so 𝔠 ≤ ℵ_{2}.

Theorem 3.1 shows that a substantial part of Gödel’s argument is correct. The rest of the paper analyzes Axioms 1-3.

4 Axiom 1 and Rectangles

In this section we consider variations of Axiom 1. We have not resolved whether some weakening of Axiom 1 is sufficient for putting a bound on the continuum, or even whether Axiom 1 can be removed altogether.

4.1 Question. Do axioms 2 and 3 together imply 𝔠 ≤ ℵ_{2}?

4.2 Remark. A diagonal argument shows that if F is a set of functions from ω_{2} to ω_{1} dominating every such function everywhere, then |F | ≥ ℵ_{3}.

The cardinal invariant 𝔡_{1} is the natural generalization of 𝔡 to ω_{1}, that is, the least κ such that there exists a set of functions from ω_{1} to ω_{1} of cardinality κ dominating every such function mod countable. Likewise, 𝔟_{1} is the least κ such that there exists a set of functions from ω_{1} to ω_{1} of cardinality κ such that no such function dominates every member of the set mod countable. In the definitions of 𝔡 and 𝔡_{1}, ‘mod finite’ and ‘mod countable’ can be replaced by ‘everywhere.’ This isn’t so for 𝔟 and 𝔟_{1}. The statement 𝔡_{1} = ℵ_{2} is a natural weakening of Axiom 1. We shall see that is is properly weaker.

4.3 Remark. Another weakening of Axiom 1 results from letting F be an eventually dominating scale. The existence of such an F with just ℵ_{2} many initial segments is easily seen to imply Axiom 1, however, since it implies 𝔡_{1} = ℵ_{2}, and a witness for Axiom 1 can be constructed replacing the initial segments of the functions in F with the rearranged members of the witness for 𝔡_{1} = ℵ_{2}.

The following proof appears in [28] with a slightly different presentation. It is essentially the same proof as Gödel’s for putting a bound on the continuum; one could make the claim that this is the natural theorem for his argument.

Theorem 4.4. ([28]) If 2^{<κ} < cof(λ) and g(λ, cof(κ)) = λ, then 2^{κ} ≤ λ.

Proof: Let 2^{κ} have the initial segment topology, and let F be as given by the fact that g(λ, cof(κ)) = λ. Let F = {f↾β : f ∈ F ∧ β < λ} and fix a wellordering E : λ → F such that for all σ ⊂ τ ∈ F, E^{−1}(σ) ≤ E^{−1}(τ).

Construct a matrix

⟨A_{α,β} ⊂ 2^{κ} : α < λ, β < cof(κ)⟩

of closed sets and a sequence

⟨B_{α} ⊂ 2^{κ} : α < λ⟩

of closed sets with the following properties.

1. B_{0} = 2^{κ}, and for α ∈ λ \ {0},

B_{α} = ⋂{A_{E^{−1}(E(α)↾η),E(α)(η)} : η ∈ dom(E(α))}.

2. For all α < λ, β < β′ < cof(κ), A_{α,β} ⊂ A_{α,β′}.

3. For all α < λ, ⋃{A_{α,β} : β < cof(κ)} = B_{α}.

4. For all α < λ, if B_{α} has cardinality λ or greater, then for each β < κ A_{α,β} is a proper subset of B_{α}.

The construction is straightforward, using the fact that intersections of closed sets are closed, and that since 2^{<κ} < cof(λ), closed sets in 2^{κ} of cardinality λ or more can be written as the union of an increasing cof(κ)-sequence of closed sets, since there must be a point which is not <λ-isolated. Given such a matrix, 2^{κ} ≤ λ as follows. For each f ∈ F, consider the sequence ⟨B_{E^{−1}(f↾η)} : η < λ⟩. The sets in this sequence must eventually be of cardinality less than λ, since 2^{κ} has a basis with 2^{<κ} < cof(λ) many members. Lastly, each element x of 2^{κ} defines a function from λ to cof(κ) by where it first appears in each column (letting the value of the function at α be 0 if x ∉ B_{α}). This function is everywhere dominated by some f ∈ F. Since f everwhere dominates the function determined by x, x is in all the B_{α}’s corresponding to f, and so x is in the set of cardinality less than or equal to λ corresponding to f restricted to some γ < λ. Thus 2^{κ} is the union of λ-many sets of cardinality less than or equal to λ, and so 2^{κ} ≤ λ.

Corollary 4.5. ∀κ < γ(g(κ^{+}, cof(κ)) = κ^{+}) implies ∀κ < γ(2^{κ} = κ^{+}).

In particular, CH + Axiom 1 implies 2^{ω_{1}} = ω_{2}.

Theorem 4.6. If CH + 2^{ω_{1}} = ω_{3} holds, then there is a forcing extension in which Axiom 1 fails, but CH and 𝔡_{1} = ℵ_{2} hold.

4.7 Remark. Similar statements are true on other cardinals. For instance, one can force 𝔡 = ℵ_{1} + “if F ⊂ ω_{1}^{ω} is such that for every g ∈ ω_{1}^{ω} there exists f ∈ F dominating g everywhere then for some α < ω_{1}, |{f↾α : f ∈ F}| ≥ ℵ_{2}”, since the second statement follows from the failure of CH.

Theorem 4.6 follows from Lemma 4.8 below. Let 𝔻^{ω_{1}} be Hechler forcing on ω_{1}. Conditions are of the form (s, f), where f ∈ ω_{1}^{ω_{1}}, s ∈ ω_{1}^{<ω_{1}} and s ⊂ f. (s, f) ≤ (t, g) iff t ⊂ s and for all α, f(α) ≥ g(α). This forcing is well known and has been used for instance in [11].

Assuming CH, which will be true in the ground model in the proof of Theorem 4.6, 𝔻^{ω_{1}} is ω_{2}-c.c. and σ-closed, so it preserves cardinals. Let 𝔻_{ω_{2}}^{ω_{1}} be the countable support iteration of 𝔻^{ω_{1}} of length ω_{2}. The following are standard facts about 𝔻_{ω_{2}}^{ω_{1}}.

Lemma 4.8. (CH)

1. 𝔻_{ω_{2}}^{ω_{1}} is σ-closed.

2. 𝔻_{ω_{2}}^{ω_{1}} is ω_{2}-c.c.

3. 𝔻_{ω_{2}}^{ω_{1}} forces 𝔡_{1} = ℵ_{2}.

Proof: Part 1 of the lemma is well known, and part 3 is trivial. We sketch a proof of part 2. Let p ∈ 𝔻_{ω_{2}}^{ω_{1}}. For α ∈ supp(p), p(α) = (ṡ_{α}^{p}, ḟ_{α}^{p}). First, we may assume that the ṡ_{α}^{p} are not names but partial functions s_{α}^{p} in the ground model. To see this, given p, construct conditions p_{n}, finite sets A_{n} and partial functions s_{α}^{n} such that • A_{n} ⊂ A_{n+1}, A_{n} ⊂ supp(p_{n}), ∪_{n<ω}A_{n} = ∪_{n<ω}supp(p_{n}),

• ∀α ∈ A_{n}, p_{n}↾α⊩ṡ_{α}^{p_{n}} = s_{α}^{n},

• p_{n+1} ≤ p_{n} ≤ p.

This is possible because 𝔻_{ω_{2}}^{ω_{1}} is σ-closed. For α ∈ ∪_{n<ω}A_{n}, let

s_{α}^{ω} = ∪{s_{α}^{n} | α ∈ A_{n}}

and define a condition p_{ω} with support ∪_{n<ω}A_{n} such that

• for all α ∈ ∪_{n<ω}A_{n} p_{ω}↾α⊩“ṡ_{α}^{p_{ω}} = s_{α}^{ω} and ḟ_{α}^{p_{ω}} ≥ ḟ_{α}^{p_{n}} everywhere.”

Then clearly p_{ω} ≤ p_{n} < p for all n. Call such p decided.

Now the rest of the proof is standard. Given an ω_{2} sequence of decided conditions, we can find a subset of size ω_{2} for which the supports form a Δsystem with root r. By CH, there are just ω_{1} many decided conditions with support r, and so our original sequence cannot have been an antichain.

5 Axiom 2 and Perfect Set Axioms

As shown in [17], [12] and Theorem 4.6, g(ω_{1}, ω) = ω_{1} implies CH, and so does not follow from axioms G1 and G2. At the time he produced his manuscript, Gödel had not realized this implication, and the use of such a scale in his proof is to derive that G_{ℵ_{1}} = F_{ℵ_{1}}. ^{1} Since an F_{ℵ_{1}} set of cardinality ℵ_{2} must contain an uncountable closed set, G_{ℵ_{1}} = F_{ℵ_{1}} implies PSP(ℵ_{2}, G_{ℵ_{1}}).

The ℵ_{3} in Axiom 2 appears to be arbitrarily chosen to link the reals to ℵ_{2}. By contrast, Axiom 3 refers only to ω_{1}, and Axiom 1 does not mention the reals at all. As far as we know, replacing ℵ_{3} with κ^{+} in Axiom 2 gives only κ as an upper bound on the continuum by the proof of Theorem 3.1. Further, Axioms 1-3 together imply that Axiom 2 is vacuously true. The weakest nonvacuous version of the axiom is PSP(𝔠, G_{ℵ_{1}}). Using this instead of Axiom 2, the proof of Theorem 3.1 gives that the cofinality of the continuum is ω_{2} or less. We haven’t resolved whether Axioms 1 and 3 plus PSP(𝔠, G_{ℵ_{1}}) is consistent with the continuum being ℵ_{ω_{1}} or ℵ_{ω_{2}}. Replacing Axiom 2 with PSP(cof(𝔠), G_{ℵ_{1}}) or ∃κ < 𝔠(PSP(κ, G_{ℵ_{1}})), the argument that 𝔠 ≤ ℵ_{2} still goes through.

Axiom 2 fails after adding ℵ_{3} many random reals.

Theorem 5.1. PSP(ℵ_{1}+κ, G_{ℵ_{1}}) is false after adding κ random reals to a model of CH.

1Briefly, the argument is that given a G_{ℵ_{1}}-representation we can write each open set as an increasing ω-sequence of closed sets, and associate each function from ω_{1} to ω to the corresponding intersection of closed sets. Each point in the intersection defines a function from ω_{1} to ω, and each intersection is eventually constant. The F_{ℵ_{1}} set then is the set of intersections which are defined by initial segments of the dominating scale and contained in the G_{ℵ_{1}} set.

We use a lemma from the proof of the Cichoń-Mokobodzki Theorem [3] to prove Theorem 5.1. This theorem says that adding random reals does not add a perfect set of random reals. During the proof the following is established.

Lemma 5.2. Let A ⊂ 2^{ω} ×2^{ω} be a Borel set such that {x ∈ 2^{ω} | A_{x} is perfect } has measure 1. Then there exists an F_{σ} null set B such that

{x ∈ 2^{ω} | ∃y ∈ B ⟨x, y⟩ ∈ A}

has measure 1.

Here A_{x} = {y | ⟨x, y⟩ ∈ A}. Except for B being F_{σ}, this is Lemma 3.2.20 of [3]. However, it is clear from the proof of Lemma 3.2.19 there that B may be taken to be F_{σ}.

Proof of Theorem 5.1 : The interesting case is when κ ≥ ω_{2}. Let C be the intersection of all G_{δ} measure one sets coded in the ground model. By CH, C is a G_{ℵ_{1}} set. Also |C| = 𝔠 in the extension since each random real belongs to C. Assume that P is a perfect set in the extension. Then P is added by adjoining one random real r, and in fact there is a Borel set A ⊂ 2^{ω} × 2^{ω} in the ground model such that P = A_{r}. Note that r ∈ {x | A_{x} is perfect} and without loss of generality we may assume that {x | A_{x} is perfect } has measure 1. By Lemma 5.2, there is an F_{σ} null set B in the ground model such that {x | ∃y ∈ B ⟨x, y⟩ ∈ A} has measure 1. Then r ∈ {x | ∃y ∈ B ⟨x, y⟩ ∈ A}, i.e., P ∩ B = A_{r} ∩ B ≠ ∅. Thus P ⊄ C.

Instead of CH, the proof of Theorem 5.1 requires just that Cof(M) = ℵ_{1}, where Cof(M) is the cardinality of the smallest basis for the meager ideal. This is so because Cof(M) = Cof(E) in ZFC ([3], Theorem 2.6.17), where E is the σ-ideal generated by the closed null sets. More generally, the proof gives that PSP(𝔠^{V [G]}, G_{Cof(M)^{V}}) fails after adding one or more random reals. Some assumption is necessary, however, since if 𝔡 > ℵ_{1} in the ground model then the forcing will preserve this, and so by Theorem 5.6 PSP(ℵ_{1}, G_{ℵ_{1}}) will hold.

5.1 Perfect set axioms and the dominating number

The following was pointed out to us by Hugh Woodin.

Proposition 5.3. There exists an ω_{1}-sequence of F_{σ} sets whose intersection has size ω_{1}.

Proof: Fix a bijection b: ω → ω × ω, and for each i < ω let b_{i}: ω^{ω} → ω^{ω} be such that b_{i}(x)(j) = x(b^{−1}(i, j)). Let ⟨a_{α}: ω → α | α < ω_{1}⟩ be a set of bijections. Let ⟨x_{α} : α < ω_{1}⟩ be such that for each infinite α < ω_{1},

{x_{β} : β < α} = {b_{i}(x_{α}) : i < ω}.

Now for α < ω_{1} and i < ω let A_{α,i} be the set of x ∈ ω^{ω} such that either x ∈ {x_{β} : β < α} or there exists j < ω such that b_{j}(x) = x_{a_{α}(i)}. Note that each A_{α,i} is an F_{σ} set, and that {x_{α} : α < ω_{1}} ⊂ ⋂{A_{α,i} : α < ω_{1}, i < ω}.

Now say that x ∈ ⋂{A_{α,i} : α < ω_{1}, i < ω}. We need to see that x is equal to some x_{α}. Since {b_{i}(x) : i < ω} is countable, there is an α such that {x_{β} : β < α} ⊄ {b_{i}(x) : i < ω}. Since x ∈ ⋂_{i<ω} A_{α,i}, then, x must be equal to some x_{β}, β < α.

Corollary 5.4. There exists an ω_{1}-sequence of F_{σ} sets whose intersection does not contain a perfect set.

Another example is given in the first section of [32]. A sequence of functions ⟨ρ_{0β} : β < ω_{1}⟩ is presented, each ρ_{0β} being an increasing function from β to ℚ ∩ (0, 1). This sequence has the property that

T (ρ_{0}) = {ρ_{0β}↾α | α ≤ β < ω_{1}}

under the extension ordering is a special Aronszajn tree. Each function ρ_{0β} induces the function x_{β} ∈ 2^{(ℚ∩(0,1))} given by the range of ρ_{0,β}. Then the set {x_{β} : β < ω_{1}} is the intersection of ω_{1} many F_{σ} sets in 2^{(ℚ∩(0,1))} as follows. For each t ∈ T (ρ_{0}), let

P_{t} = {x ⊂ ℚ ∩ (0, 1) | x ∩ sup(range(t)) = range(t)}.

Each P_{t} is a perfect subset of 2^{(ℚ∩(0,1))}. For each β < ω_{1}, let

G_{β} = ⋂{P̄_{t} | lev_{T} (t) = β} \ {x_{α} : α < β},

where P̄_{t} is the complement of P_{t}. Then ⟨G_{α} : α < ω_{1}⟩ is an increasing sequence of G_{δ} subsets of 2^{(ℚ∩(0,1))}, and since there are no cofinal paths through T (ρ_{0}),

⋃_{α<ω_{1}} G_{α} = 2^{(ℚ∩(0,1))} \ {x_{β} : β < ω_{1}}.

Lastly, it is shown in [32] that {x_{β} : β < ω_{1}} is of universal measure zero, and so cannot contain a perfect set.

Lemma 5.5. Every analytic set of reals can be written as the union of a family of compact sets of size at most 𝔡.

Proof: Let f : ω^{ω} → ℝ be continuous with range(f) = A. For each x ∈ ω^{ω}, let C_{x} = {y ∈ ω^{ω} | ∀n < ω y(n) ≤ x(n)}. Then each C_{x} is compact, and so each D_{x} = f[C_{x}] is compact as well. If D ⊂ ω^{ω} is a dominating family of size 𝔡, then {D_{x} : x ∈ D} is a family of 𝔡-many compact sets whose union is A.

The following theorem subsumes the well known fact that Cov(M) ≥ ℵ_{2} implies PSP(ℵ_{1}, G_{ℵ_{1}}).

Theorem 5.6. For any cardinal κ, 𝔡 > κ ⇔ PSP(ℵ_{1}, G_{κ}).

Proof: For the reverse direction, by Proposition 5.3, all we need to see is that every F_{σ} set is in G_{𝔡}. By Lemma 5.5, every co-analytic set is in G_{𝔡}.

For the other direction, we work in 2^{ω}. Define T ⊂ 2^{<ω} by

s ∈ T ⇔ [s] ∩ ⋂_{α<κ} U_{α} ≠ ∅,

where {U_{α} : α < κ} is a sequence of open sets such that ⋂_{α<κ} U_{α} is uncountable, and [s] indicates the set of extensions of s. Note that each U_{α} is open dense in T, i.e., for all α < κ, s ∈ T there is a t ∈ T such that s ⊂ t and [t] ⊂ U_{α}. Also, ⋂_{α<κ} U_{α} ⊂ [T] and \overline{⋂_{α<κ} U_{α}} = [T]. Since ⋂_{α<κ} U_{α} is uncountable, [T] contains a perfect set, so we can assume that T is a perfect tree.

Choose recursively

{x_{σ} ∈ ⋂_{α<κ} U_{α} : σ ∈ ω^{<ω}}, {k_{σ,n} ∈ ω : σ ∈ ω^{<ω}, n ∈ ω}

such that

• each k_{σ,n} is the largest integer k such that x_{σ^{⌢}⟨n⟩}↾k = x_{σ}↾k,

• for each σ the k_{σ,n}’s form an increasing sequence,

• for all σ, n, k_{σ^{⌢}⟨n⟩,0} > k_{σ,n}.

Then the sequences x_{σ^{⌢}⟨n⟩}↾(k_{σ,n} + 1) form a perfect tree. We now use 𝔡 > κ to find a perfect subtree all of whose branches are in ⋂_{α<κ} U_{α}.

For each α < κ, define ϕ_{α} : ω^{<ω} → ω by

ϕ_{α}(σ) = min{k | [x_{σ}↾k] ⊂ U_{α}}.

Now let M be a model of set theory of size κ containing everything so far, and assume f ∈ ω^{ω} is unbounded over M, and that for all n, f(n + 1) ≥ f(n) + 2.

For α < κ, let g_{α}(n) = max{ϕ_{α}(σ) | |σ| = n and ∀i < n σ(i) ≤ f(i) + 1}. We claim that for all α there are infinitely many n such that f(n) > g_{α}(n). To see this, assume that for all n ≥ n_{0}, f(n) ≤ g_{α}(n). Define recursively ḡ_{α} ∈ ω^{ω} by

• ḡ_{α}↾n_{0} = f↾n_{0},

• ḡ_{α}(n) = max{ϕ_{α}(σ) | |σ| = n and ∀i < n σ(i) ≤ ḡ_{α}(i) + 1}.

Note that ḡ_{α}(n_{0}) = g_{α}(n_{0}). Then ḡ_{α} ∈ M, so there is a minimal n > n_{0} with f(n) > ḡ_{α}(n). Since f(i) ≤ ḡ_{α}(i) for all i < n, we have g_{α}(n) ≤ ḡ_{α}(n), a contradiction. This proves the claim.

Next, define recursively

{s_{t} : t ∈ 2^{<ω}} ⊂ T, {σ_{t} ∈ ω^{<ω} : t ∈ 2^{<ω}}

such that 1. σ_{⟨⟩} = ⟨⟩, s_{⟨⟩} = ⟨⟩,

2. σ_{t^{⌢}⟨0⟩} = σ_{t}^{⌢}⟨f(|t|)⟩, σ_{t^{⌢}⟨1⟩} = σ_{t}^{⌢}⟨f(|t|) + 1⟩,

3. s_{t^{⌢}⟨0⟩} = x_{σ_{t^{⌢}⟨0⟩}}↾(k_{σ_{t},f(|t|)} + 1), s_{t^{⌢}⟨1⟩} = x_{σ_{t^{⌢}⟨1⟩}}↾(k_{σ_{t},f(|t|)+1} + 1),

We have that s_{t} ⊂ s_{t^{⌢}⟨i⟩} and s_{t^{⌢}⟨0⟩} ≠ s_{t^{⌢}⟨1⟩}. This means that

P = { ⋃_{i<ω} s_{h↾i} : h ∈ 2^{ω}}

is a perfect subset of T. We check P ⊂ ⋂_{α<κ} U_{α} by showing that for all i < ω, α < κ, if f(i) > g_{α}(i) and t ∈ 2^{i}, then [s_{t^{⌢}⟨0⟩}], [s_{t^{⌢}⟨1⟩}] ⊂ U_{α}. This follows from the fact that for all j < i, σ_{t}(j) ≤ f(j) + 1. By the definition of g_{α}, then, f(i) > ϕ_{α}(σ_{t}), so we are done.

For the following, recall that a set of reals is dense in itself if it has no isolated points.

Corollary 5.7. 𝔡 is equal to each of the following.

1. The least κ such that there is an uncountable G_{κ} set which does not contain a perfect set.

2. The least κ such that there is a G_{κ} set which is dense in itself and does not contain a perfect set.

Proof: Call the first κ_{1} and the second κ_{2}. Theorem 5.6 says just that κ_{1} = 𝔡. Since every uncountable G_{κ} set contains a G_{κ} set which is dense in itself, κ_{1} ≥ κ_{2}. That κ_{2} ≥ 𝔡 follows from a straightforward generalization of the proof of Theorem 5.6.

5.8 Conjecture. PSP(𝔡, G_{𝔡}) is false.

If Conjecture 5.8 is correct, then the next step is to analyze the axioms PSP(𝔡^{+}, G_{𝔡}) and PSP(𝔡^{++}, G_{𝔡}) (see [30]).

As we shall see, PSP(ℵ_{3}, G_{ℵ_{1}}) does not follow from 𝔡_{1} = ℵ_{2} and 𝔡 = ℵ_{1}. Furthermore, Axioms 1 and 3 together do not imply a bound on the continuum. Theorem 5.9 also shows that Axioms 1-3 don’t imply CH.

Theorem 5.9. Forcing to add ω_{1} many Hechler reals with finite support preserves Axiom 1 and makes 𝔡 = Cov(SN) = ℵ_{1}, and so forces Axiom 3.

To see that Cov(SN) = ℵ_{1} after adding ω_{1}-many Hechler reals, note that Hechler forcing makes the reals of the ground model f-coverable for every f : ω → ℚ^{+} in the ground model. Actually, the Hechler reals are not needed; since the iteration is by finite support, Cohen reals are added, and Cohen forcing also makes the ground model reals f-coverable for every f in the ground model. For Hechler reals, and many other kinds of reals, something stronger is true, that finite support iterations of any length with uncountable cofinality force Cov(SN) = ℵ_{1}. This argument is much more difficult, see [25] or Theorem 8.4.5 of [3].

5.2 Perfect set axioms in the iterated Sacks model

Sacks forcing 𝕊 is the set of perfect subtrees of 2^{<ω} ordered by inclusion. We let 𝕊_{α} be the α-length iteration of 𝕊 with countable support. For p ∈ 𝕊_{α}, supp(p) is the support of p.

We let B_{ℵ_{1}} denote the sets which can be represented as an intersection of ℵ_{1} many Borel sets.

Theorem 5.10. (CH) If G ⊂ 𝕊_{ω_{2}} is V -generic, then V [G] ⊧ P SP (ℵ_{2}, B_{ℵ_{1}}).

It is well known (see [2]) that 𝔡 = Non(M) = ℵ_{1} holds in V [G] as above, and so PSP(ℵ_{1}, G_{ℵ_{1}}) fails there - see Theorem 5.6.

Given a set of reals A in B_{ℵ_{1}} and a forcing extension V [G], we let A^{V [G]} be the interpretation of A in V [G].

Theorem 5.11. Let A ⊂ 2^{ω} be a B_{ℵ_{1}} set in V. If G ⊂ 𝕊_{ω_{2}} is V -generic, then either A^{V [G]} = A or there is a perfect set P ∈ V [G] such that P ⊂ A^{V [G]}.

The idea behind Theorem 5.11 is contained in the one-step argument.

Proposition 5.12. Let A ⊂ 2^{ω} be a B_{ℵ_{1}} set in V. If G is 𝕊-generic over V then either A^{V [G]} = A or there is a perfect set P ∈ V [G] such that P ⊂ A^{V [G]}.

Proof: Assume that the first alternative fails. Let ḟ be an 𝕊-name for a member of 2^{ω} such that 1_{𝕊}⊩ḟ ∈ A^{V [Ġ]} \ A. Given S ∈ 𝕊, it is straightforward to construct T ≤ S, a perfect set P and a homeomorphism F : [T] → P such that

T ⊩ḟ = F (ṡ),

where ṡ is the canonical name for the 𝕊-generic real (see [27]). Now work in V [G], assuming T ∈ G. From [27], we have that every new real is 𝕊-generic. Also note that for each perfect tree coded in V, there is a perfect subtree coded in V [G] all of whose branches are new reals; this is easy to see and always true when new reals are added. So T contains a perfect subtree T′, all of whose branches are Sacks-generic. Now P′ = F′′[T′] is a perfect subset of P and we claim that P′ ⊂ A^{V [G]}. For indeed, if x ∈ P′, then x = F (s) for some Sacks generic real s ∈ [T′]. Let G′ be the corresponding generic filter. So x = F (s) = F (ṡ_{G′}) = ḟ_{G′}. Since s ∈ [T′] ⊂ [T], T ∈ G′ follows. As T ⊩ḟ = F (ṡ) ∈ A^{V [Ġ]}, x ∈ A^{V [G]} follows.

5.13 Remark. The pointclass B_{ℵ_{1}} can be increased to the class of ω_{1}-Borel sets as defined in [33] in the statement of Proposition 5.12. We don’t know if this is true for Theorems 5.10 and 5.11. Roughly, the ω_{1}-Borel sets are those sets of reals which have descriptions of size ℵ_{1}. The issue is that unlike for B_{ℵ_{1}} sets, the statement that a perfect set is contained in a given ω_{1}-Borel set is not necessarily upwards absolute; if one real is added to a model of CH, for example, then the reals of the ground model are an F_{ℵ_{1}} set not containing a perfect set, even though they trivially contain a perfect set in the ground model. The absoluteness of the statement that a given perfect set is contained in a given B_{ℵ_{1}} set is key to the proofs in this section.

Ciesielski and Pawlikowski have recently [10] produced a shorter proof of Theorem 5.11, using their axiom CPA_{prism}. In that paper they also prove the following theorem, refuting a conjecture in an earlier version of this paper (and negatively answering a question whose positive answer implied the conjecture).

Theorem 5.14. [10] If CH holds and G ⊂ 𝕊_{ω_{2}} is V -generic, then B_{ℵ_{1}} ≠ F_{ℵ_{1}} in V [G].

It is easy to see that B_{ℵ_{1}} = F_{ℵ_{1}} is equivalent to the statement G_{ℵ_{1}} = F_{ℵ_{1}}, which Gödel mistakenly claimed to derive from G1 and G2. That G_{ℵ_{1}} = F_{ℵ_{1}} follows trivially from CH, and implies ¬PSP(ℵ_{1}, G_{ℵ_{1}}) and PSP(ℵ_{2}, G_{ℵ_{1}}). To see ¬PSP(ℵ_{1}, F_{ℵ_{1}}), note that ZFC implies the existence of an F_{ℵ_{1}} set of cardinality ℵ_{1} which does not contain a perfect set. The statement PSP(ℵ_{2}, F_{ℵ_{1}}) follows from the fact that uncountable closed sets contain perfect sets.

The following questions remain open for lack of models that would show consistency.

5.15 Question. Does G_{ℵ_{1}} = F_{ℵ_{1}} imply CH?

5.16 Question. Does 𝔡 = ℵ_{1}∧ PSP(ℵ_{2}, G_{ℵ_{1}}) imply 𝔠 ≤ ℵ_{2}?

Proof of Theorem 5.11: By Proposition 5.12, we need to consider only limit stages α where α has cofinality ω. Given a condition p_{0} ∈ 𝕊_{α} and a name ḟ for a member of 2^{ω}, we shall construct:

Step 1. a condition p ≤ p_{0} and a perfect tree T such that p⊩_{α}ḟ ∈ [T] in a canonical way,

Step 2. a canonical name Ṡ such that p⊩_{α}“Ṡ ⊂ T, Ṡ is a perfect tree, and

[Ṡ] ⊂ V [Ġ_{α}] \ ⋃_{β<α} V [Ġ_{β}], ”

i.e., all branches have the same constructibility degree as ḟ.

Then given a name ġ and a condition q_{0} ≤ p such that q_{0}⊩_{α}ġ ∈ [Ṡ], we construct:

Step 3. a condition q ≤ q_{0} and a perfect tree U such that q⊩_{α}ġ ∈ [U] in a canonical way, and a condition r ≤ p such that canonically r⊩_{α}ḟ ∈ [U], and further, whenever G_{α} is generic, q ∈ G_{α}, then there is G′_{α} generic, r ∈ G′_{α} such that

ġ_{G_{α}} = ḟ_{G′_{α}}.

More explicitly, there will be a canonical isomorphism π : 𝕊_{α}↾q → 𝕊_{α}↾r with π(q) = r such that whenever G_{α} is generic, q ∈ G_{α}, then ġ_{G_{α}} = ḟ_{π[G_{α}]}.

From Step 3 we then have that p forces that all branches of Ṡ are generic in the same sense as ḟ, so whatever p forces about ḟ will be true about all members of [Ṡ], so we will be done.

Step 1: We construct a fusion sequence ⟨p_{n} : n < ω⟩. The given condition is p_{0}; p will be the result of the fusion. We also construct for each σ ∈ 2^{<ω} a sequence S_{|σ|,σ} ∈ 2^{<ω} and a condition p_{|σ|}⟨σ⟩ ≤ p_{|σ|} such that • {S_{|σ|,σ} | σ ∈ 2^{<ω}} forms a perfect tree, with extension and incompatibility

in accordance with the corresponding σ’s,

• for each n, {p_{n}⟨σ⟩ | σ ∈ 2^{n}} forms a maximal antichain below p_{n},

• for each σ ∈ 2^{<ω}, p_{|σ|}⟨σ⟩⊩S_{|σ|,σ} ⊂ ḟ.

Then

T = {s ∈ 2^{<ω} | ∃σ ∈ 2^{<ω}(s ⊂ S_{|σ|,σ})}

will be the desired tree.

Fix a function F : ω → ω such that

(*) F^{−1}({n}) is infinite for all n.

(**) n < m ⇒ min(F^{−1}(n)) < min(F^{−1}(m)).

Note then that F (n) ≤ n for all n. Our construction will also fix ordinals α_{n} such that supp(p) = {α_{n} : n ∈ ω}. We suppress the bookkeeping of which α_{n}’s arise when, except for the stipulation that α_{n} ∈ supp(p_{n}) for each n. The fusion condition will be that for each α_{n} and each m ≥ min{i | F (i) = n}, 1_{𝕊_{αn}} forces that the first |{i ≤ m | F (i) = n}| splitting levels of p_{m}(α_{n}) and p_{m+1}(α_{n}) will be the same. We will use the following sets, where n ∈ ω and γ < α :

A_{n}^{γ} = {i < n | α_{F (i)} ≤ γ}, B_{n}^{γ} = {i < n | α_{F (i)} < γ}

(so A_{n}^{γ} \ B_{n}^{γ} = {i < n | α_{F (i)} = γ}). During our construction, we will produce sequences S_{n,σ↾A_{n}^{γ}}^{γ} ∈ 2^{<ω}, where n ∈ ω, σ ∈ 2^{n} and γ ∈ {α_{F (i)} : i < n}, such that

(a) for γ ∈ {α_{F (i)} : i < n} and σ: B_{n}^{γ} → 2, there exists (p_{n}↾γ)⟨σ⟩ ≤ p_{n}↾γ such that (p_{n}↾γ)⟨σ⟩ forces

• {S_{n,τ}^{γ} | σ ⊂ τ ∧ dom(τ) = A_{n}^{γ}} ⊂ p_{n}(γ),

• ∀S ∈ p_{n}(γ)∃τ ∈ 2^{A_{n}^{γ}}(σ ⊂ τ ∧ (S ⊂ S_{n,τ}^{γ} ∨ S_{n,τ}^{γ} ⊂ S)),

(b) fixing n ∈ ω and γ ∈ {α_{F (i)} : i < n}, for all distinct σ, σ′ ∈ 2^{A_{n}^{γ}}, S_{n,σ}^{γ} and S_{n,σ′}^{γ} are incompatible,

(c) fixing n ∈ ω and γ ∈ {α_{F (i)} : i < n}, for all σ ∈ 2^{A_{n+1}^{γ}},

S_{n,σ↾A_{n}^{γ}}^{γ} ⊂ S_{n+1,σ}^{γ}.

Therefore, if (p_{n}↾γ)⟨σ⟩ ∈ G_{γ} then the S_{n,τ}^{γ} with σ ⊂ τ canonically define the |A_{n}^{γ}\B_{n}^{γ}|-th splitting level of p_{n}(γ). Note that if α_{F (n)} > γ, then for all σ ∈ 2^{n+1} we may choose

S_{n+1,σ↾A_{n+1}^{γ}}^{γ} = S_{n,σ↾A_{n}^{γ}}^{γ}.

The (p_{n}↾γ)⟨σ⟩ will satisfy the following recursively defined conditions for γ ∈ {α_{F (i)} : i < n} and σ with dom(σ) = B_{n}^{γ}. The same construction will be used in Step 3 (without repeating the details).

• If γ = min{α_{F (i)} : i < n}, then B_{n}^{γ} = ∅ and (p_{n}↾γ)⟨σ⟩ = p_{n}↾γ⟨⟩ = p_{n}↾γ.

• Assume (p_{n}↾γ)⟨σ↾B_{n}^{γ}⟩ is defined for some γ ∈ {α_{F (i)} : i < n} and σ with dom(σ) = A_{n}^{γ}. Let δ = min{α_{F (i)} | i < n ∧ α_{F (i)} > γ}. Note that B_{n}^{δ} = A_{n}^{γ}. Then

(p_{n}↾δ)⟨σ⟩ = (p_{n}↾γ)⟨σ↾B_{n}^{γ}⟩^{⌢}(p_{n}(γ)_{S_{n,σ}^{γ}})^{⌢}p_{n}↾[γ + 1, δ),

where as usual 1_{𝕊_{α}}⊩ṗ_{s} = {ṫ ∈ ṗ | s ⊂ ṫ∨ ṫ ⊂ s}. This makes sense because by (b) above we have indeed that (p_{n}↾γ)⟨σ↾B_{n}^{γ}⟩⊩S_{n,σ}^{γ} ∈ p_{n}(γ).

Similarly, p_{n}⟨σ⟩ = (p_{n}↾γ)⟨σ↾B_{n}^{γ}⟩^{⌢}(p_{n}(γ)_{S_{n,σ}^{γ}})^{⌢}p_{n}↾[γ + 1, α), where

γ = max{α_{F (i)} : i < n}.

Note that in this case A_{n}^{γ} = n. The key point is that

{(p_{n}↾γ)⟨σ⟩ | σ ∈ 2^{B_{n}^{γ}}}

forms a maximal antichain below p_{n}↾γ, and the same holds for {p_{n}⟨σ⟩ | σ ∈ 2^{n}} and p_{n}. Further, if γ < δ and σ ∈ 2^{B_{n}^{γ}} is a subfunction of τ ∈ 2^{B_{n}^{δ}} (or 2^{n}), then (p_{n}↾δ)⟨τ⟩ (or p_{n}⟨τ⟩) ≤ (p_{n}↾γ)⟨σ⟩.

Now for the details of the construction. In the case n = 0, put S_{n,σ} = σ = ⟨⟩, and let all (p_{0}↾γ)⟨⟩ = p_{0}↾γ, p_{0}⟨⟩ = p_{0}. Then all the conditions are satisfied trivially. Given the construction for n, construct for n + 1 as follows. Let δ = α_{F (n)}; since for all i < n α_{i} ∈ supp(p_{i}) this is defined (not all α_{i} have been). Choose, for each σ ∈ 2^{B_{n+1}^{δ}}, p_{σ}^{∗∗} ≤ (p_{n}↾δ)⟨σ⟩ and S_{n+1,τ}^{δ} for τ ⊃ σ, τ ∈ 2^{A_{n+1}^{δ}} such that p_{σ}^{∗∗}⊩S_{n+1,τ}^{δ} ∈ p_{n}(δ), and such that

• for distinct τ, the S_{n+1,τ}^{δ} are incompatible,

• in case δ ∈ {α_{F (i)} : i < n}, p_{σ}^{∗∗}⊩S_{n,τ↾A_{n}^{δ}}^{δ} ⊂ S_{n+1,τ}^{δ}.

Now recursively for each

γ ∈ {α_{F (i)} > δ | i < n},

we produce

S_{n+1,σ}^{γ}

for σ ∈ 2^{A_{n+1}^{γ}}, as follows.

Say that we have chosen for γ, and want to choose for

γ′ = min({α_{F (i)} : i < n} \ γ + 1).

For j ∈ {0, 1} fix G_{γ′}^{j} generic such that (p_{n}↾γ′)⟨ϑ^{⌢}⟨j⟩⟩ ∈ G_{γ′}^{j}, where dom(ϑ) = B_{n}^{γ′} = A_{n}^{γ}, τ ⊂ ϑ. Then look at S_{n,χ}^{γ′}, χ ⊃ ϑ, dom(χ) = A_{n}^{γ′}. These form a finite maximal antichain in p_{n}(γ′). So, in the models V [G_{γ′}^{0}] and V [G_{γ′}^{1}] we may find

S_{n+1,χ^{⌢}⟨0⟩}^{γ′}, S_{n+1,χ^{⌢}⟨1⟩}^{γ′}

which are incompatible such that S_{n,χ}^{γ′} ⊂ S_{n+1,χ^{⌢}⟨j⟩}^{γ′} ∈ p_{n}(γ′) holds in V [G_{γ′}^{j}].

Having chosen the

S_{n+1,ϑ↾A_{n+1}^{γ}}^{γ}

we now choose p_{n+1}. This choice induces our choices of the (p_{n+1}↾γ)⟨σ⟩ and p_{n+1}⟨σ⟩. For σ ∈ 2^{<ω}, and q ∈ 𝕊_{α}, let ϕ(σ, q) be the statement that for each γ ∈ {α_{F (i)} : i < |σ|}, q↾γ is consistent with

S_{|σ|,σ↾A_{|σ|}^{γ}}^{γ}

being an initial segment of the stem of q(γ). The fusion requirement is maintained by requiring that ϕ(σ, p_{n+1}) holds for each σ ∈ 2^{n+1}. We have chosen the

S_{n+1,σ↾A_{n+1}^{γ}}^{γ}

so that ϕ(σ, p_{n}) always holds. Further, these statements are mutually incompatible. Let q_{σ} ≤ p_{n} force that

S_{n+1,σ↾A_{n+1}^{γ}}^{γ}

is an initial segment of the stem of p_{n}(γ), and furthermore extend the q_{σ} so that they decide incompatible initial segments S_{n+1,σ} of ḟ. Then by the same recursive procedure as for choosing the

S_{n+1,σ↾A_{n+1}^{γ}}^{γ}

we can choose p_{n+1} so that the q_{σ} (= p_{n+1}⟨σ⟩) form a maximal antichain below p_{n+1} as desired. This completes the standard argument and the construction for n + 1.

This completes Step 1.

For Step 2, we first give a definition of S = Ṡ_{G_{α}} in the generic extension. That is, suppose G_{α} is 𝕊_{α}-generic, with p ∈ G_{α}. Then f = ḟ_{G_{α}} ∈ T, and there is a unique y ∈ 2^{ω} such that f = ḟ_{G_{α}} ⊃ S_{n,y↾n} for all n. Let {l_{n} : n < ω} be such that {α_{l_{n}} : n ∈ ω} is strictly increasing and converges to α. Choose the l_{n}’s so that α_{l_{n}} > α_{i} for all i < l_{n} - this will be useful in Step 3. Let Z be the set of z ∈ 2^{ω} such that

• ∀n∀j ∈ F^{−1}({l_{n}}) z(min(F^{−1}({l_{n}}) \ (j + 1))) = y(j),

• ∀j ∉ F^{−1}({l_{n} : n ∈ ω}) z(j) = y(j).

Note that for each n, z can take any value at min{i | F (i) = l_{n}}. Let

S = {g | ∃z ∈ Z∀n S_{n,z↾n} ⊂ g}.

S is a perfect subset of T, so there a name Ṡ such that Ṡ_{G_{α}} = S. Also note the following (still in V [G_{α}]): if g ∈ S then in V [g] we can reconstruct z ∈ Z such that ∀nS_{n,z↾n} ⊂ g. From z we can reconstruct y, and from y we can reconstruct f. So f ∈ V [g]. This means that all branches of S are reals which arise only in V [G_{α}].

On the other hand, by [23] we have that exactly one new constructibility degree, above all the previous degrees, arises in V [G_{α}]. So f, S and all the branches of S have the same constructibility degree. This completes Step 2.

Step 3: First, fix a name ġ and a condition q_{0} ≤ p such that q_{0}⊩ġ ∈ [Ṡ]. As above, we have a name ẏ such that p⊩ḟ = ⋃_{n∈ω} S_{n,ẏ↾n}. This gives us a name ż such that q_{0}⊩ġ = ⋃_{n∈ω} S_{n,ż↾n}, where ż is forced to be in the set Z defined from ẏ as above.

We construct a fusion sequence ⟨q_{n} : n < ω⟩, where q_{0} is the given condition. The result of the fusion will be q. The construction of the tree U will be induced by the construction, for each σ ∈ 2^{<ω}, of a sequence T_{|σ|,σ} ∈ 2^{<ω} and a condition q_{|σ|}⟨σ⟩ ≤ q_{|σ|} such that

• {T_{|σ|,σ} | σ ∈ 2^{<ω}} forms a perfect tree, with extension and incompatibility in accordance with the corresponding σ’s,

• each q_{|σ|}⟨σ⟩⊩T_{|σ|,σ} ⊂ ż,

• for each n, {q_{n}⟨σ⟩ | σ ∈ 2^{n}} forms a maximal antichain below q_{n}.

• the length of T_{|σ|,σ} depends only on n (and will denoted by k_{n}).

Then U = {t ∈ 2^{<ω} | ∃n ∈ ω, σ ∈ 2^{n}(t ⊂ S_{k_{n},T_{n,σ}})} will be as desired. Along the way we also fix ordinals β_{n} (n ∈ ω) such that supp(q) = {β_{n} : n ∈ ω}. Note that {α_{n} : n ∈ ω} ⊂ {β_{n} : n ∈ ω}. We stipulate that β_{n} ≤ α_{n} for all n, and also that if β_{n} ∈ {α_{i} : i ∈ ω} then there exists i ≤ n such that β_{n} = α_{i}. We will use the following sets, where γ < α and n ∈ ω:

C_{n}^{γ} = {i < n | β_{F (i)} ≤ γ}, D_{n}^{γ} = {i < n | β_{F (i)} < γ},

(so C_{n}^{γ} \ D_{n}^{γ} = {i < n | β_{F (i)} = γ}).

Additionally, we produce

i. sequences T_{n,σ↾C_{n}^{γ}}^{γ} ∈ 2^{<ω} (of length ≥ |C_{n}^{γ}|), where n ∈ ω, σ ∈ 2^{n}, and γ ∈ {β_{F (i)} : i < n}

ii. U_{n,σ↾C_{n}^{γ}}^{γ} ∈ 2^{<ω}, where n ∈ ω, σ ∈ 2^{n}, and γ ∈ {β_{F (i)} : i < n}

such that

a. for all n ∈ ω, γ ∈ {β_{F (i)} : i < n},

• for all distinct σ ∈ 2^{C_{n}^{γ}}, the T_{n,σ}^{γ} are all pairwise incompatible,

• for all σ ∈ 2^{C_{n+1}^{γ}}, T_{n,σ↾C_{n}^{γ}}^{γ} ⊂ T_{n+1,σ}^{γ},

b. for all n ∈ ω and γ ∈ {β_{F (i)} : i < n}, • for all distinct σ ∈ 2^{C_{n}^{γ}}, the U_{n,σ}^{γ} are all pairwise incompatible,

• for all σ ∈ 2^{C_{n+1}^{γ}}, U_{n,σ↾C_{n}^{γ}}^{γ} ⊂ U_{n+1,σ}^{γ},

c. for all n ∈ ω, σ ∈ 2^{n} and γ ∈ {β_{F (i)} : i < n} ∩ {α_{i} : i ∈ ω},

U_{n,σ↾C_{n}^{γ}}^{γ} = S_{k_{n},T_{n,σ}↾A_{kn}^{γ}}^{γ} .

d. for all n ∈ ω, γ ∈ {β_{F (i)} : i < n} and σ ∈ 2^{D_{n}^{γ}}, there exists (q_{n}↾γ)⟨σ⟩ ≤ q_{n} such that (q_{n}↾γ)⟨σ⟩ forces

• {T_{n,τ}^{γ} | σ ⊂ τ ∧ dom(τ) = C_{n}^{γ}} ⊂ q_{n}(γ),

• ∀T ∈ q_{n}(γ)∃τ ∈ 2^{C_{n}^{γ}}(σ ⊂ τ ∧ (T ⊂ T_{n,τ}^{γ} ∨ T_{n,τ}^{γ} ⊂ T)),

i.e., each T_{n,τ}^{γ} canonically defines the |C_{n}^{γ} \ D_{n}^{γ}|-th splitting level of q_{n}(γ), so together they define a finite maximal antichain in q_{n}(γ). In particular, if β_{F (n)} > γ, then for all σ ∈ 2^{n+1}

T_{n+1,σ↾C_{n+1}^{γ}}^{γ} = T_{n,σ↾C_{n}^{γ}}^{γ}.

As specified by (d), the sets q_{n}↾γ⟨σ⟩ and q_{n}⟨σ⟩ are built up from the T_{n,σ}^{γ}’s in the same way that the sets p_{n}↾γ⟨σ⟩ and p_{n}⟨σ⟩ were built from the S_{n,σ}^{γ}’s in Step 1, using {β_{i} : i < ω} in place of {α_{i} : i < ω}. Furthermore, the condition r is constructed in the same way from sets r_{n}↾γ⟨σ⟩ and r_{n}⟨σ⟩ built in the same way from the U_{n,σ}^{γ}’s. Conditions (a) and (b) then induce the isomorphism π between 𝕊_{α}↾q and 𝕊_{α}↾r. The remaining point, that ġ_{G_{α}} = ḟ_{G′_{α}} whenever G_{α} ⊂ 𝕊_{α} is generic with q ∈ G_{α}, follows from condition (c) and the fact that each q_{n}⟨σ⟩⊩T_{|σ|,σ} ⊂ ż.

The construction requires one more condition:

e. if n ∈ ω, σ, σ̄ ∈ 2^{n}, γ ≥ β_{F (i)} is the minimal ordinal in

{β_{F (i)} : i ≤ n} ∩ {α_{j} : j < ω},

and σ↾C_{n}^{γ} = σ̄↾C_{n}^{γ}, then there is a j ∈ A_{k_{n+1}}^{γ} \ (A_{k_{n}}^{γ} ∪ B_{k_{n+1}}^{γ}) such that T_{n+1,σ^{⌢}⟨0⟩}(j) ≠ T_{n+1,σ̄^{⌢}⟨1⟩}(j).

Now for the details. By the argument for Step 1, we can construct the T_{n,σ↾C_{n}^{γ}}^{γ}’s to satisfy (a) and (d). We may further assume by augmenting the previous argument that the sequences T_{n,σ} (σ ∈ 2^{n}) all have the same length k_{n}. To satisfy (e), we add another condition to the step n → n + 1 of the construction. Fix n ∈ ω, σ, σ̄ ∈ 2^{n}. Let δ = β_{F (n)} and let γ ≥ β_{F (i)} be the least ordinal in {β_{F (i)} : i ≤ n} ∩ {α_{j} : j < ω}, if such an ordinal exists. Let G_{γ} be generic, with

(q_{n}↾γ)⟨σ↾D_{n}^{γ}⟩ = (q_{n}↾γ)⟨σ̄↾D_{n}^{γ}⟩ ∈ G_{γ}.

Now consider

(q_{n}(γ))_{T_{n,σ↾C^{γ}_{n}}^{γ}}

in V [G_{γ}]. Since q_{n} ≤ p,

(q_{n}(γ))_{T_{n,σ↾C^{γ}_{n}}^{γ}} ≤ p(γ).

By the construction of p, p(γ) is the set of s for which there exists k, τ such that

• dom(τ) = A_{k}^{γ},

• s ⊂ S_{k,τ}^{γ},

• ∀j ∈ ω(α_{F (j)} < γ ⇒ τ(j) = H(j)),

where the function H : {i ∈ ω | α_{F (i)} < γ} → 2 is canonically given by the generic G_{γ} as in Step 1, i.e.,

H = ⋃{ρ: B_{m}^{γ} → 2 | m ∈ ω, (p_{m}↾γ)⟨ρ⟩ ∈ G_{γ}}.

Fix ī ∈ ω such that γ = α_{F (ī)}. Still arguing in V [G_{γ}], we can find extensions q̄, ¯ q̄ of

(q_{n}(γ))_{T_{n,σ↾C^{γ}_{n}}^{γ}}

such that for some j ∈ F^{−1}({F (ī)}) \ k_{n} and some τ̄, ¯ τ̄ ∈ 2^{A_{k_{n+1}}^{γ}} (assuming that we have chosen k_{n+1} to be large enough) we have τ̄(j) ≠ ¯ τ̄(j) and stem(q̄) = S_{k_{n+1},τ̄}^{γ}, stem(¯q̄) = S_{k_{n+1},¯τ̄}^{γ}. By the definitions of ẏ in Step 2 and (b) in Step 1, this means q̄ and ¯ q̄ force different values to ẏ(j), so they force different values to ż(j) or to ż(min(F^{−1}({F (ī)}) \ (j + 1))). In either case, by letting

T_{n+1,σ^{⌢}⟨0⟩↾D_{n+1}^{γ}}^{γ}, T_{n+1,σ̄^{⌢}⟨1⟩↾D_{n+1}^{γ}}^{γ}

extend the sequences S_{k_{n+1},τ̄}^{γ} and S_{k_{n+1},¯τ̄}^{γ} respectively, we ensure that T_{n+1,σ^{⌢}⟨0⟩} and T_{n+1,σ̄^{⌢}⟨1⟩} are distinct at some

j ∈ A_{k_{n+1}}^{γ} \ (A_{k_{n}}^{γ} ∪ B_{k_{n+1}}^{γ}).

The construction for Step 3 is essentially finished. We just need to see that conditions (b) and (c) worked out. First note that in the case where γ ∉ {α_{i} : i ∈ ω} we can arbitrarily choose U_{n,σ↾C_{n}^{γ}}^{γ} to satisfy (b) because clause (c) is void. So assume γ ∈ {α_{i} : i ∈ ω}. We first check for each n that the requirement in (c) is well-defined. That is, we show that for all γ ∈ {β_{F (i)} : i < n} ∩ {α_{i} : i ∈ ω}, if σ, σ̄ ∈ 2^{n} satisfy σ↾C_{n}^{γ} = σ̄↾C_{n}^{γ}, then T_{n,σ}↾A_{k_{n}}^{γ} = T_{n,σ̄}↾A_{k_{n}}^{γ}.

For n = 0 there is nothing to show. For n + 1 ∈ ω, fix

γ ∈ {β_{F (i)} : i < n + 1} ∩ {α_{i} : i ∈ ω}.

Say γ = β_{F (i_{0})}. Assume that σ, σ̄ ∈ 2^{n+1} are such that σ↾C_{n+1}^{γ} = σ̄↾C_{n+1}^{γ}. Let j ∈ A_{k_{n+1}}^{γ}. Then α_{F (j)} ≤ γ. Also note (trivially) that

(q_{n+1}↾γ + 1)⟨σ↾C_{n+1}^{γ}⟩ = (q_{n+1}↾γ + 1)⟨σ̄↾C_{n+1}^{γ}⟩

and that

q_{n+1}⟨σ⟩⊩T_{n+1,σ} ⊂ ż, q_{n+1}⟨σ̄⟩⊩T_{n+1,σ̄} ⊂ ż,

so they decide ż(j). However, by Steps 1 and 2, if two conditions below p force different values to ż(j), then

• in case j ∈ F^{−1}({m}), where m ∉ {ln̄ : n̄ ∈ ω}, they force different values to ẏ(j), which means the conditions are forced by 1_{𝕊_{α_{F (j)}}} to take different values at the α_{F (j)}th stage of the iteration, which is not the case for q_{n+1}⟨σ⟩ and q_{n+1}⟨σ̄⟩

• in case j ∈ F^{−1}({ln̄}) for some n̄, if j ≠ min(F^{−1}({ln̄})), they still force different values to some ẏ(j̄), where j̄ < j and F (j̄) = ln̄ = F (j), so the argument reduces to the previous case

• in case j = min(F^{−1}({ln̄})) for some n̄, we have β_{F (j)} ≤ α_{F (j)} ≤ γ = β_{F (i_{0})}, so j ≤ i_{0} by (**) in Step 1, and the choice of the l_{n}’s in Step 2. Let i_{1} ≤ i_{0} be minimal such that β_{F (i_{1})} ≥ α_{F (j)}. We still have j ≤ i_{1},

for the same reason, and also that C_{i_{1}+1}^{β_{F (i_{1})}} = i_{1} + 1. Also, C_{i_{1}+1}^{β_{F (i_{1})}} ⊂ C_{n+1}^{γ} because β_{F (i_{1})} ≤ γ. Let τ = σ↾(i_{1} + 1) = σ̄↾(i_{1} + 1). Then

T_{i_{1}+1,τ} = T_{n+1,σ}↾k_{i_{1}+1} = T_{n+1,σ̄}↾k_{i_{1}+1},

and since j ≤ i_{1} ≤ k_{i_{1}} < k_{i_{1}+1}, they agree at j.

This verifies (c). Finally, we check (b). The inclusion relation is immediate. We need to check incompatibility, that is, that if σ↾C_{n}^{γ} = σ̄↾C_{n}^{γ} and n ∈ C_{n+1}^{γ} (so that σ^{⌢}⟨0⟩↾C_{n+1}^{γ} ≠ σ̄^{⌢}⟨1⟩↾C_{n+1}^{γ}), then U_{n+1,σ^{⌢}⟨0⟩↾C_{n+1}^{γ}}^{γ} and U_{n+1,σ̄^{⌢}⟨1⟩↾C_{n+1}^{γ}}^{γ}

are incompatible. By (c) it suffices to show that

T_{n+1,σ^{⌢}⟨0⟩}↾A_{k_{n+1}}^{γ} ≠ T_{n+1,σ̄^{⌢}⟨1⟩}↾A_{k_{n+1}}^{γ}.

This follows by induction.

For T_{0,⟨⟩}↾A_{k_{0}}^{γ} (the case n = −1) there is nothing to show. Given that this holds for n, we argue for n + 1. Now β_{F (n)} ≤ γ, since n ∈ C_{n+1}^{γ}. Also, by the assumption in (b), γ = β_{F (i)}, for some i ≤ n. Let γ′ be the least member of {β_{F (i)} : i ≤ n} ∩ {α_{j} : j ∈ ω} (recall that γ is in this intersection). By (e) there is j ∈ A_{k_{n+1}}^{γ′} \ A_{k_{n}}^{γ′} such that

T_{n+1,σ^{⌢}⟨0⟩}(j) ≠ T_{n+1,σ̄^{⌢}⟨1⟩}(j),

so we are done.

This finishes Step 3, and the proof.

6 Axiom 3

It is easy to see that Axiom 3 holds if 𝔡 = Cov(SN) = ℵ_{1}. We show here that Axiom 3 implies 𝔡 = ℵ_{1}. Axiom 3 as stated trivially implies Cov(N) = ℵ_{1}, and in the second subsection we show that this would still hold even if we restricted Axiom 3 to sets of measure zero.

6.1 Remark. As pointed out in [30], Axiom 2 + 𝔡 = ℵ_{1} + Cov(SN) ≤ ℵ_{2} trivially implies that 𝔠 ≤ ℵ_{2}. This follows from the fact that each strong measure zero set is contained in a G_{𝔡} set of strong measure zero, and so by Axiom 2 must have cardinality less than or equal to ℵ_{2} since it can’t contain a perfect set. By the same reasoning, 𝔡 = ℵ_{1}, Cov(SN) = ℵ_{1} and PSP(ℵ_{2}, G_{ℵ_{1}}) together imply CH.

Axiom 3 implies that there exists an ω_{1}-sequence of functions from ω to ℝ^{+} such that no G_{ℵ_{1}} set which is f-coverable for every f in the sequence contains a perfect set. This follows from the fact that H has no decreasing ω_{2} sequences, and so there is an ω_{1} sequence of elements of H with no lower bound in H. Any perfect set coverable by every function in the sequence would be a counterexample to Axiom 3. Since perfect sets cannot have strong measure zero, the existence of such a sequence follows from 𝔡 = ℵ_{1}. The converse also holds.

Theorem 6.2. If there exists a κ-sequence of functions from ω to ℝ^{+} such that no G_{κ} set which is f-coverable for every f in the sequence contains a perfect set then 𝔡 ≤ κ.

Proof: We prove the contrapositive. Assume that 𝔡 > κ. Given a sequence ⟨f_{α} : α < κ⟩ of functions from ω to ℝ^{+}, for each α < κ define f′_{α} : ω → ℝ^{+} by letting

f′_{α}(n) = min{f_{α}(m) : m ∈ [2^{n+1} − 2, 2^{n+2} − 3]}.

Using 𝔡 > κ, let g : ω → ℝ^{+} be such that for all α < κ {n ∈ ω | g(n) < f′_{α}(n)} is infinite. Now build a binary tree of intervals such that the members of the nth level (not counting the root as a node these are the (2^{n+1} − 2)th to (2^{n+2} − 3)th nodes of the tree) are disjoint and of diameter less than g(n). Let P be the set of reals arising from paths through this tree, and note that P is a perfect set. For each α, then, we have infinitely many n such that P can be covered by a sequence of intervals of diameters as prescribed by f_{α}↾[2^{n+1} − 2, 2^{n+2} − 3], so P is f_{α}-coverable.

In certain cirumstances the requirement that we decompose into an increasing sequence of smaller sets is not restrictive. These circumstances do not always hold, though, see Theorem 6.37.

Lemma 6.3. Assume 𝔡 = ℵ_{1} and let A be a set of reals such that for any countable set G of functions g: ω → ℝ^{+} for which A is not g-coverable there is an h: ω → ℝ^{+} such that A is not h-coverable and h ≥ g mod-finite for all g ∈ G. If A = ⋃_{α<ω_{1}} B_{α}, where each B_{α} ⊲ A, then there is an increasing sequence ⟨D_{α} : α < ω_{1}⟩ such that A = ⋃_{α<ω_{1}} D_{α} and each D_{α} ⊲ A.

Proof: We want to write the union of the B_{α}’s as an increasing union of sets smaller than A. We may assume that each B_{α} is a G_{δ}-set, and so by Lemma 5.5 each B_{α} is a union of compact sets ⟨C_{αζ} : ζ < ω_{1}⟩. We show that each D_{β} ⊲ A, where D_{β} = ⋃{C_{αζ} : α, ζ < β}. Since the C_{αζ}’s are compact, it suffices to find for each β < ω_{1} a function h_{β}: ω → ω such that A is not h_{β}-coverable but each C_{αζ}, α, ζ < β, is. Our assumption on A gives us such an h_{β}.

6.1 Cardinal invariants and Axiom 3

For f : ω → ℝ^{+}, Cov(f) is the least κ such that there is a κ-sequence of f-sets whose union is the reals. In this context we say that f is nontrivial if ∑_{n∈ω} f(n) is finite, since otherwise Cov(f) = 1. We define two more cardinal invariants.

6.4 Definition. The least cardinal κ such that for some nontrivial f ∈ (ℝ^{+})^{ω}, Cov(f) = κ is denoted 𝔪𝔠 (minCov). The least cardinal κ such that for all nontrivial f ∈ (ℝ^{+})^{ω}, Cov(f) ≤ κ is denoted 𝔰𝔠 (supCov).

Note that 𝔰𝔠 ≥ 𝔪𝔠 ≥ Cov(N) trivially.

6.5 Remark. If 𝔰𝔠 = ℵ_{1}, the ω_{1}-many f-coverable sets can be taken to be an increasing sequence (even if 𝔡 > ℵ_{1}). This follows from the fact that for any f : ω → ℝ^{+} there is a g : ω → ℝ^{+} such that any countable union of g-coverable sets is f-coverable. Assuming that f is strictly decreasing, any g such that

g(n) ≤ f (\frac{(n + 1)(n + 2)}{2})

for all n suffices.

Axiom 3 then follows from 𝔡 = 𝔰𝔠 = ℵ_{1}.

6.6 Remark. Unlike the case for 𝔡 = Cov(SN) = ℵ_{1}, we know of no argument other than our version of Gödel’s argument which proves 𝔠 ≤ ℵ_{2} from our first two axioms plus 𝔡 = 𝔰𝔠 = ℵ_{1}. However, if we replace Axiom 3 by 𝔡 = 𝔰𝔠 = ℵ_{1} in Theorem 3.1, then we can replace Axiom 1 with 𝔡_{1} = ℵ_{2}, which unlike Axiom 1 does follow from G1+G2. The point is that using 𝔡 = Cov(SN) = ℵ_{1} we can carry out Gödel’s construction so that each descending sequence of B_{α}’s becomes strong measure zero in at most ω_{1} many steps, so each real is in a strong measure zero set corresponding to some member of a given witness to the value of 𝔡_{1}.

In this section, we show the following, among other things.

• 𝔪𝔠 and 𝔰𝔠 can be characterized in terms of covering numbers for trees.

• Cov(N) = ℵ_{1} does not imply 𝔰𝔠 = ℵ_{1}.

• 𝔟 > Cov(N) implies 𝔪𝔠 = Cov(N).

• 𝔰𝔠 = ℵ_{1} does not imply (Non(M) = ℵ_{1} ∨ Cov(SN) = ℵ_{1}).

• 𝔡 = 𝔰𝔠 = ℵ_{1} does not imply Cov(SN) ≤ ℵ_{2}.

We also leave several questions open.

Many arguments about 𝔪𝔠 and 𝔰𝔠 are easier to carry out in terms of covering numbers for trees. First, we need to relate the two types of covering. Given s ∈ 2^{<ω}, let x_{s} = Σ_{i<|s|} \frac{s(i)}{2^{i}}, and let I_{s} = (x_{s^{⌢}⟨0⟩}, x_{s^{⌢}⟨1⟩}). Then, minus the countable dense set

D = \{x_{s} : s ∈ 2^{<ω}\} = \{\frac{m}{2^{n}} : m < 2^{n} ∈ ω\},

we can identify the unit interval with 2^{ω} by identifying each real a with the unique f ∈ 2^{ω} such that a ∈ I_{f↾n} for all n < ω, and thus [s] with I_{s}. The key point is the following.

Lemma 6.7. Let I be an interval contained in (0, 1) of width r. Let n be the largest integer such that 2^{−n} ≥ r. Then there exist s_{0}, s_{1} ∈ 2^{n} such that I ⊂ [s_{0}] ∪ [s_{1}].

Proof: Given I, let n_{0} < ω be least such that there exists d = \frac{m}{2^{n_{0}}} ∈ D ∩ I. Note that 2^{−n_{0}} ≥ r and that d is unique. Let s ∈ 2^{<ω} be such that d = x_{s}. If n_{0} = n, then we can let s_{0} = s_{1} = s. Otherwise, d splits I into two pieces. Let s_{0} be the extension of s^{⌢}⟨0⟩ of length n which takes value 1 on every integer in the extension of the domain, and let s_{1} be the extension of s^{⌢}⟨1⟩ of length n which takes value 0 on every integer in the extension of the domain. Then I ⊂ [s_{0}] ∪ [s_{1}].

6.8 Definition. Given a function g: ω → ω, say that A ⊂ 2^{ω} is a g-selection if A is of the form ⋂_{m<ω} ⋃{[s_{n}] : m < n < ω} where each s_{n} ∈ 2^{g(n)}. We let T Cov(g) be the least κ such that there is a set of g-selections of size κ with union 2^{ω}.

In this context, g is nontrivial if ∑_{n∈ω} \frac{g(n)}{2^{n}} is finite.

Lemma 6.9. For every g: ω → ω there is an f: ω → ℝ^{+} such that every gselection is coverable by two f-sets and every f-set is coverable by a g-selection.

For every f: ω → ℝ^{+} there is a g: ω → ω such that every f-set is coverable by two g-selections and every g-selection is coverable by two f-sets.

Proof: For the first part, we may assume that g is nondecreasing. Then we define f by letting f(n) = 2^{−g(2n+1)}. Then given any g-selection, for each n we can cover the (2n + 1)th interval by one of size f(n), and the (2n)th (for n > 0) interval by one of size f(n − 1). Since every point of the g-selection appears infinitely often in either the odd intevervals or the even ones, these two f-sets cover the given g-selection. That every f-set is coverable by a g-selection follows from Lemma 6.7.

For the second part, given f: ω → ℝ^{+}, define g by letting g(n) be the greatest m such that 2^{−m} ≥ f(n). Then every f-set can be covered by two g-selections by Lemma 6.7, and every g-selection can be covered by two f-sets since g(n) ≤ 2f(n). .

Corollary 6.10. Let Q ⊂ P ⊂ (0, 1). Then there is a f : ω → ℝ^{+} such that Q is f-coverable but P is not contained in a countable union of f-sets if and only if there is a g: ω → ω such that Q is contained in a g-selection but P is not contained in a countable union of f-selections.

This shows that the spectrum of values Cov(f) for nontrivial f: ω → ℝ^{+} is the same as the spectrum of values T Cov(g) for g: ω → ω. In particular, we have

Corollary 6.11.

𝔪𝔠 = inf{T Cov(g) | g ∈ ω^{ω} nontrivial}.
𝔰𝔠 = sup{T Cov(g) | g ∈ ω^{ω} nontrivial}.

It is a standard fact that every continuous map between metric compacta is uniformly continuous. This leads to the following observations.

Lemma 6.12. If P ⊂ 2^{ω} is a perfect set, then for any continuous 1-1, onto function g : 2^{ω} → P there is a nondecreasing function h : ℝ → ℝ such that for all f : ω → ℝ and A ⊂ P, if A is (h ∘ f)-coverable then g^{−1}[A] is f-coverable.

Corollary 6.13. If there exists a perfect set P such that for every f : ω → ω P is a union of ℵ_{1} many f-coverable sets, then 𝔰𝔠 = ℵ_{1}.

To see that Cov(N) = ℵ_{1} does not imply 𝔰𝔠 = ℵ_{1} we use the following characterization of Non(M).

Theorem 6.14. ([3], Theorem 2.4.7) Non(M) is equal to the least κ such that there is a set F ⊂ ω^{ω} of cardinality κ such that for every g ∈ ω^{ω} there is an f ∈ F such that f(n) = g(n) for infinitely many n.

Theorem 6.15. 𝔰𝔠 ≤ Non(M).

Proof: Applying Theorem 6.14, let F ⊂ ω^{ω} of cardinality Non(M) be such that for all g ∈ ω^{ω} there is an f ∈ F such that g and f agree on an infinite set. Let h : ω → ℚ^{+}, and let {A_{i} : i < ω} be a basis of intervals for ℝ. For f ∈ F, let X_{f} be the h-set defined by letting O_{n}^{f} be A_{f(n)} if the diameter of A_{f(n)} is less than h(n), and ∅ otherwise.

Given a real x, define a function g_{x} ∈ ω^{ω} by letting g_{x}(n) be the least m such that the diameter of A_{m} is less than h(n) and x ∈ A_{m}. Then letting f ∈ F agree with g_{x} on an infinite set, we have that x ∈ X_{f}.

Theorem 6.16. If 𝔡 < Non(M) then Non(M) = 𝔰𝔠.

Proof: By Theorem 6.15 we have that 𝔰𝔠 ≤ Non(M).

Let μ be the Lebesgue measure on ω^{ω}. Assume that 𝔡, 𝔰𝔠 < Non(M). Let M be an elementary substructure of a sufficiently large H(θ) of cardinality max{𝔡, 𝔰𝔠}. By Theorem 6.14 there is a function g ∈ ω^{ω} such that for all f ∈ ω^{ω} ∩ M, f(n) = g(n) for only finitely many n. By the elementarity and cardinality of M, there is an f ∈ ω^{ω} ∩ M such that for all n f(n) > g(n). For each n ∈ ω, let h(n) ∈ ℚ^{+} be smaller than

min{μ(σ^{∗}f, τ^{∗}0) | σ ≠ τ ∈ ω^{n} ∧ ∀i < n σ(i) ≤ τ(i) < f(i)},

where σ^{∗}f and τ^{∗}0 are the members of ω^{ω} extending σ and τ such that for i ≥ n, σ^{∗}f(i) = f(i) and τ^{∗}0(i) = 0. Then since |M| ≥ 𝔰𝔠, there is an h-set defined by a sequence of intervals ⟨O_{i} : i < ω⟩ in M such that g ∈ ⋂_{n<ω} ⋃_{i≥n} O_{i}. By the definition of h, though, each O_{i} can intersect at most one set of the form

{r ∈ ω^{ω} | r↾i = σ ∧ ∀j < ω r(j) < σ^{∗}f(j)}

for some σ ∈ ω^{i}. Then letting σ_{i} be the unique such σ ∈ ω^{i} if it exists, and letting t(i) = σ_{i+1}(i), we have that t and g agree on an infinite set, which is a contradiction since t ∈ M.

The following theorem is an improvement of a result in [4].

Theorem 6.17. ([9]) If GCH holds, then for any regular cardinal κ, there is a forcing which preserves cardinals and makes the following hold.

1. 𝔡 = Cov(N) = ℵ_{1}.

2. Non(M) = κ.

Theorems 6.16 and 6.17 give us the following.

Corollary 6.18. Cov(N) = ω_{1} does not imply 𝔰𝔠 = ℵ_{1}.

Theorem 6.19. If 𝔟 > Cov(N), then 𝔪𝔠 = Cov(N).

Theorem 6.19 follows from Lemmas 6.23 and 6.24 below. We will use the following definitions from [3], relating 𝔪𝔠 to another type of covering number for trees. Assume that H ∈ ω^{ω} is such that ∑_{n<ω} \frac{1}{H(n)} is finite - we will call such H nontrivial as well. Let

C_{H} = \{s ∈ ([ω]^{<ω})^{ω} | ∑_{n<ω} \frac{|s(n)|}{H(n)} < ∞\}

and

X_{H} = ∏_{n<ω} H(n) = {x ∈ ω^{ω} | ∀n < ω x(n) < H(n)}.

Then

Cov(C_{H}) = min{|A| | A ⊂ C_{H} ∧ ∀x ∈ X_{H}∃s ∈ A∃^{∞}n(x(n) ∈ s(n))}.

Note that we can identity X_{H} with the interval [0, 1] (minus countably many points) by first diving [0, 1] into H(0) many equal intervals, then dividing ech of these into H(1) many, and so on. As a consequence, Cov(C_{H}) ≥ Cov(N). Bartoszyński has shown the following.

Theorem 6.20. ([3], Theorem 2.5.12) If Cov(N) < 𝔟 then

Cov(N) = min{Cov(C_{H}) : H ∈ ω^{ω} nontrivial }.

6.21 Question. Is it consistent that Cov(N) < Cov(C_{H}) for all nontrivial H ∈ ω^{ω}?

6.22 Conjecture. It is consistent to have Cov(N) = 𝔡 = ℵ_{1} and 𝔪𝔠 = 𝔰𝔠 = ℵ_{2}.

The proof of the following is essentially the same as the proof of Theorem 2.5.12 in [3]

Lemma 6.23. For every nontrivial f ∈ (ℝ^{+})^{ω} there is a nontrivial H ∈ ω^{ω} such that Cov(C_{H}) ≤ Cov(f).

Lemma 6.24. Assume that 𝔟 > Cov(C_{H}). Then there is a nontrivial f ∈ (ℝ^{+})^{ω} such that Cov(C_{H}) ≥ Cov(f).

Proof: Let A ⊂ C_{H} witness 𝔟 > Cov(C_{H}), and canonically identify X_{H} with [0, 1]. For s ∈ A, let g_{s} ∈ ω^{ω} be such that g_{s}(k) is least such that

∑_{n≥g_{s}(k)} \frac{|s(n)|}{H(n)} < \frac{1}{2^{k}} .

Fix g such that for all s ∈ A {n | g_{s}(n) ≥ g(n)} is finite. For each k < ω, let

l_{k} = \frac{1}{∏_{i<g(k+1)} H(i)}

and let

t_{k} = \frac{∏_{i<g(k+1)} H(i)}{2^{k}}.

Now define f ∈ (ℚ^{+})^{ω} so that for each k, f takes the value l_{k} exactly t_{k} times. Then f is nontrivial. It remains to see that each of the sets represented by a member of A is f-coverable.

For each s ∈ A, for cofinitely many k ∈ ω,

(∗∗) ∑_{n=g(k)}^{g(k+1)−1} \frac{|s(n)|}{H(n)} < \frac{1}{2^{k}} .

For such s and k, we can think of ⋃_{n=g(k)}^{g(k+1)−1} s(n) as representing

∑_{n=g(k)}^{g(k+1)−1} (|s(n)| · ∏_{i∈(g(k+1)\{n})} H(i))

many pairwise disjoint (or identical) intervals, all of which have length l_{k}. By (∗∗), the collection of these intervals must have size less than t_{k}.

A set of reals X is strongly meager [2] if for each null set Y,

{x + y | ∃x ∈ X, y ∈ Y } ≠ ℝ.

We let SM denote the class of strongly meager sets. These sets are in fact meager [3].

Theorem 6.25. There exists a nontrivial f : ω → ℚ^{+} such that Cov(f) ≤ Non(SM).

Proof: Let X be not strongly meager, and Y a null set such that X +Y = ℝ. Then there is a nontrivial f ∈ (ℚ^{+})^{ω} such that Y is f-coverable. For any real x, x + Y = {x + y | y ∈ Y } is also f-coverable.

The following chart, where larger (≥, since if CH holds they are all equal) invariants point to smaller ones, summarizes the relationships between some of the invariants mentioned in this section.

[figure: Cov(N); 𝔪𝔠; 𝔰𝔠; Non(SM); Cov(SN); Non(M)]

6.26 Question. Is it consistent to have, with 𝔠 arbitrarily large, for each cardinal κ ≤ 𝔠 a nontrivial f such that Cov(f) = κ?

Although 𝔰𝔠 = ℵ_{1} follows from each of Non(M) = ℵ_{1} and Cov(SN) = ℵ_{1}, the following shows that it implies neither, as 𝔟 = ℵ_{2} implies Non(M) = ℵ_{2} and the Borel Conjecture implies that Cov(SN) = 𝔠.

Theorem 6.27. After adding ω_{2} many Mathias (or Laver) reals to a model of CH, the following hold.

1. 𝔟 = ℵ_{2}.

2. The Borel Conjecture.

3. 𝔰𝔠 = ℵ_{1},

Proof: The first two consequences are well known ([7], [22]). For the third, the key property is the following standard fact. Let M be an intermediate model in the iteration, and τ ∈ M a name in the rest of the iteration for a function from ω to ω. Let ⟨x_{n} : n ∈ ω⟩ ∈ M be a sequence of finite subsets of ω and p a condition forcing that each τ(n) will be in x_{n}. Then there exists q ≤ p and a sequence of finite sets ⟨y_{n} ∈ [x_{n}]^{2^{n^{2}}} : n < ω⟩ such that for all n q⊩τ(n) ∈ y_{n}. Now, given a name σ for a subset of ω, and a decreasing function g: ω → ℝ^{+} in an intermediate model M, for each n ∈ ω let m_{n} be such that

\frac{1}{2^{m_{n}}} < g (∑_{i=0}^{n} 2^{i^{2}}) ,

and let τ be a name for a function from ω to ω such that τ(n) is a code for σ↾m_{n}. Let x_{n} be the set of codes for the members of P(m_{n}). Then applying the key fact, we can shrink to a condition allowing just 2^{n^{2}} possibilities for each σ↾m_{n}. Listing the decoded versions of these possibilities, we get a description of a g-set which has the realization of σ as a member. Since CH holds in each intermediate model, then, there is an ω_{1}-sequence of g-sets covering the reals in the final model, and so 𝔰𝔠 = ℵ_{1} holds there.

Gödel’s proof shows that Axioms 1 and 3 together imply that Cov(SN) ≤ ℵ_{2}. By the theorem below, Cov(SN) ≤ ℵ_{2} does not follow from 𝔰𝔠 = ℵ_{1} plus 𝔡 = ℵ_{1}.

Theorem 6.28. Let κ ≥ ℵ_{2} have uncountable cofinality. After forcing to add κ many simultaneous Sacks reals to a model of CH with countable support, 𝔰𝔠 = 𝔡 = ℵ_{1} and Cov(SN) = κ.

Proof: We give a proof that Cov(SN) = κ. The other parts are standard. For A ⊂ κ, let 𝕊_{A} be the countable support product of the copies of Sacks forcing (𝕊) with index in A. Since 𝕊_{κ} is ω^{ω}-bounding, 𝔡 = ℵ_{1}, and so every strong measure zero set is contained in a G_{ℵ_{1}} strong measure zero set. So it suffices to consider such sets C. Every such C is coded in some extension via 𝕊_{A_{C}}, where A_{C} ⊂ κ, |A_{C}| ≤ ℵ_{1}. Hence it suffices to prove that if α ∉ A_{C}, then s_{α}, the Sacks real added by 𝕊_{{α}}, does not belong to C.

This, however, is easy. Working in the ground model, let p ∈ 𝕊_{κ} be any condition. Let p_{α} be its αth coordinate. Let f ∈ (ℝ^{+})^{ω} be a function such that whenever {I_{n}^{i} : n ∈ ω, i < n} is a set of intervals in 2^{ω} such that each I_{n}^{i} has length ≤ f(n), then [p_{α}] \ {y ∈ 2^{ω} | ∃^{∞}⟨n, i⟩(y ∈ I_{n}^{i})} contains a perfect set.

Let Ċ be the 𝕊_{A_{C}} -name for C. It is forced to be contained in an f-coverable G_{δ} set, say Ḋ. By the Sacks property of 𝕊_{A_{C}}, there are q ≤ p↾A_{C} and a set of intervals {I_{n}^{i} : n ∈ ω, i < n} such that I_{n}^{i} has length less than or equal f(n), and such that

q⊩Ḋ ⊂ {y ∈ 2^{ω} | ∃^{∞}⟨n, i⟩(y ∈ I_{n}^{i})}.

Now find q_{α} ≤ p_{α} such that [q_{α}] ∩ {y ∈ 2^{ω} | ∃^{∞}⟨n, i⟩(y ∈ I_{n}^{i})} = ∅. Then the condition r defined by • r_{α} = q_{α}

• r↾A_{C} = q

• r↾κ \ ({α} ∪ A_{C}) = p↾κ \ ({α} ∪ A_{C})

clearly forces that ṡ_{α} ∉ Ḋ. Hence r⊩ṡ_{α} ∉ Ċ, as required.

6.29 Question. After adding κ ≥ ω_{2} many simultaneous Sacks reals to a model of CH, does PSP(ℵ_{2}, G_{ℵ_{1}}) hold? Is every strong measure zero set of cardinality ℵ_{1} or less?

6.30 Remark. For many ω^{ω}-bounding forcings, every strong measure zero set has cardinality ℵ_{1} or less after an iteration of length ω_{2}. This holds for the ‘infinitely often equal reals’ forcing, but fails after the corresponding product forcing (see [3]).

6.31 Remark. It is also shown in [2] that after adding an arbitrary number of random reals to a model of CH, 𝔡 = ℵ_{1} but all strong measure zero sets have cardinality ℵ_{1} or less. In this model, 𝔰𝔠 > ℵ_{1}.

6.32 Remark. Adding uncountably many Cohen reals preserves Axioms 1 and 2 and forces 𝔰𝔠 = ℵ_{1}, so together they do not imply a bound on the continuum.

6.2 Axiom 3 and decompositions

Axiom 3 clearly follows from 𝔡 = 𝔰𝔠 = ℵ_{1}, but we would like to see whether it is in fact weaker. As we have defined it, Axiom 3 implies Cov(N) = ℵ_{1}. Theorem 6.36 below shows that for a certain class of perfect sets, if some P in this class can be written as a union of 𝔡 many sets Q ⊲ P, then Cov(N) ≤ 𝔡. Therefore, Cov(N) = ℵ_{1} is a consequence of any decomposition scheme along the lines of Gödel’s orignal proof.

6.33 Definition. A perfect set P is uniformly perfect if P is the set of paths through a tree T ⊂ 2^{<ω} such that on each level of T either all or none of the nodes split.

6.34 Remark. If T is the subset of 2^{<ω} whose members take the value 0 on every even member of their domains, then T represents a uniformly perfect set P of measure 0.

For a finite branching tree T, a subtree S of T has measure 0 in T if lim_{n→∞} \frac{|A_{n}∩S|}{2^{n}} = 0, where A_{n} is the set of nth splitting nodes of T. The idea behind this definition is that if a tree T ⊂ 2^{<ω} represents a perfect set P, then P has measure zero if and only if lim_{n→∞} \frac{|2^{n}∩T|}{2^{n}} = 0.

Lemma 6.35. Let Q ⊲ P be perfect sets, with P uniformly perfect. Then the image of Q under the canonical bijection between P and 2^{ω} has measure 0.

Proof: Citing Corollary 6.10, we work in terms of coverings of the trees representing Q and P. Let S, T ⊂ 2^{<ω} represent Q and P respectively, and let g ∈ ω^{ω} be such that some g-selection covers T, but S cannot be covered by inifinitely many g-selections. We want to see that lim_{n→∞} \frac{|2^{n}∩S|}{|2^{n}∩T|} = 0. Notice that the limit exists since the values are nonincreasing.

Say that there is some m such that for all n \frac{|2^{n}∩S|}{|2^{n}∩T|} ≥ 2^{−m}. For each n < ω, let A_{n} indicate the first level of T to have size 2^{n}. Fix t̄ ∈ A_{m}. We will show that for any set of finite sequences ⟨s_{i} : i < ω⟩ defining a g-selection covering S, there is a g-selection ⟨t_{i} : i < ω⟩ covering T ∩ [t̄], where [t] denotes the set of all extensions in T of the sequence [t].

Since P is compact, we can choose integers {n_{i}, j_{i} : i < ω} such that for all i

2^{n_{i}} ∩ S ⊂ ⋃{[s_{j}] : j_{i} ≤ j < j_{i+1}}.

For each i < ω we will pick t_{i} of the same length as s_{i}, in such a way that for all i < ω,

[t̄] ∩ 2^{n_{i}} ∩ T ⊂ ⋃{[t_{j}] : j_{i} ≤ j < j_{i+1}}.

We have the following by induction on n: if ⟨p_{i} : i < k⟩ is a set of finite sequences such that

∑_{i<k} |{t ∈ A_{n} | p_{i} ⊂ t}| ≥ 2^{r}

for some integer r ≤ n, then for any t^{∗} ∈ A_{n−r} there is a set ⟨q_{i} : i < k⟩ of sequences such that length(p_{i}) = length(q_{i}) for all i < k, and such that

{t ∈ A_{n} | ∃i < k q_{i} ⊂ t} = {t ∈ A_{n} | t^{∗} ⊂ t}.

For the induction step to n, note that either the number of p_{i}’s in A_{n} is even, or

∑_{i<k} |{t ∈ A_{n} | p_{i} ⊂ t}| ≥ 2^{r}

holds even if we remove one such p_{i}. Then by pairing off the p_{i}’s in A_{n−1} we can replace them with shorter sequences and apply the induction hypothesis.

So by the fact that \frac{|2^{n}∩S|}{|2^{n}∩T|} ≥ 2^{−m} for all n, we can choose the t_{i}’s as desired.

Lemmas 5.5, 6.12 and 6.35 give the following.

Theorem 6.36. (𝔡 = ℵ_{1}) If there exists a uniformly perfect set P contained in a set ⋃_{α<ω_{1}} Q_{α}, where each Q_{α} ⊲ P then Cov(N) = ℵ_{1}.

Proof: Say that P and Q_{α} (α < ω_{1}) are as in the statement of the theorem. Since each Q_{α} ⊲ P as witnessed by a G_{δ} set covering Q_{α}, we may assume that each Q_{α} is G_{δ}. By Lemma 5.5, each Q_{α} is a union of ω_{1}-many perfect sets, so by Lemmas 6.35 and 6.12 Cov(N) = ℵ_{1}.

We would like to know whether the condition in Lemma 6.35 that P is uniformly perfect is necessary.

Certain strenghtenings of Axiom 3 do imply that 𝔪𝔠 = ℵ_{1}.

Theorem 6.37. If the set of functions H in Axiom 3 is required to be linearly ordered under mod-finite domination, then this strengthened version of the axiom implies 𝔪𝔠 = ℵ_{1}.

Proof: Fix H as in the statement of Axiom 3. If 𝔪𝔠 > ℵ_{1}, then there is a sequence ⟨(A_{i}, f_{i}) ∈ P(ℝ) × H : i < ω⟩ such that each A_{i} is f_{i}-coverable but not f_{j}-coverable for any j < i. To construct such a sequence, using Theorem 6.36, let ⟨(B_{α}, g_{α}) ∈ P(ℝ) × H : α < ω_{1}⟩ be such that each g_{α} is nontrivial, each B_{α} is a g_{α}-set and ⋃_{α<ω_{1}} B_{α} = ℝ. Let A_{0} = B_{0} and f_{0} = g_{0}. Then given (A_{j}, f_{j}) ∈ P(ℝ) (j < i), define h: ω → ℝ^{+} by letting h(ai+b) = f_{b}(a) whenever a, b ∈ ω with b < i. Using 𝔪𝔠 > ℵ_{1}, some B_{α} is not h-coverable, so we can let A_{i} = B_{α} and f_{i} = g_{α}.

Now by Lemma 5.5, by shrinking the A_{i}’s if necessary, we can assume that they are all compact, and that there are disjoint intervals I_{i} (i ∈ ω) such that A_{i} ⊂ I_{i}.

Let D = ⋃_{i<ω} A_{i}. Now if H is linearly ordered by mod-finite domination, then D is h-coverable for all h ∈ H not dominated mod-finite by some f_{i}. But if D can be written as an increasing union of ω_{1} many sets which are each f_{i} for some integer i, then there is a fixed integer i such that D can be covered by ω_{1}-many f_{i}-coverable sets. But D was constructed to make this impossible.

There are many other questions one could ask in this area, especially : does Axiom 3 imply 𝔰𝔠 = ω_{1}?

7 Appendix : Chart of Models

Model:  1  2  3  4  5  6  7
Axiom 1  t  t  t  t  t  t  F
Axiom 2  t  T  F  F  T  T  ?
Axiom 3  t  F  F  T  F  T  T
PSP(ℵ_{1}, G_{ℵ_{1}})  F  T  f  F  T  F  F
PSP(ℵ_{2}, G_{ℵ_{1}})  t  T  F  F  T  T  ?
𝔡 = ℵ_{1}  t  F  T  T  F  T  T
Cov(SN) = ℵ_{1}  t  T  F  T  F  F  F
𝔰𝔠 = ℵ_{1}  t  T  F  T  T  T  T
SN ⊂ [ℝ]^{ℵ_{1}}  t  F  T  F  T  T  T?

T and F correspond to true and false, t and f to trivially true and trivially false. Question marks indicate open questions or, if accompanied by T or F, conjectures. The models listed are as follows, where each forcing is conducted over a model of GCH, and ‘many’ means ≥ ℵ_{3}, so models 2, 3, 4 and 7 do not satisfy 𝔠 ≤ ℵ_{2}.

1. Ground model.

2. Adding many Cohen reals.

3. Adding many random reals.

4. Adding many reals by c.c.c. forcing, followed by ω_{1} Hechler reals.

5. Adding ω_{2} Mathias reals.

6. Adding ω_{2} Sacks reals.

7. Adding many Sacks reals simultaneously.

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Graduate School of Science and Technology Kobe University Kobe 657-8501 Japan

brendle@pascal.seg.kobe-u.ac.jp Department of Mathematics and Statistics Miami University Oxford, OH 45056 USA

larsonpb@muohio.edu

Université Paris VII - C.N.R.S. UPRESA 7056 Paris 75251, Cedex 05 France

stevo@math.jussieu.fr