A MAD Q-set

Arnold W. Miller^{1}

Abstract

A MAD (maximal almost disjoint) family is an infinite subset A of the infinite subsets of ω = {0, 1, 2, …} such that any two elements of A intersect in a finite set and every infinite subset of ω meets some element of A in an infinite set. A Q-set is an uncountable set of reals such that every subset is a relative G_{δ} set. It is shown that it is relatively consistent with ZFC that there exists a MAD family which is also a Q-set in the topology in inherits a subset of P (ω) = 2^{ω}.

In this paper we answer a question of Hrušák by showing that it is consistent that there exists a maximal almost disjoint family A ⊆ [ω]^{ω} which is also a Q-set. The reference Hrušák [6] contains some related problems. A topological space is a Q-set iff every subset is a G_{δ}-set. His reason for asking this question was because in a certain argument involving a topological space Ψ(A) built from a MAD family it would have been helpful to assume that a MAD family cannot be a Q-set. Szeptycki [11] contains some results on van Douwen’s Ψ and also on Q-sets.

Our construction is similar to that in Fleissner and Miller [3] where a Qset is obtained which is concentrated on the rationals. In Judah and Shelah [7] it is shown consistent to have a Q-set while at the same time 𝔟 = 𝔡 = ω_{1}. Their Q-set forcing has the Sack’s property. Their forcing is also used in Nowik and Weiss [10] to construct a Q-set with certain properties and also Gruenhage and Koszmider [4] to construct a topological space with certain properties. In our model as in [3] we have that 𝔡 = 𝔠 = ω_{2} and 𝔟 = ω_{1}.

In Dow [2] and Brendle [1] a type of Q-set forcing is used which preserves towers (so 𝔭 = ω_{1}) which generalizes Hechler dominating real forcing, and 𝔟 = 𝔡 = 𝔠.

Theorem 1 It is relative consistent with ZFC, that there exists a MAD family A ⊆ [ω]^{ω} which is also a Q-set.

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 Steprans who directed the special program in set theory and analysis.

Mathematics Subject Classification 2000: 03E35

Fundamenta Mathematicae, 178(2003), 271-281.

Proof

We begin by forcing a generic MAD family and then we iterate our Q-set forcing to make the generic MAD into a Q-set. The difficulty is to ensure the family stays maximal.

Define: Let ℙ be the usual poset for forcing a MAD family: (p, q) ∈ ℙ iff

1. p : F → 2^{N} for some finite F ⊆ ω_{1} and N < ω (write F = dom(p) and N = N_{p})

2. q a partial function from a subset of [F]^{2} into N

3. if q(α, β) = n, then for every i with n ≤ i < N either p(α)(i) = 0 or p(β)(i) = 0.

The uniformity N of lengths in condition (2) is not strictly necessary but it will be convenient and would occur on a dense set anyway.

Define: (p_{1}, q_{1}) ≤ (p_{2}, q_{2}) iff

1. dom(p_{1}) ⊇ dom(p_{2}),

2. p_{1}(α) ⊇ p_{2}(α) for all α ∈ dom(p_{2}), and

3. q_{1} ⊇ q_{2}.

Intuitively, we are describing a family {a_{α} ⊆ ω : α < ω_{1}} as follows:

1. p(α) = s means (i ∈ a_{α} iff s(i) = 1) for i < |s|

2. q(α, β) = n promises that a_{α} ∩ a_{β} ⊆ n.

Note that (p_{1}, q_{1}) and (p_{2}, q_{2}) are compatible iff there exists p_{3} ≤ p_{1}, p_{2} such that (p_{3}, q_{1} ∪ q_{2}) is in ℙ.

This forcing is due to Hechler [5]. For G ℙ-generic over M define

x_{α}^{G} = ⋃{p(α) : ∃q (p, q) ∈ G}

Let X = {x_{α}^{G} : α < ω_{1}} and let A = {a_{α} ⊆ ω : α < ω_{1}} where each x_{α} is the characteristic function of a_{α}, i.e.

a_{α} = {n : x_{α}(n) = 1}.

The following lemma is due to Hechler.

Lemma 2 ℙ is ccc. If G is ℙ-generic over M, then in M[G] the set A is a maximal almost disjoint family of infinite subsets of ω.

We will in a sense need to reprove this lemma since we will show that after our new version of Q-set forcing our generic family still remains a maximal almost disjoint family. The idea of the argument is that given a name τ for some infinite subset of ω, we find an α which is not involved with deciding n ∈ τ for any n. Then we get a contradiction by swapping the value of x_{α}(n) = 0 to x′_{α}(n) = 1 while still forcing n ∈ τ. In the usual Q-set forcing while the condition forcing n ∈ τ doesn’t directly talk about x_{α}, it may decide that [s] ⊆ U_{n} where the other condition says x_{α} ∉ U_{n}. These conditions may become inconsistent when we change to x′_{α} because it might be that s ⊆ x′_{α} even though s is not a subset of x_{α}.

A new Q-set forcing

The following is to motivate our definition of ℙ ∗ ℚ̃. It would be the definition of the new Q-set forcing in the model M[G] where G is ℙ-generic.

Define. For x ∈ 2^{ω},s ∈ 2^{<ω}, k < ω

swap(x, s, k) = {y ∈ 2^{ω} : s ⊆ y, |{i ≥ |s| : y(i) ≠ x(i)}| ≤ k}.

Note that swap(x, s, k) is a countable closed subset of [s]. It contains x if s ⊆ x. Also swap(x, ⟨⟩, 0) = {x}.

Suppose we are given X ⊆ 2^{ω} such that for all x ≠ y ∈ X there are infinitely many n with x(n) ≠ y(n). For Y ⊆ X define ℚ(X, Y) as follows:

r ∈ ℚ(X, Y) iff r is a finite subset of

{(n, s) : n < ω, s ∈ 2^{<ω}} ∪ {(n, (x, t, k)) : x ∈ Y, t ∈ 2^{<ω}, n, k < ω}

subject to the condition:

if (n, s) ∈ r and (n, (x, t, k)) ∈ r, then [s] ∩ swap(x, t, k) = ∅.

The ordering is by inclusion r_{1} ≤ r_{2} iff r_{1} ⊇ r_{2}. The meaning of these conditions is

1. (n, s) means “[s] ⊆ U_{n}”

2. (n, (x, t, k)) means “swap(x, t, k) ∩ U_{n} = ∅”

Now suppose G is ℚ(X, Y)-generic over a model N. Define

U_{n}^{G} = ⋃{[s] : ∃r ∈ G (n, s) ∈ r}

An easy genericity argument shows that

X ∩ ⋂_{n<ω} U_{n}^{G} = X \ Y

To see this suppose y ∈ Y and r is any condition. Let n be sufficiently large so as to not appear in r at all. Then let r′ = r ∪ {(n, (y, ⟨⟩, 0))} and note that

r′|⊢y ∉ U_{n}.

On the otherhand let y ∈ X \Y, r be any condition, and n < ω be arbitrary. Since y is infinitely often different from any element of X mentioned in r (they must come from Y), we can find l < ω so that

[y ↾ l] ∩ swap(x, s, k) = ∅

for any (n, (x, s, k)) ∈ r. Now we let r′ = r ∪ {(n, y ↾ l)}, then

r′|⊢y ∈ U_{n}.

Next we describe the ordering ℙ ∗ ℚ̃ which is a basic building block of our iteration. If G is ℙ-generic over M then ℚ̃^{G} is essentially the same as ℚ(X, X).

Define. ((p, q), r) ∈ ℙ ∗ ℚ̃ iff

1. (p, q) ∈ ℙ

2. r is a finite subset of the union of

{(n, t) : n < N_{p}, t ∈ 2^{<N_{p}}}

and

{(n, (α, s, k)) : α ∈ dom(p), s ∈ 2^{<N_{p}}, n, k < N_{p}}

3. if (n, (α, s, k)) ∈ r and (n, t) ∈ r, then either s and t are incomparable or s ⊆ t and

|{i : |s| ≤ i < |t|, t(i) ≠ p(α)(i)}| > k.

Condition (3) guarantees that for any x ∈ 2^{ω} such that x ⊇ p(α) that

swap(x, s, k) ∩ [t] = ∅.

The ordering is given by

((p_{1}, q_{1}), r_{1}) ≤ ((p_{2}, q_{2}), r_{2}) iff (p_{1}, q_{1}) ≤ (p_{2}, q_{2}) and r_{1} ⊇ r_{2}. Note that ((p_{1}, q_{1}), r_{1}) and ((p_{2}, q_{2}), r_{2}) are compatible iff there exists p_{3} ≤ p_{1}, p_{2} such that ((p_{3}, q_{1} ∪ q_{2}), r_{1} ∪ r_{2}) is a condition.

The ω_{2} iteration.

Our iteration can be described as a suborder of the product

ℙ × ∑_{α<ω_{2}} 𝔼

where 𝔼 is the set of all finite subsets of

{(n, t) : n < ω, t ∈ 2^{<ω}} ∪ {(n, (α, s, k)) : α ∈ ω_{1}, s ∈ 2^{<ω}, n, k < ω}

and ∑_{α<ω_{2}} 𝔼 is the set of all r : ω_{2} → 𝔼 such that r(δ) is trivial (i.e., the empty set) for all but finitely many δ.

By induction on β ≤ ω_{2} define

ℙ_{β} ⊆ ℙ × ∑_{α<β} 𝔼

as follows:

Define. ℙ_{0} = ℙ, and suppose that we have defined ℙ_{β} and we are also given a ℙ_{β} name Y_{β} for a subset of ω_{1}, i.e.,

|⊢_{β}Y_{β} ⊆ ω_{1}

Define. ((p, q), r) ∈ ℙ_{β+1} iff

1. ((p, q), r ↾ β) ∈ ℙ_{β},

2. ((p, q), r(β)) ∈ ℙ ∗ ℚ̃

3. ((p, q), r ↾ β)|⊢_{β}α ∈ Y_{β}

whenever (n, (α, s, k)) ∈ r(β) for some n, s, k, α.

For limit ordinals λ ≤ ω_{2} we define ((p, q), r) ∈ ℙ_{λ} iff for all β < λ we have ((p, q), r ↾ β) ∈ ℙ_{β} and for all but finitely many β < λ we have that r(β) is the trivial condition (i.e. empty set).

Since the iteration of ccc forcing is ccc all of these forcings are ccc. To see this directly we can argue as follows: Standard arguments using Δ systems show that ℙ_{β} has precalibre ω_{1}, i.e., any ω_{1} sequence of conditions contain an ω_{1} subsequence which is centered. Start with ((p_{α}, q_{α}), r_{α}) ∈ ℙ_{β} for α < ω_{1}. We can find an uncountable Σ ⊆ ω_{1} and finite sets F and H and N < ω so that

1. N_{α} = N for all α ∈ Σ,

2. dom(p_{α}) ∩ dom(p_{β}) = F for α ≠ β ∈ Σ,

3. dom(r_{α}) ∩ dom(r_{β}) = H for α ≠ β ∈ Σ,

4. p_{α} ↾ F are all the same for α ∈ Σ,

5. q_{α} ↾ [F]^{2} are all the same for α ∈ Σ, and

6. r_{α} ↾ H are all the same with respect to {(n, s) : n < ω, s ∈ 2^{<ω}} for α ∈ Σ.

Then any two (or even finite subset) of them are compatible.

Assuming that the ground model satisfies the GCH by the usual book keeping argument we can arrange things so that for any Y ⊆ ω_{1} which appears in M[G_{ω_{2}}] there will be a name for it in the list Y_{α} for some α < ω_{2}. The simplest way to do this is to take

{⟨Z_{β}^{α} : β < ω_{1}⟩ : α < ω_{2}}

which lists all ω_{1} sequences of countable subsets of ℙ × ∑_{α<ω_{2}} 𝔼 with ω_{2} repetitions and then define

Y_{α} = {⟨p, β̌⟩ : β < ω_{1}, p ∈ Z_{β}^{α} ∩ ℙ_{α}}

If we define

x_{α} = ⋃{s ∈ 2^{<ω} : ∃((p, q), r) ∈ G, s = p(α)} and X = {x_{α} : α < ω_{1}} then X will be the characteristic functions of an almost disjoint family A = {a_{α} : α < ω_{1}}. Furthermore if we define the open sets

U_{n}^{β} = ∪{[s] : ∃((p, q), r) ∈ G (n, s) ∈ r(β)}

then by the usual genericity argument

⋂_{n<ω} U_{n}^{β} ∩ X = {x_{α} : α ∉ Y_{β}^{G}}

and so X will be a Q-set.

The nontrivial part of our argument is to prove that A remains a maximal almost disjoint family. So let τ be a name for a counterexample, i.e., suppose

((p_{0}, q_{0}), r_{0})|⊢τ ∈ [ω]^{ω} and ∀α < ω_{1} τ ∩ a_{α} is finite .

Let Σ ⊆ ℙ_{ω_{2}} be a countable set of conditions extending ((p_{0}, q_{0}), r_{0}) such that for any n ∈ ω Σ contains a maximal antichain beneath ((p_{0}, q_{0}), r_{0}) which decides n ∈ τ. Let α_{0} < ω_{1} be any ordinal not mentioned in any condition from Σ. We show a_{α_{0}} ∩ τ is infinite.

Suppose for contradiction that we have ((p_{1}, q_{1}), r_{1}) ≤ ((p_{0}, q_{0}), r_{0}), and N_{1} < ω such that

((p_{1}, q_{1}), r_{1})|⊢τ ∩ a_{α_{0}} ⊆ N_{1}

Without loss of generality we may assume that N_{1} = N_{p_{1}}. By tacking on strings of zeros to the conditions in p_{1} we may assume that every integer occurring in r is bounded by N_{1} − 2 (and not just as required by N_{1}). Let

F = {β : {α_{0}, β} ∈ dom(q_{1})}

Define r′ ⊇ r_{1} as follows:

r′(δ) = r_{1}(δ) ∪ {(n, (α_{0}, t′, k + 1)) : (n, (α_{0}, t, k)) ∈ r_{1}(δ), t′ ∈ A_{δ,n}, t′ ⊇ t}

for each δ where

A_{δ,n} = {t′ ∈ 2^{N_{1}−1} : t′ is incomparable with all s such that (n, s) ∈ r_{1}(δ)}

Note that ((p_{1}, q_{1}), r′) is a valid condition because α_{0} is forced into Y_{δ} and t′ incomparable with all s which might be a problem. Let G be a generic filter containing ((p_{1}, q_{1}), r′). Since τ^{G} is almost disjoint from each a_{β}^{G} and infinite, there exists some n_{0} ∈ τ^{G} with n_{0} > N_{1} and n_{0} ∉ a_{β}^{G} for all β ∈ F. Let ((p_{2}, q_{2}), r_{2}) ∈ Σ ∩ G be so that

((p_{2}, q_{2}), r_{2})|⊢n_{0} ∈ τ.

Since it is from Σ it does not mention α_{0}.

Let ((p^{∗}, q^{∗}), r^{∗}) ∈ G be stronger than both ((p_{1}, q_{1}), r′) and (p_{2}, q_{2}), r_{2}) and such that N^{∗} > n_{0}. Note that ((p^{∗}, q_{1} ∪ q_{2}), r′ ∪ r_{2}) is a valid condition. Any γ that needs to be forced into some Y_{β} is already forced in by either ((p_{1}, q_{1}), r′ ↾ β) or ((p_{2}, q_{2}), r_{2} ↾ β).

If p^{∗}(α_{0})(n_{0}) = 1 then we already have a contradiction and there is nothing to prove. So assume not, and define p′ to be exactly the same as p^{∗} except p′(α_{0})(n_{0}) = 1.

Claim. ((p′, q_{1} ∪ q_{2}), r ∪ r_{2}) is a valid condition, extending both ((p_{1}, q_{1}), r_{1}) and ((p_{2}, q_{2}), r_{2}).

Proof: Note that we have dropped the extra conditions from r′, these were put there just to prove this Claim. The fact that p′ extends both p_{1} and p_{2} uses that n_{0} > N_{p_{1}} = N_{1} and α_{0} is not in the domain of p_{2}. Similarly since q_{2} does not mention α_{0}, so if {α_{0}, β} ∈ dom(q_{1} ∪ q_{2}), then β ∈ F and we know that p^{∗}(β)(n) = 0 for each β ∈ F. So making p′(α_{0})(n) = 1 does not violate any promises of disjointedness made in q_{1} ∪ q_{2}. So we have that (p′, q_{1} ∪ q_{2}) ∈ ℙ.

Now fix δ and we must check that

((p′, q_{1} ∪ q_{2}), r_{1}(δ) ∪ r_{2}(δ)) ∈ ℙ ∗ ℚ̃

We need to check condition (3)

(3) if (n, (α, s, k)), (n, t) ∈ r_{1}(δ) ∪ r_{2}(δ) then either s and t are incomparable or s ⊆ t and

|{i : |s| ≤ i < |t|, t(i) ≠ p′(α)(i)}| > k

Suppose it fails. It can only fail if the α = α_{0} and since r_{2} does not mention α_{0} it must be that (n, (α_{0}, s, k)) ∈ r_{1}(δ) and (n, t) ∈ r_{2}(δ). Also it must be that s and t are comparable with s ⊆ t but

|{i : |s| ≤ i < |t|, t(i) ≠ p′(α_{0})(i)}| ≤ k

Note also that |t| > n_{0} > N_{1} because otherwise

{i : |s| ≤ i < |t|, t(i) ≠ p′(α_{0})(i)} = {i : |s| ≤ i < |t|, t(i) ≠ p^{∗}(α_{0})(i)} But then

|{i : |s| ≤ i < |t|, t(i) ≠ p^{∗}(α_{0})(i)}| > k

because ((p^{∗}, q_{1} ∪ q_{2}), r_{1}(δ) ∪ r_{2}(δ)) ∈ ℙ ∗ ℚ̃.

Now let t′ = t ↾ (N_{1} − 1).

Case 1. t′ is comparable with some s′ such that (n, s′) ∈ r_{1}(δ).

Recall that every integer occurring in r_{1}(δ) is bounded by N_{1} − 1. So it must be that s′ ⊆ t′ but intuitively this is easy because r_{1}(δ) is already asserting [s′] ⊆ U_{n}^{δ} and this implies [t] ⊆ U_{n}^{δ}. More formally, s′ ⊆ t′ and therefore s′ and s are both initial strings of t′ and so comparable, but then we know:

|{i : |s| ≤ i < |s′| < N_{1}, s′(i) ≠ p_{1}(α_{0})(i)}| > k

But this is still true for p′ since we have not changed it below N_{1}.

Case 2. t′ ∈ A_{δ,n} and so we added (α_{0}, t′, k + 1) to r′(δ).

But remember ((p^{∗}, q_{1} ∪ q_{2}), r′ ∪ r_{2}) is a valid condition, which means that

|{i : N_{1} ≤ i < |t|, t(i) ≠ p^{∗}(α_{0})(i)}| > k + 1

but k < N_{1}−2 and p^{∗}(α_{0}) agrees with p′(α_{0}) except at exactly one coordinate so

|{i : |s| < N ≤ i < |t|, t(i) ≠ p′(α_{0})(i)}| > k

This proves that ((p′, q_{1} ∪ q_{2}), r_{1}(δ) ∪ r_{2}(δ)) ∈ ℙ ∗ ℚ̃ for every δ.

Finally we must show that

((p′, q_{1} ∪ q_{2}), (r_{1} ∪ r_{2}) ↾ β)|⊢_{β}γ ∈ Y_{β}

whenever (n, (γ, s, k)) ∈ (r_{1} ∪ r_{2})(β) for some n, s, k. But by induction

((p′, q_{1} ∪ q_{2}), (r_{1} ∪ r_{2}) ↾ β)

extends both ((p_{1}, q_{1}), r_{1} ↾ β) and ((p_{2}, q_{2}), r_{2} ↾ β), one of which does the required forcing.

This proves the Claim. The theorem now follows from the contradiction that

((p_{1}, q_{1}), r_{1})|⊢τ ∩ a_{α_{0}} ⊆ N_{1}
((p_{2}, q_{2}), r_{2})|⊢n_{0} ∈ τ

[figure: N_{1}; n_{0}; s; t′; t; N^{∗}; [t] ⊆ U_{n} in r_{2}(δ); swap(x_{α_{0}}, s, k) ∩ U_{n} = ∅ in r_{1}(δ); swap(x_{α_{0}}, t′, N_{1} − 1) ∩ U_{n} = ∅ in r′(δ)]

Figure 1: The swap

where n_{0} > N_{1} and

((p′, q_{1} ∪ q_{2}), r_{1} ∪ r_{2})|⊢n_{0} ∈ a_{α_{0}}.

QED

Remark. The usual Q-set forcing kills the maximality of an almost disjoint family X. To see this suppose {x_{n} : n < ω} ⊆ X and conditions are finite consistent sets of sentences of the form: “[s] ⊆ U_{n}” or “x ∉ U_{n}” where x ∈ X \ {x_{n} : n < ω}. So when we force we get a G_{δ} set so that

∩_{n<ω}U_{n} ∩ X = {x_{n} : n < ω}.

In the generic extension we can find {k_{n} : n < ω} increasing so that

k_{n+1} ∉ ∪_{i<n}x_{i} and {y ⊆ ω : k_{n+1} ∈ y} ⊆ ⋂_{i<n} U_{i}.

Why? Given p find k_{n+1} > k_{n} not in any x_{i} for i < n or in any x mentioned in p and put

p′ = p ∪ {[s_{1}] ⊆ U_{i} : i < n, s ∈ 2^{k_{n+1}−1}}

But then {k_{n} : n < ω} is almost disjoint from all elements of X.

Remark. Since there are perfect almost disjoint families, eg.,

A = {{x ↾ n : n < ω} : x ∈ 2^{ω}} ⊂ P (2^{<ω})

there are always MAD families of arbitrarily large Borel order. Obviously a Q-set cannot have cardinality continuum, however a σ-set can.

Define. X ⊆ 2^{ω} is a σ-set iff for every Borel set B ⊆ 2^{ω} there exists a G_{δ} set G such that B ∩ X = G ∩ X.

A Sierpinski set is an example of σ-set (Poprougenko, see Miller [9]).

Theorem 3 It is consistent with any cardinal arithmetic that there exists a MAD σ-set of size the continuum.

Proof This is an easy modification of the argument of the main theorem. Taking any countable transitive model M first force a generic MAD of size continuum, then do a finite support iteration of length continuum to make it into a σ-set. QED

Remark. H.Woodin, see Larson [8], has shown that if there exists a measurable Woodin cardinal κ, and V and V [G] are both models of CH where V [G] is a generic extension using a partial order of size less than κ, then V and V [G] model exactly the same Σ_{1}^{2} sentences. The existence of a MAD σ-set is a Σ_{1}^{2} sentence. It follows that

CH + there exists a measurable Woodin cardinal implies there is a MAD σ-set.

It is virtually certain that MAD σ-sets have nothing to do with large cardinals, so we have the conjecture:

Conjecture 4 CH implies there exists a MAD σ-set.

Theorem 5 The generic MAD set A = {a_{α} : α < ω} is concentrated on {a_{n} : n < ω}, i.e., every open set containing {a_{n} : n < ω} contains all but countably many elements of A.

Proof Let M be a countable standard model of ZFC and G be ℙ-generic over M. Working in M suppose

|⊢{a_{n} : n < ω} ⊆ U an open set.

Let Σ ⊆ ℙ be countable so that for every s ∈ 2^{<ω} there exist a maximal antichain in Σ which decides “[s] ⊆ U”.

Claim. |⊢a_{α} ∈ U for any α larger than any mentioned in Σ. Proof Suppose not and let (p, q)|⊢a_{α} ∉ U. Choose some n so that n is not in the domain of p. Let p′ = p ∪ {(n, s)} where (α, s) ∈ p and let

q′ = q ∪ ⋃{({n, β}, k) : ({α, β}, k) ∈ q}

so (p′, q′) ≤ (p, q) and it says the same things about a_{n} and a_{α}. There exists (p̂, q̂) ∈ Σ compatible with (p′, q′) such that Np̂ > N_{p} and

(p̂, q̂)|⊢[x_{n} ↾ Np̂] ⊆ U

Let (p^{∗}, q^{∗}) extend both (p′, q′) and (p̂, q̂). Change p^{∗} to r with same domain but r(α) = p^{∗}(n) and other coordinate all the same. But then (r, q′ ∪ q̂) is a common extension of both (p′, q′) and (p̂, q̂). And this is a contradiction.

This proves the Claim and Theorem. QED

Theorem 6 CH implies exists a MAD family which is concentrated on the finite subsets of ω and is a λ-set (i.e., every countable subset is a relative G_{δ}.

Proof It is easy to construct a MAD family {a_{α} : α < ω_{1}} so that if f_{α} : ω → a_{α} is the strictly increasing enumeration of α, then for every α < β we have that f_{α} <^{∗} f_{β} and for every g ∈ ω^{ω} there exists α < ω_{1} such that g ≤^{∗} f_{α}, i.e., they form a scale. Rothberger (see Miller [9]) showed that any well-ordered subset of (ω^{ω}, ≤^{∗}) is a λ-set and that any ω_{1} -ordered unbounded set is concentrated on the rationals. QED

The same large cardinal results lead to the following conjecture:

Conjecture 7 CH implies there exists a MAD family which is concentrated on a countable subset of itself.

Paul Szeptycki pointed out that the Q-set forcing using in Theorem 1 can be used to prove the following:

Theorem 8 It is relatively consistent that there exists a Q-set X ⊆ [ω]^{ω} satisfying the property that for every a ∈ [ω]^{ω} for all but countably many x ∈ X we have that |x ∩ a| = |x \ a| = ω, ie X is a strong splitting family.

Proof We replace ℙ by the Cohen real partial order, i.e., just drop the q’s from the (p, q). We use the same ℙ ∗ ℚ̃. Note that in the basic argument for p′ we could have flipped p′(α_{0})(n_{0}) = 1 − p^{∗}(α_{0})(n_{0}) and α_{0} could be any α < ω_{1} not mentioned in Σ. QED

Allan Dow asked if it is possible to have X = {x_{α} ∈ 2^{ω} : α < ω_{1}} and Y = {y_{α} ∈ 2^{ω} : α < ω_{1}} such that x_{α} =^{∗} y_{α} for every α < ω_{1}, and X a Q-set and Y not a Q-set. The answer is yes, in fact, we force something stronger:

Theorem 9 It is relatively to consistent to have X = {x_{α} ∈ 2^{ω} : α < ω_{1}} and {y_{n} ∈ 2^{ω} : n < ω} such that x_{n} =^{∗} y_{n} for every n < ω, and X a Q-set concentrated on {y_{n} : n < ω} (hence Y = {y_{n} : n < ω} ∪ {x_{α} : ω ≤ α < ω_{1}} is not a Q-set).

Proof This is a variant of the forcing used in Fleissner-Miller [3]. In that forcing we start by adding an ω_{1} batch of Cohen reals X = {x_{α} ∈ 2^{ω} : α < ω_{1}}. The Q-set forcing is modified to always allow statements of the form e ∉ U_{n}^{α} where e is any eventually zero element of 2^{ω}. With this modification the Q-set X is shown to be concentrated on the eventually zero elements of 2^{ω}.

We modify this slightly as follows. Let y_{n} be defined by y_{n}(m) = x_{n}(m) except y_{n}(n) = 1 − x_{n}(n), i.e. we change only one coordinate. Now in the inductive construction of ℙ_{α} we use the y_{n}’s instead of the eventually zero reals, i.e. if x ∉ U_{m}^{α} ∈ p(α), then either x = y_{n} for some n or p ↾ α|⊢x ∈ Y_{α}.

The rest of the argument is the same as in [3]. QED

I am not sure how to do Theorem 9 with a MAD family. This would be interesting because it would show that the Ψ space does not determine whether or not you have a Q-set.

References

[1] Brendle, Jorg, Dow’s principle and Q-sets. Canad. Math. Bull. 42 (1999), no. 1, 13–24.

[2] Dow, Alan, On compact separable radial spaces. Canad. Math. Bull. 40 (1997), no. 4, 422–432.

[3] Fleissner, William G.; Miller, Arnold W., On Q-sets. Proc. Amer. Math. Soc. 78 (1980), no. 2, 280–284.

[4] Gruenhage, Gary; Koszmider, Piotr, The Arkhangelski-Tall problem: a consistent counterexample. Fund. Math. 149 (1996), no. 2, 143–166.

[5] Hechler, Stephen H., Short complete nested sequences in βN\N and small maximal almost-disjoint families. General Topology and Appl. 2 (1972), 139–149.

[6] Hrušák, Michael, MAD families and the rationals. Comment. Math. Univ. Carolin. 42 (2001), no. 2, 345–352.

[7] Judah, Haim; Shelah, Saharon, Q-sets, Sierpiński sets, and rapid filters. Proc. Amer. Math. Soc. 111 (1991), no. 3, 821–832

[8] Larson, Paul, Notes on the stationary tower, to appear.

[9] Miller, Arnold W., Special subsets of the real line. Handbook of settheoretic topology, 201–233, North-Holland, Amsterdam, 1984.

[10] Nowik, Andrzej; Weiss, Tomasz, Not every Q-set is perfectly meager in the transitive sense. Proc. Amer. Math. Soc. 128 (2000), no. 10, 3017–3024.

[11] Szeptycki, Paul, Countable metacompactness in Ψ-spaces. Proc. Amer. Math. Soc. 120 (1994), no. 4, 1241–1246.

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