A Nonhereditary Borel-cover γ-set

Arnold W. Miller^{1}

Abstract

In this paper we prove that if there is a Borel-cover γ-set of cardinality the continuum, then there is one which is not hereditary.

In this paper we answer some of the questions raised by Bartoszyński and Tsaban [1] concerning hereditary properties of sets defined by certain Borel covering properties.

Define. An ω-cover of a set X is a family of sets such that every finite subset of X is included in an element of the cover but X itself is not in the family.

Define. A γ-cover of a set X is an infinite family of sets such that every element of X is in all but finitely many elements of the family.

Define. A set X is called a Borel-cover γ-set iff every countable ω-cover of X by Borel sets contains a γ-cover.

These concepts were introduced by Gerlits and Nagy [5] for open covers. Being a Borel-cover γ-set is equivalent to saying that for any ω-sequence

of countable Borel ω-covers of X we can choose one element from each and get a γ-cover of X – this is denoted S_{1}(B_{Ω}, B_{Γ}). The equivalence was proved by Gerlitz and Nagy [5] for open covers but the proof works also for Borel covers as was noted in Scheepers and Tsaban [8]:

Let B_{n} be Borel ω-covers of X. Since

{U ∩ V : U ∈ U, V ∈ V}

is an ω-cover if U and V are, we may assume that B_{n+1} refines B_{n}. Let x_{n} for n < ω be distinct elements of X and let

B = {A \ {x_{n}} : n < ω, A ∈ B_{n}}

It is easy to check that B is an ω-cover of X. Now let C be a γ-subcover of B. Note that for any fixed n at most finitely many of the elements of C can be of the form A \ {x_{n}}. By refining C we may assume at most one thing is taken from each B_{n} and since they are refining we can fatten up C to take exactly one element of each B_{n}.

1Thanks to the Fields Institute for Research in Mathematical Sciences at the University of Toronto for their support during the time this paper was written and to Juris Steprāns who directed the special program in set theory and analysis.

Mathematics Subject Classification 2000: 03E50; 03E17

Define. A family of subsets of X, U is a τ cover of X iff every element of X is in infinitely many elements of U and for every x, y ∈ X at least one of the sets

{U ∈ U : x ∈ U, y ∉ U} or {U ∈ U : x ∉ U, y ∈ U}

is finite.

Clearly any γ-cover is a τ-cover. These covers were introduced in Tsaban [9].

Theorem 1 Suppose there is a Borel-cover γ-set of size the continuum. Then there is a Borel-cover γ-set X and subset Y of X which is not a Borelcover γ-set. In fact, there is an open ω-cover of Y with no τ-subcover.

Proof For X ⊂ P (ω) let

X̃ = {ω \ a : a ∈ X}

be the dual of X, i.e., the set of complements of elements of X. Let P ⊆ [ω]^{ω} be a perfect set of independent subsets of ω. This means that for every disjoint pair F_{1}, F_{2} of finite subsets of P the set

⋂ F_{1} ∩ ⋂ F̃_{2} is infinite.

Such a set was first constructed by Fichtenholtz, Kantorovich, and Hausdorff, see Kunen [6]. To construct one, let Q = {(n, s) : n ∈ ω, s ⊆ P (n)}. Define A_{x} = {(n, s) : x ∩ n ∈ s} for each x ⊆ ω and

P = {A_{x} : x ⊆ ω} ⊆ P (Q) = 2^{Q}.

Let Z ⊆ P be a Borel-cover γ-set of cardinality the continuum.

Claim. Z ∪ Z̃ is a Borel-cover γ-set.

proof: Let {B_{n} : n < ω} be a Borel ω-cover of Z∪Z̃. Then it is easy to see that {(B_{n} ∩ B̃_{n}) : n < ω} is an ω-cover of Z. This is because if (F ∪ F̃) ⊆ B_{n} then F ⊆ (B_{n} ∩ B̃_{n}).

Since Z is a Borel-cover γ-set there exists an a ∈ [ω]^{ω} such that

{(B_{n} ∩ B̃_{n}) : n ∈ a}

is a γ-cover of Z. But then it is also a γ-cover of Z̃. This proves the Claim.

Let X = Z ∪ Z̃ and to pick Y ⊆ X as required we will choose for each a ∈ Z to put either a ∈ Y or (ω \ a) ∈ Y (but not both). Since Z was a subset of P and P was independent we will have that the intersection of any finite subset of Y is infinite. In particular,

U = {U_{n} : n ∈ ω} where U_{n} = {a ⊆ ω : n ∈ a}

is an ω-cover of Y. But {U_{n} : n ∈ b} is a γ-cover of Y iff b ⊆^{∗} a for every a ∈ Y. But this is easy to defeat. Using that Z has cardinality the continuum let Z = {a_{α} : α < 𝔠} and let [ω]^{ω} = {b_{α} : α < 𝔠}. For each α if b_{α} ⊆^{∗} a_{α} put (ω \ a_{α}) into Y and otherwise put a_{α} into Y.

To construct Y so that U has no τ-subcovers can be done by using two elements a_{0}, a_{1} of Z for each b ∈ [ω]^{ω}. First note that the set {U_{n} : n ∈ b} is a τ-cover of Y iff b meets every element of Y in an infinite set and for every two elements a_{0}, a_{1} of Y either (a_{0} ∩ b) ⊆^{∗} a_{1} or (a_{1} ∩ b) ⊆^{∗} a_{0}.

Notation: a^{(0)} = a and a^{(1)} = ω \ a.

Claim. There exists i, j in {0, 1} such that (a) b ∩ a_{i}^{(j)} is finite or

(b) both b ∩ a_{0}^{(i)} ∩ a_{1}^{(1−j)} and b ∩ a_{0}^{(1−i)} ∩ a_{1}^{(j)} are infinite.

proof: Assume case (a) fails for all i, j in {0, 1}. The four sets a_{0}^{(i)} ∩ a_{1}^{(j)} partition ω into infinite sets since a_{0} and a_{1} are independent. If all four meet b in an infinite set then we are done. So assume that b ∩ a_{0}^{(i)} ∩ a_{1}^{(j)} is finite for some i, j. But since b ∩ a_{0}^{(i)} is infinite it must be that b ∩ a_{0}^{(i)} ∩ a_{1}^{(1−j)} is infinite. A similar argument shows b ∩ a_{0}^{(1−i)} ∩ a_{1}^{(j)} is infinite. This proves the Claim.

To kill off the possibility of b giving a τ-subcover we put a_{i}^{(j)} into Y in case (a) or put both a_{0}^{(i)}, a_{1}^{(j)} into Y in case (b). This proves the Theorem. QED

Remark. Tsaban points out the following corollary of our result. In Problem 7.9 of Bukovský, Recław, and Repický [3] it is asked whether every γset of reals which is also a σ-set is a hereditary γ-set. It is shown in Scheepers and Tsaban [8] that every Borel-cover γ-set (more generally S_{1}(B_{Γ}, B_{Γ})-set) is a σ-set. Hence the answer to the problem is no.

The following result is due to Brendle [2]. Our proof is a slight modification of a result of Todorčević – see Theorem 4.1 of Galvin and Miller [4] and is perhaps simpler.

Theorem 2 (Brendle) Assume CH. Then there exists a Borel-cover γ-set of size ω_{1}.

Proof The idea is to construct an Aronszajn tree T ⊆ ω^{<ω_{1}} labeled by perfect sets.

We construct perfect subtrees p_{s} ⊆ 2^{<ω} for s ∈ T and X_{s} ⊆ [p_{s}] countable dense sets such that

1. if s ⊆ t then p_{s} ⊇ p_{t},

2. if s and t are incomparable then [p_{s}] ∩ [p_{t}] = ∅,

3. if α < β < ω_{1} and n < ω then for every s ∈ T_{α} there exists t ∈ T_{β} with s ⊆ t and p_{s} ∩ 2^{n} = p_{t} ∩ 2^{n}, and

4. for every sequence (B_{n} : n < ω) of Borel subsets of 2^{ω} there exists α < ω_{1} such that either for some finite F ⊆ ∪{X_{s} : s ∈ T_{≤α+1}} no B_{n} covers F or there exists an a ∈ [ω]^{ω} such that {B_{n} : n ∈ a} is a γ-cover of

∪{[p_{s}] : s ∈ T_{≤α+1}} ∪ ⋃{X_{s} : s ∈ T_{≤α}}

After the construction is completed we will let X = ∪{X_{s} : s ∈ T }. The last item guarantees that X will be a Borel-cover γ-set. The first three items are simply to guarantee that our construction can continue at limit levels. To do the last item we use the following Lemma.

Define for any perfect tree p and s ∈ p,

p⟨s⟩ = {t ∈ p : t ⊆ s or s ⊆ t}

Lemma 3 Suppose ⟨p_{n} : n < ω⟩ are perfect trees and (B_{n} : n < ω) is a sequence of countable Borel ω-covers of ∪_{n<ω}[p_{n}]. Then there exists perfect pairwise disjoint subtrees q_{n} ⊆ p_{n} and {B_{n} ∈ B_{n} : n < ω} which is a γ-cover of ∪_{n<ω}[q_{n}].

Proof We can begin by refining the p_{n}’s so that [p_{n}] are pairwise disjoint. So we may as well assume this to begin with. Also since Borel sets have the (relative) property of Baire with respect to each perfect set, by passing to perfect subsets we may assume that each of our Borel covers is an open cover.

Note that for finite sequences (k_{i} : i < n) and (t_{i} ∈ p_{k_{i}} : i < n), there exists a U_{n} ∈ B_{n} and r_{i} ⊇ t_{i} with r_{i} ∈ p_{k_{i}} such that

∪_{i<n}[p_{k_{i}}⟨r_{i}⟩] ⊆ U_{n}

Using this observation it is easy to construct a fusion sequence which produces the q_{n} and the required γ-cover. This proves the Lemma. QED

Let T_{α} = T ∩ ω^{α}.

In our construction of the tree we start by assuming that {B_{α} : α < ω_{1}} is a list containing all countable families of Borel subsets of 2^{ω}. At limit ordinals α, we use the usual fusion arguments to produce p_{s} for s ∈ T_{α}. We take care of condition (4) as follows.

Suppose by induction we have already constructed: (p_{s} : s ∈ T_{α}) and (X_{s} : s ∈ T_{<α}).

To obtain condition (4) let {x_{n} : n < ω} = ∪{X_{s} : s ∈ T_{<α}}, and define

B_{α}^{n} = {B ∈ B_{α} : {x_{i} : i < n} ⊆ B}

If some B_{α}^{n} is not an ω-cover of ∪{[p_{s}] : s ∈ T_{α}}, then there exists a finite subset of

{x_{n} : n < ω} ∪ ⋃{[p_{s}] : s ∈ T_{α}}

which is not covered by any B ∈ B_{α}. In this case, we choose X_{s} ⊆ p_{s} so that this finite set is included in ∪{X_{s} : s ∈ T_{≤α}}. We then choose p_{sn} so that p_{sn} ∩ 2^{n} = p_{s} ∩ 2^{n}, p_{sn} ⊆ p_{s}, and [p_{sn}] for n < ω are pairwise disjoint. We don’t need to worry about B_{α} because it cannot be an ω-cover of X.

So we may assume each B_{α}^{n} is an ω-cover of ∪{[p_{s}] : s ∈ T_{α}}. Apply the Lemma to the sequence

(p_{s}⟨t⟩ : s ∈ T_{α} and t ∈ p_{s})

Then for each s ∈ T_{α} and n < ω let

p_{sn} = ∪{q_{s,t} : t ∈ 2^{n} ∩ p_{s}}

In this case we can take each X_{sn} ⊆ [p_{sn}] to be an arbitrary countable dense subset. This proves Theorem 2. QED

Theorem 4 is probably known but we include its proof here for completeness.

Theorem 4 Suppose that M is any countable standard model of ZFC. Then there exists a ccc poset ℙ in M of size continuum such if G is any ℙ-filter generic over M, then X = M ∩ 2^{ω} in M is a Borel-cover γ-set in M[G]. Note that forcing with ℙ does not change the size of the continuum in M[G].

Proof

This is really a corollary of result noted by Gerlitz and Nagy [5] that assuming MA (or even just MA(σ-centered)) that every set X of size less than continuum is a γ-set.

Let {B_{n} : n < ω} be an ω-cover of X. For each x ∈ X let

a_{x} = {n : x ∈ B_{n}}.

The family {a_{x} : x ∈ X} has the finite intersection property. So there is a well-known ccc poset of size |X| (see Kunen [7]) which adds an infinite a ∈ [ω]^{ω} such that a ⊆^{∗} a_{x} for each x ∈ X. Then {B_{n} : n ∈ a} is a γ-cover of X. To obtain the model M[G] simply iterate continuum many times, with the usual dovetailing argument to take care of all sequences of Borel sets in M[G]. QED

Question 5 Does MA imply there exists a Borel-cover γ-set of size the continuum?

The theorems in this section show that it is consistent that the classes S_{1}(B_{Ω}, B_{Γ}), S_{1}(B_{Ω}, B_{T}), and S_{fin}(B_{Ω}, B_{T}) are not hereditary.

I don’t know about the other classes in Bartoszyński and Tsaban [1], for example:

Question 6 Is the class S_{1}(B_{Ω}, B_{Ω}) hereditary?

For the definitions of these classes see [1].

References

[1] Bartoszyński, Tomek; Tsaban, Boaz; Hereditary topological diagonalizations and the Menger-Hurewicz conjectures, Proceedings of the AMS, to appear. http://arXiv.org/abs/math/0208224

[2] Brendle, Jörg; Generic constructions of small sets of reals. Topology Appl. 71 (1996), no. 2, 125–147.

[3] Bukovský, Lev; Recław, Ireneusz; Repický, Miroslav; Spaces not distinguishing convergences of real-valued functions. Topology Appl. 112 (2001), no. 1, 13–40.

[4] Galvin, Fred; Miller, Arnold W.; γ-sets and other singular sets of real numbers. Topology Appl. 17 (1984), no. 2, 145–155.

[5] Gerlits, J.; Nagy, Zs.; Some properties of C(X). I. Topology Appl. 14 (1982), no. 2, 151–161.

[6] Kunen, Kenneth; Ultrafilters and independent sets. Trans. Amer. Math. Soc. 172 (1972), 299–306.

[7] Kunen, Kenneth; Set theory. An introduction to independence proofs. Studies in Logic and the Foundations of Mathematics, 102. North-Holland Publishing Co., Amsterdam-New York, 1980.

[8] Scheepers, Marion; Tsaban, Boaz; The combinatorics of Borel covers. Topology Appl. 121 (2002), no. 3, 357–382.

[9] Tsaban, Boaz; A topological interpretation of 𝔱. Real Anal. Exchange 24 (1998/99), no. 1, 391–404.

Arnold W. Miller miller@math.wisc.edu http://www.math.wisc.edu/∼miller University of Wisconsin-Madison Department of Mathematics, Van Vleck Hall 480 Lincoln Drive Madison, Wisconsin 53706-1388