The cofinality of the strong measure zero ideal for κ inaccessible Johannes Philipp Schürz ∗ Abstract We investigate the cofinality of the strong measure zero ideal for κ inaccessible, and show that it is independent of the size of 2κ. 1 Introduction In this paper we continue to investigate the strong measure zero sets on 2κ for κ at least inaccessible (see [Sch19]). Halko generalized the notion of strong measure zero to uncountable cardinals as fol- lows (see [Hal96]): Definition 1.1. Let X ⊂ 2κ. We say X has strong measure zero iff κ f(i) [ 8f 2 κ 9(ηi)i<κ : 8i < κ ηi 2 2 ^ X ⊂ [ηi]: i<κ We shall denote this by X 2 SN . The following is an easy fact: Fact 1.2. SN is a κ-closed, proper ideal on 2κ which contains all singletons. Therefore, the following cardinal characteristics are well defined: Definition 1.3. [ add(SN ) := minfjFj: F ⊂ SN ^ F 2= SN g arXiv:2012.08576v1 [math.LO] 15 Dec 2020 [ cov(SN ) := minfjFj: F ⊂ SN ^ F = 2κg non(SN ) := minfjXj: X ⊂ 2κ ^ X2 = SN g cof(SN ) := minfjFj: F ⊂ SN ^ 8X 2 SN 9Y 2 F X ⊂ Y g ∗supported by FWF project I3081 1 In [Yor02], Yorioka introduced the so-called Yorioka ideals approximating the ideal of strong measure zero sets on 2!. We will generalize this notion to κ and use it to investigate cof(SN ). Eventually, we shall show that cof(SN ) < cκ, cof(SN ) = cκ as well as cof(SN ) > cκ are all consistent relative to ZFC. In the last chapter we first generalize the Galvin–Mycielski–Solovay theorem (see [BJ95]) to κ inaccessible. This result was originally proven by Wohofsky (see [Woh]). Then we follow Pawlikowski [Paw90] and show the relative consistency of cov(SN ) < add(Mκ) for κ strongly unfoldable (see Definition 4.3). Strong measure zero sets for κ regular uncountable have also been studied in [Hal96] and [HS01]. More questions have been stated in [KLLS16]. Last but not least, I would like to thank my advisor Martin Goldstern for the very helpful comments during the preparation of this paper. 2 Prerequisites We start with several definitions: Definition 2.1. Let f; g 2 κκ and f strictly increasing. • We define the order on κκ as follows: f g iff 8α < κ 9β < κ 8i ≥ β : g(i) ≥ f(iα). Here iα is defined using ordinal arithmetic. <κ κ • For σ 2 (2 ) define gσ as follows: gσ(i) := dom(σ(i)). <κ κ κ T S • For σ 2 (2 ) define Y (σ) ⊆ 2 as follows: Y (σ) := i<κ j≥i [σ(j)]. <κ κ <κ κ • Define S(f) ⊆ (2 ) as follows: S(f) := fσ 2 (2 ) : f gσg. • Define A ⊆ 2κ to be f-small iff there exists σ 2 S(f) such that A ⊆ Y (σ). • Set I(f) := fA ⊆ 2κ : A is f-smallg. Definition 2.2. Let f 2 κκ be strictly increasing and let σ 2 S(f). For every α < κ α α α we define Mσ to be the minimal ordinal ≥ 2 such that 8i ≥ Mσ : gσ(i) ≥ f(i ). Lemma 2.3. I(f) forms a κ-closed ideal. Proof. I(f) is obviously closed under subsets. So we must show that it is also closed under κ-unions. Let (Ak)k<κ be a family of f-small sets and (σk)k<κ such that Ak ⊆ Y (σk). We shall S find a τ 2 S(f) such that k<κ Y (σk) ⊆ Y (τ). For every k < κ let gk := gσk and let M α := M α for ever α < κ. Define m := supfM 3·α : j; α ≤ kg. k σk k j We need the following definitions for i ≥ m0: 2 P P • Define c(i) > 0 such that m0 + j<c(i) j · mj ≤ i < m0 + j≤c(i) j · mj. P • Define d(i) := m0 + j<c(i) j · mj. • Define a(i) and b(i) such that i − d(i) = c(i) · a(i) + b(i) with b(i) < c(i). P • Define e(i) := j<c(i) mj. The following are immediate consequences for i > m0: P 0 • i ≥ m0 + j<c0 j · mj ) c(i) ≥ c • a(i) < mc(i) • 0 ≤ b(i) < c(i) ≤ e(i). • d(i) ≤ c(i) · e(i) (show by induction) • i = d(i) + c(i) · a(i) + b(i) < c(i) · e(i) + c(i) · a(i) + c(i) = c(i) · (e(i) + a(i) + 1) ≤ ((e(i) + a(i)) · (e(i) + a(i) + 1) ≤ ((e(i) + a(i))3 • 8k < κ 81l < κ 9i < κ: e(i) + a(i) = l ^ b(i) = k P The last statement can be deduced as follows: Given k < κ choose l ≥ mj, so P P P j≤k there is a c~ > k such that mj ≤ l < mj. Define i := m0 + j · mj +c ~ · P j<c~ j≤c~ P j<c~ (l − mj) + k. Then c(i) =c ~. (This follows because mc~ ≥ (l − mj) + 1 and j<c~ P P j<c~ P so c~ · mc~ > c~ · (l − mj) + k.) Hence d(i) = m0 + j · mj, a(i) = l − mj, Pj<c~ j<c~ j<c~ b(i) = k and e(i) = j<c~ mj. Hence e(i) + a(i) = l. Figure 1: Definition of τ Now we can define τ: If i ≥ m0 set τ(i) = σb(i)(e(i) + a(i)). Else set τ(i) = h i. We must show that τ has the required properties. 3 S First let us show that k<κ Y (σk) ⊆ Y (τ). Let x 2 Y (σk) for some k < κ. If l < κ is large enough, then there exists i < κ such that τ(i) = σk(l). Hence x 2 Y (τ). P Now we must show that τ 2 S(f). Let α < κ be arbitrary. If i ≥ m0 + j≤α j·mj (hence c(i) > α), then the following (in-)equalities hold: τ(i) = σb(i)(e(i) + a(i)) by definition, 3·α 3·α gb(i)(e(i) + a(i)) ≥ f((e(i) + a(i)) ) since e(i) + a(i) ≥ Mb(i) and b(i); α < c(i), and f((e(i) + a(i))3·α) ≥ f(iα) since f is strictly increasing and i ≤ (e(i) + a(i))3. Hence α gτ (i) ≥ f(i ). The following fact is straightforward: T Fact 2.4. SN = f2κκ I(f). The next lemma implicitly shows that comeager sets cannot be strong measure zero: Lemma 2.5. Let A ⊆ 2κ be comeager. Then there exists f 2 κκ such that for every f-small set B the set A n B is non-empty. Proof. Let A be comeager. We shall show that A contains a perfect set, hence there exists f 2 κκ such that A is not f-small: Let P ⊂ 2κ be a perfect set, and let T ⊂ 2<κ be a perfect tree such that [T ] = P . Pick a function f 2 κκ such that f(α) > supfdom(t): t is in the αth splitting level of T g. Let σ 2 F(f) be arbitrary. Then an x 2 [T ] n Y (σ) can be constructed by induction. T Now assume that A = i<κ Di+1 where Di+1 are open dense and decreasing. We induc- tively construct a perfect tree T ⊂ 2<κ such that no branches die out and [T ] ⊂ A: • Set T0 := fth ig where th i := h i. 0 i0 • If i = i + 1 assume inductively that Ti0 = ftη : η 2 2 g has already been defined T 0 and for every tη 2 Ti0 it holds that [tη] ⊂ j<i0 Dj+1. For every tη 2 Ti0 find tη . tη 0 0_ 0 i such that [tη] ⊂ Di. Set tη_hii := tη hii and Ti := ftη0 : η 2 2 g. γ S γ • If γ is a limit and η 2 2 define tη := 0 tη0 and set Tγ := ftη : η 2 2 g. Then T η / η we can deduce that [tη] ⊂ j<γ Dj+1 for every tη 2 Tγ. S It follows from the construction that (the downward closure of) T := i<κ Ti is a perfect tree and that [T ] ⊂ A. Conversely, the following fact holds true: Fact 2.6. For every f 2 κ strictly increasing there exists a comeager A ⊂ 2κ such that A 2 I(f). κ + Lemma 2.7. Assume CH, i.e. j2 j = κ , and let (fα)α<κ+ be a κ-scale such that fα is β β<κ+ strictly increasing. Then there exists a matrix (Aα)α<κ+ with the following properties: + β κ • 8α; β < κ : Aα ⊆ 2 is comeager and fα-small. 4 0 + 0 β β0 • 8α; β; β < κ : β ≤ β ) Aα ⊆ Aα . + κ + β • 8α < κ 8fα-small B ⊆ 2 9β < κ : B ⊆ Aα. + κ T 0 • 8α < κ 8fα-small B ⊆ 2 : α > 0 ) γ<α Aγ n B 6= ;. This means that for + T 0 every α < κ the set γ<α Aγ is not fα-small. β + + Proof. We shall construct Aα by a lexicographic induction on (α; β) 2 κ × κ : Assume β β<κ+ T 0 that (Aγ )γ<α have already been defined. Since γ<α Aγ is comeager, there exists f such T 0 that γ<α Aγ is not f-small by Lemma 2.5. W.l.o.g. let f = fα. Choose some τ0 2 S(fα) 0 such that Y (τ0) is comeager and set Aα := Y (τ0).