Luzin and anti-Luzin almost disjoint families by

MATHEMATICAE 158 (1998)

Luzin and anti-Luzin almost disjoint families

by

Judith Roitman (Lawrence, Kan.) and Lajos Soukup (Budapest)

Abstract. Under MA

every uncountable almost disjoint family is either anti-Luzin or has an uncountable Luzin subfamily. This fails under CH. Related properties are also investigated.

0. Introduction. This paper concerns two combinatorial properties of almost disjoint families, Luzin and anti-Luzin, along with some variants.

Let A be an uncountable almost disjoint family on a countable set W . Definition 0.1. A is Luzin iff there is an enumeration A = {a

: α < ω

} such that for all w ∈ [W ]

and α < ω

, {β < α : a

∩ a

⊂ w} is finite.

Definition 0.2. A is anti-Luzin iff for all B ∈ [A]

there exist C, D ∈ [B]

such that S

C ∩ S

D is finite.

The notion of “Luzin” is somewhat standard. The notion of “anti-Luzin”

is new.

Luzin almost disjoint families are an analogue of Hausdorff gaps: they exist (via a diagonal construction) in ZFC, and because of the finitary nature of the definition the “Luzin” property is upwards absolute. Perhaps because of this similarity, they have been called Luzin gaps.

Anti-Luzin families also exist in ZFC, as canonical objects: any uncount- able set of branches of a countable tree T is an anti-Luzin family on T .

Luzin and anti-Luzin families are hereditary in the following sense: un- countable subfamilies of Luzin (respectively anti-Luzin) almost disjoint fam- ilies are Luzin (respectively anti-Luzin).

1991 Mathematics Subject Classification: 54G20, 03E35, 04A20, 06E99.

Key words and phrases: Luzin, almost disjoint family.

Research of J. Roitman partially supported by the University of Kansas General Research Fund grant # 3058.

Research of L. Soukup supported by the Hungarian National Foundation for Scientific Research grant # 16391.

[51]

Definition 0.3. A is near-Luzin iff for all C, D ∈ [A]

, S C ∩ S

D is infinite.

We will rely heavily on the following obvious statement: A is anti-Luzin iff it has no uncountable near-Luzin subfamily.

Near-Luzin first appeared in [HJ], where it was called ω

-full: near-Luzin families give rise to compact Hausdorff spaces in which the intersection of any two uncountable open sets is non-empty.

Claim 0.4. If A is Luzin then it is near-Luzin.

P r o o f. Fix an arbitrary enumeration {a

: α < ω

} of A. Suppose A is not near-Luzin. Then there are uncountable C, D ⊂ A with S

C ∩ S D = w finite. So for all c ∈ C and d ∈ D, c ∩ d ⊂ w. There is α with a

∈ C and {a

∈ D : β < α} infinite. Hence A is not Luzin.

Corollary 0.5. An anti-Luzin family is not Luzin.

While superficially Corollary 0.5 does not reverse (a disjoint union of a Luzin and an anti-Luzin family is neither Luzin nor anti-Luzin), does it reverse in any deep sense? In particular, must any uncountable almost disjoint family which does not embed one embed the other? The answer is yes and no.

Theorem 0.6. Assume MA

. Every uncountable almost disjoint family is either anti-Luzin or contains an uncountable Luzin subfamily.

Theorem 0.7. Assume

. There is an uncountable almost disjoint family which contains no uncountable anti-Luzin and no uncountable Luzin subfam- ilies.

Here

is the following weakening of CH: There is a family S ⊂ [ω

]

of size ω

so that every uncountable subset of ω

contains a set in S.

Theorem 0.6 says that under MA

, almost disjoint families have a lot of structure. Theorem 0.7 says that under

they do not. This used to be what one would expect, but recent work on iterating totally proper forcing while preserving CH has changed our expectations. In particular, Abraham and Todorˇcevi´c showed the consistency of “CH + all (ω

, ω

)-gaps contain an uncountable Hausdorff subgap”. If Luzin almost disjoint families were combinatorially similar to Hausdorff gaps the conclusion of Theorem 0.6 would also be consistent with CH. Thus Theorem 0.7 destroys the parallel between Luzin almost disjoint families and Hausdorff gaps.

In Section 1 we prove Theorem 0.6, in Section 2 we prove Theorem 0.7, in the rest of the paper we explore some of the fine combinatorial structure of these notions.

Conventions. In this paper almost disjoint families are collections of

infinite sets whose pairwise intersections are finite; the superscript “

” means

“mod finite”; all trees grow upward; and properties are listed consecutively no matter what theorem, lemma, or definition they occur in, so if there is a reference to property 17 the reader can easily find it.

Acknowledgments. We would like to express our thanks to the referee for his (her) numerous valuable suggestions and comments.

1. Proof of Theorem 0.6. Assume MA

. Let A = {a

: α < ω

} be an almost disjoint family which is not anti-Luzin. By moving to a possibly smaller subfamily, we may assume A is near-Luzin.

Let P be the set [ω

]

under the following partial order:

p ≤ q iff p ⊃ q and ∀α ∈ p \ q ∀β ∈ q if α < β then a

∩ a

6⊂ k

where k

= max S

{a

∩ a

: {α, β} ∈ [q]

}.

Remark 1.1. If β > sup p then p ∪ {β} ≤ p.

A quick ad hoc definition: for s, t ⊂ ON, s  t iff sup s < inf t.

Remark 1.2. If r, s, t are disjoint, r  s ∪ t, p = r ∪ s, q = r ∪ t and k

= k

= k, then p and q are compatible iff for all α ∈ s and β ∈ t, a

∩ a

6⊂ k.

By Remark 1.1, enough sets are dense so that if P is ccc then, by MA

, there is a generic filter G so that {a

: α ∈ S

G} is Luzin. Remark 1.2 will be used to show ccc.

Lemma 1.3. Suppose n, k ∈ ω and let E ∈ [[ω

]

]

be pairwise disjoint.

Then there are s 6= t ∈ E with a

∩ a

6⊂ k for all α ∈ s and β ∈ t.

To prove Lemma 1.3 we need

Sublemma 1.3.1. If S, T are uncountable disjoint subsets of ω

and k < ω then there are S

∈ [S]

, T

∈ [T ]

and k

> k with k

∈ a

∩ a

for all α ∈ S

and β ∈ T

.

P r o o f. Otherwise for all k

> k either S(k

) = {α ∈ S : k

∈ a

} is countable or T (k

) = {α ∈ T : k

∈ a

} is countable. Let S

= S \ S

{S(k

) : k

> k and S(k

) is countable}. Let T

= T \ S

{T (k

) : k

> k and T (k

) is countable}. Then S

and T

are uncountable and S

α∈S⁰⁰

(a

\ k) ∩ S

β∈T⁰⁰

(a

\ k) = ∅, contradicting our hypothesis on A.

Proof of Lemma 1.3 . For s ∈ E we write s = {α(s, i) : i < n}. Let

{(i

, j

) : m < n

} enumerate n × n. Using Sublemma 1.3.1 iteratively, at

each stage m we have uncountable disjoint subsets of E, E(m) and F (m),

with E(m) ⊃ E(m + 1) and F (m) ⊃ F (m + 1), and k

> k with k

∈

a

∩ a

for all s ∈ E(m) and t ∈ F (m). But then for α ∈ s ∈

E(n

− 1) and β ∈ t ∈ F (n

− 1), a

∩ a

6⊂ k.

Lemma 1.4. P is ccc.

P r o o f. Given an uncountable subset F of P we may without loss of generality assume that for some n, F ⊂ [ω

]

; F is a ∆-system with root r so that each p ∈ F has the form r ∪ s

; there is some k with k = k

for all p ∈ F ; and E = {s

: p ∈ F } is well-ordered by . By Lemma 1.3 there are p 6= q ∈ F so that for all α ∈ s

and β ∈ s

, a

∩ a

6⊂ k. By Remark 1.2, p and q are compatible.

2. Proof of Theorem 0.7. In this section we prove Theorem 0.7. In the next section we will give a stronger version, but the combinatorics is sufficiently complicated that it makes sense to give the weaker proof first and then show how to improve it.

Let {S

: α < ω

} be a

-sequence, i.e. each S

⊂ α and every uncount- able X ⊂ ω

contains some S

.

The family A = {f

: α < ω

} will be a subset of ℘(ω × ω), where each f

is a function from ω to ω.

We require:

1. If α 6= γ then {i : f

(i) = f

(i)} is finite.

2. If β ≤ α then f

∩ S

γ∈Sβ

f

is infinite.

3. If β ≤ α then there is n such that {γ ∈ S

: f

∩ f

⊂ n × n} is infinite.

Property 1 makes A almost disjoint and property 2 makes it near-Luzin (hence prevents an uncountable anti-Luzin subfamily). Finally, property 3 prevents an uncountable Luzin subfamily. Indeed, let {f

: ν < ω

} be an enumeration of an uncountable subfamily of A. Pick β < ω

with S

⊂ {α

: ν < ω

}. Let I = {ν : α

∈ S

} and fix µ < ω

such that α

≥ β and µ > sup I. Then, by 3, there is m ∈ ω such that {ν < µ : f

∩f

⊂ m×m}

is infinite, which contradicts the requirement formulated in Definition 0.1.

So {f

: ν < ω

} is not Luzin.

Some preliminaries:

Definition 2.1. A set F ⊂ ω × ω is fat iff lim sup

|π

F | = ω, where π

F = {j : (n, j) ∈ F }.

Notice that if there is a finite family of functions U with F ⊂

S

U then F is not fat, and fat sets are infinite.

Lemma 2.2. If C is an infinite almost disjoint family of functions from ω to ω, then S

C is fat.

P r o o f. Fix m < ω. Let G ∈ [C]

. There is r so that if f 6= g ∈ G then f ¹[r, ω) ∩ g¹[r, ω) = ∅. So m < lim sup

|π

S

G|.

Definition 2.3. Let C be a countable almost disjoint collection of func- tions from ω to ω. A finite partial function σ from ω to ω is C-free iff ∃

g ∈ C g ∩ σ = ∅.

Lemma 2.4. Let C be a countable collection of functions from ω to ω.

There is a function s : ω → ω + 1 so that if σ is a finite partial function from ω to ω with σ(i) 6= s(i) for all i ∈ dom σ, then σ is C-free.

Such an s is called C-tight.

P r o o f. Consider C as a subset of the compact space (ω + 1)

. Let s be an accumulation point of C in (ω + 1)

. If σ is a finite partial function from ω to ω with s ∩ σ = ∅ then s 6∈ {f : f ∩ σ 6= ∅} which is closed, so C \ {f : f ∩ σ 6= ∅} is infinite.

The construction is a straightforward induction, given the following Lemma 2.5. Let C be a countable almost disjoint collection of functions from ω to ω, let C

⊂ C for each n < ω, and let F

be fat for each n < ω.

Then there is a function f : ω → ω so that:

4. {f } ∪ C is almost disjoint.

5. For each n, f ∩ F

is infinite.

6. ∀n ∃m

{g ∈ C

: f ∩ g ⊂ m

× {f (i) : i < m

}} is infinite.

The family {f

: α < ω

} will be constructed recursively in ω

steps.

Assume that {f

: α < β} is already constructed. Fix an enumeration {β

: n < ω} of β. Let C = {f

: α < β}, C

= {f

: α ∈ S

} and F

= S

C

. Now we can apply Lemma 2.5 to get f

as f .

In the next section, we will need to deal with many more fat sets, which is why 2.5 is stated in its current generality.

Proof of Lemma 2.5. Let C = {g

: i < ω}. At stage j we construct a finite set U

(“U ” is short for “used up”) of functions in C where U

⊂ U

; we define m

≥ j, and define f on (m

, m

].

So suppose we are at stage j. We know m

for each k < j, f ¹m

+ 1, U

, and, for each k < j, we have a C

-tight s

. Our induction hypothesis is that

∀k < j f ¹(m

, m

] ∩ s

¹(m

, m

] = ∅.

Let s

be C

-tight. Since F

is fat, for k ≤ j there is some r

> m

with r

< r

, and some (r

, t

) ∈ F

\ S

U

. Let m

= r

. Let r

= m

.

For k ≤ j, let f (r

) = t

. (This is towards property 5.)

For i ∈ (r

, r

) with i 6= r

for each k, let f (i) be any m 6∈ {g(i) :

i ∈ U

} ∪ {s

(i) : k ≤ j}. (This is towards properties 4 and 6.)

For each k ≤ j,

f ¹(m

, m

] ∩ s

¹(m

, m

] = ∅

so f ¹(m

, m

] is C

-free, for each k ≤ j. Hence, for each k ≤ j there is g

∈ C

\ U

with

g

∩ f ¹(m

, m

] = ∅.

Let

U

= U

∪ {g

: k ≤ j} ∪ {g

}.

Property 4 is satisfied: If n ≤ j and i ∈ (m

, m

] then g

(i) 6= f (i), so f ∩ g

⊂ m

× {f (i) : i < m

}.

Property 5 is satisfied: f (r

) = t

for all j ≥ n and (r

, t

) ∈ F

, so f ∩ F

is infinite.

Property 6 is satisfied: If j ≥ n then f¹(m

, m

]∩g

= ∅, so g

∩f ⊂ m

× {f (i) : i < m

}.

3. A strengthening of Theorem 0.7. In this section we strengthen Theorem 0.7.

Definition 3.1. An uncountable almost disjoint family A is strongly near-Luzin iff, for every C

, . . . , C

∈ [A]

, T

i≤n

S C

is infinite.

Strongly near-Luzin families appear in [JN], where they are called strong Luzin families. They cannot exist under MA + ¬ CH. The following theorem shows that they need not be Luzin.

Theorem 3.2. Assume

. There is an uncountable almost disjoint family which is strongly near-Luzin, but has no uncountable Luzin subfamilies.

The proof is somewhat like that of Theorem 0.7, but the combinatorics is more complicated, so complicated that we will invoke elementary submodels to avoid stating it explicitly.

So let {S

: α < ω

} be a

-sequence. We begin by strengthening pro- perty 2 to

7. If β

, . . . , β

< α and sup S

∪ (β

+ 1) < inf S

for each i, then f

∩ T

i≤n

S

γ∈S_βi

f

is infinite.

We will be done if the sequence of f

’s has properties 1, 7, and 3. Indeed,

as we have seen in the proof of Theorem 0.7, property 3 implies that A

does not contain an uncountable Luzin subfamily. Property 1 yields that

A is almost disjoint. So we need to show that if property 7 holds then A

is strongly near-Luzin. Let n ∈ ω and I

, . . . , I

∈ [ω

]

. Since {S

:

α < ω

} is a

-sequence we can find β

< β

< . . . < β

∈ ω

such

that S

⊂ I

and sup S

∪ (β

+ 1) < min S

. Then, by property 7,

T

i<n

S {f

: γ ∈ I

} ⊃ T

i<n

S {f

: γ ∈ S

} is infinite, which was to be proved.

In applying 2.5 in the previous section we had the luxury of knowing that each S

C

was fat. But an intersection of fat sets need not be fat. So we must ensure that the following property holds:

8. If β T

, . . . , β

< α and sup S

∪ (β

+ 1) < inf S

for each i, then

i≤n

S

γ∈S_βi

f

is fat.

Property 8 allows us to construct a family in which property 7 holds.

How will we build a family in which property 8 holds?

Definition 3.4. For F ⊂ ω × ω and E ∈ [ω]

let π

F = F ∩ (E × ω).

To get property 8 to hold, we need to start with enough fat C’s, then have enough E’s so that the resulting π

C’s are fat, iterate the process. . . Rather than try to define the precise combinatorics of “enough”, we take advantage of elementary submodels which provide all the fat sets we need.

Along with constructing our sequence of functions f

, then, we will con- struct a sequence of large enough countable elementary submodels {N

: α < ω

} where

9. f

, S

, {N

: β ≤ α} ∈ N

for each α, {N

: β < α} ⊂ N

, and {S

: α < ω

} ∈ N

.

Further requirements are:

1. If α 6= γ then {i : f

(i) = f

(i)} is finite.

10. If C ∈ N

is fat and E = {n : (n, f

(n)) ∈ C} then π

C is fat.

11. If C ∈ N

is fat, then f

∩ C is infinite.

12. If S ∈ N

∩ [α]

then there is m ∈ ω such that {γ ∈ S : f

∩ f

⊂ m × m} is infinite.

Note that property 11 follows from property 10.

As before, property 1 yields that {f

: α < ω

} is almost disjoint. It remains to show that properties 10 and 11 imply property 8 (which implies property 7, which implies strongly near-Luzin), and property 12 implies there are no uncountable Luzin subfamilies.

Lemma 3.5. Suppose A = {f

: α < ω

} and {N

: α < ω

} satisfy 1, 9, 10, 11 and 12. Then

(a) A has property 8.

(b) A has no uncountable Luzin subfamily.

P r o o f. (a) We show by induction on n that if β

< . . . < β

and sup S

< inf S

for each i then T

i≤n

S

γ∈S_βi

f

is fat. So suppose β

<

. . . < β

, sup S

< inf S

for all i, and C = T

i<n

S

γ∈S_βi

f

is fat.

Let α < inf S

with β

∈ N

. Then C ∈ N

. Let {γ

: j < ω} ⊂

S

with γ

< γ

. Define {E

: j < ω} and {C

: j < ω} as follows:

E

= {k : (k, f

(k)) ∈ C}, C

= π

C, E

= {k : (k, f

(k)) ∈ C

}, C

= π

C

. By property 9, C

, E

∈ N

for all i. By property 10, each C

is fat.

But then, by property 11, each C

∩ f

is infinite, so since A is almost disjoint and C

⊃ C

, C ∩ S

{f

: j < ω} is fat.

(b) Given an uncountable subfamily B of A and an enumeration B = {g

: α < ω

}, where g

= f

, there are α ≤ β < ω

so that S

⊂ ran φ∩β and φ

β = ran φ ∩ β. Let δ = φ(β) ≥ β. Note that α ∈ N

⊂ N

⊂ N

, so, by property 12, there is m such that {γ ∈ S

: f

∩ f

⊂ m × m} is infinite.

Since g

= f

and φ

S

⊂ β it follows that B is not Luzin.

Now notice that the construction used in the proof of 2.5 easily adapts to a construction of a family with properties 1, 9, 11, and 12. To get property 10, r

is required to satisfy

|π

C

| > j, which can be done because C

is fat.

4. Trees and anti-Luzin families. The canonical example of an anti- Luzin family is a set of branches of a countable perfect tree, i.e. a countable tree such that there are two incomparable nodes above every node. What about the reverse? Must every anti-Luzin family look like the branches of a tree?

Definition 4.1. An uncountable almost disjoint family A is a tree family iff there is a tree ordering T = ( S

A, ≺) so that for every a ∈ A there is a branch b of T with a =

b.

We will show that under CH + “there exists a Suslin line” there is an anti-Luzin family which contains no uncountable tree families.

Question 4.2. Is there (under ZFC alone) an anti-Luzin family which contains no uncountable tree families?

We do not know the answer to this question, but we have a related MA + ¬CH result.

Definition 4.3. An almost disjoint family A is a hidden tree family iff for some infinite T ⊂ S

A the set {a ∩ T : a ∈ A} is a tree family.

Hidden tree families need not be anti-Luzin. For example, let A = {a

: α < ω

} be a tree family on the set of even integers, and let B = {b

: α < ω

} be Luzin on the set of odd integers. Then {a

∪ b

: α < ω

} is both Luzin and a hidden tree family.

In fact, under MA + ¬CH all uncountable almost disjoint families of size

< 2

are hidden tree families.

Theorem 4.4. Assume MA(precaliber ω

). Then every uncountable al- most disjoint family of size < 2

on ω is a hidden tree family.

P r o o f. We define p ∈ P: p = (T

, ≺

, A

, h

) where

• T

∈ [ω]

, (T

, ≺

) is a finite tree,

• A

⊂ A is finite,

• h

: A

→ ω is a function,

• each (a \ h

(a)) ∩ T

is linearly ordered by ≺

,

• ∀a ∈ A

∀n ∈ a ∩ (T

\ h

(a)) ∀k ∈ T

\ h

(a) if k ≺

n then k ∈ a.

We define p ≤ q iff

(a) (T

, ≺

) is an initial subtree of (T

, ≺

), (b) A

⊂ A

,

(c) h

⊂ h

.

P is easily seen to have precaliber ω

.

Subclaim 4.4.1. For each a ∈ A, the set D

= {p ∈ P : a ∈ A

} is dense in P.

P r o o f. Assume that a 6∈ A

. Let q = (T

, ≺

, A

∪ {a}, h

∪ (a, n)), where n > max T

. Then q ∈ P because a ∩ (T

\ h

(a)) = ∅. The relation q ≤ p is clear.

Subclaim 4.4.2. For each n ∈ ω and a ∈ A,

D

= {p ∈ P : a ∈ A

and a ∩ (T

\ n) 6= ∅}

is dense in P.

P r o o f. Let p ∈ P. By Subclaim 4.4.1 we can assume that a ∈ A

. Let k ∈ a \ [

(A

\ {a}) \ max{T

∪ n + 1}.

Let q = (T

∪ {k}, ≺

, A

, h

), where ≺

⊂ ≺

and l ≺

k for each l ∈ a ∩ (T

\ h

(a)). Since k 6∈ a

for a

∈ A

\ {a} and a ∩ (T

\ h

(A) is linearly ordered by ≺

we have q ∈ P and clearly q ≤ p.

Let

D = {D

: a ∈ A, n ∈ ω}.

By MA(precaliber ω

) we have a D-generic filter G. Then T = (T, ≺) =  [

p∈G

T

, [

p∈G

≺



witnesses that A is a hidden tree family: taking h = S

p∈G

h

we see that

(T ∩ a) \ h(a) is a tail of a branch of T .

In contrast we get under CH + “there exists a Suslin line” an anti-Luzin family which has no uncountable hidden tree families. In fact, we get some- thing stronger.

Definition 4.5. An uncountable almost disjoint family A is a weak tree family iff there is a tree ordering T = ( S

A, ≺) and a 1-1 function φ : A → Br(T ) (here Br(T ) is the set of branches of T ) where range φ is pairwise disjoint, and each a ⊂

φ(a).

A is a hidden weak tree family iff, for some T , {a ∩ T : a ∈ A} is a weak tree family.

Weak tree families appeared in [V] where they are called neat families.

Velickovic proved the following result (Lemma 2.3 of [V]): Assume MA

. If A ⊂ [ω]

is an almost disjoint family then there is an uncountable family B ⊂ A and a partition b = b

∪ b

for each b ∈ B such that B

= {b

: b ∈ B}

is a weak tree family for i ∈ 2.

Remark. One can consider the following weakening of the notion of weak tree families. An uncountable almost disjoint family A is a very weak tree family iff there are a tree ordering T = ( S

A, ≺) and a function φ : A → [Br(T )]

such that the range of φ is pairwise disjoint and each a ⊂

S

φ(a).

A is a hidden very weak tree family iff, for some T , {a ∩ T : a ∈ A} is a very weak tree family.

However, as observed by the referee, a hidden very weak tree family can be split into countably many hidden tree families: for every element x of A fix a node of the tree such that above this node x is covered by a single branch, and split A accordingly.

Theorem 4.6. Assume CH + “there exists a Suslin line”. Then there is an uncountable anti-Luzin almost disjoint family which contains no uncount- able hidden weak tree families.

First, a quick lemma.

Lemma 4.7. Let T

be Aronszajn, and B an uncountable set of branches of T

so that no two elements of B have the same order type. Then there are incompatible elements s, t ∈ T

so that {b ∈ B : s ∈ b} and {b ∈ B : t ∈ b}

are uncountable.

P r o o f. By contraposition, suppose B is a set of branches of T

with different order types so that if s and t are incompatible then either B(s) = {b ∈ B : s ∈ b} is countable or B(t) = {b ∈ B : t ∈ b} is countable. Then S = {s : B(s) is uncountable} forms a chain, hence is countable. So there is α with T

(α) ∩ S = ∅, where T

(α) is the set of elements of T

of height α. But all but countably many elements of B are in S

s∈T^∗(α)

B(s), so B is

countable.

Proof of Theorem 4.6. Let T

be a Suslin tree so that every element has successors at arbitrarily high levels, and for each t ∈ T

construct c

∈ [ω]

so that if s < t then c

⊃

c

and if s, t are not comparable then c

∩ c

=

∅.

Let B be an uncountable set of branches of T

so that no two elements of B have the same order type and every element of S

B is in uncountably many branches of B.

Let B = {b

: α < ω

}. We will define A = {a

: α ∈ ω

} where:

13. For all s ∈ b

, c

⊃

a

. A will clearly be almost disjoint.

Let {T

= (T

, ≺

) : α < ω

} enumerate all perfect trees whose under- lying set is some infinite subset of ω.

We further require:

14. For all β < α either for some s ∈ b

, c

∩ T

is contained, mod finite, in a branch of T

, or a

∩ T

is not a subset, mod finite, of a branch of T

.

The family A is constructed recursively in ω

steps. In the αth step we apply Lemma 4.8 below to get a

.

Lemma 4.8. Suppose {a

: β < α} has property 13. Then there is a set a with properties 13 and 14.

P r o o f. Rather than describe the proof as an induction, we will (equiv- alently) use the Rasiowa–Sikorski lemma (see [K, Theorem 2.21]), defining a countable set of forcing conditions and countably many dense sets so that any generic filter meeting the dense sets gives rise to the desired object.

The partial order is as follows: P consists of all pairs p = (a

, b

) where a

is a finite subset of ω and b

is a finite subset of b

. The order is as follows: p ≤ q iff a

⊃ a

, b

⊃ b

, and a

\ a

⊂ T

t∈bq

c

.

Clearly {p : |a

| > n} and {p : s ∈ b

} are dense for each n < ω and s ∈ b

, so if G is a filter meeting each of these dense sets then S

p∈G

a

has property 13.

Towards property 14, fix T = T

and T = T

. We may assume that for every t ∈ b

, c

∩ T is not a subset, mod finite, of a branch of T . For each n < ω define

D(T , n) = {p : a

\ n contains T -incomparable elements}.

We show that D(T , n) is dense for each n.

Fix q 6∈ D(T , n). Let c = T

t∈bq

c

. Since c ∩ T \ n is not a subset, mod finite, of a branch of T , there are two ≺

-incompatible elements, ξ and η, of T ∩ (c \ n). Set a

= a

∪ {ξ, η} and b

= b

. Then p ∈ D(T , n) and p ≤ q.

If, for all n, G meets D(T , n), S

p∈G

a

∩ T will not be a subset, mod

finite, of any branch of T .

The following two lemmas, once proved, will complete the proof of The- orem 4.6.

Lemma 4.9. If property 14 holds, then A has no uncountable hidden weak tree families.

P r o o f. This is where we use the fact that T

is Suslin.

So suppose B is an uncountable subset of A, T ⊂ ω is infinite, and C = {a ∩ T : a ∈ B} is a collection of infinite sets. We show that C is not a weak tree family.

Let T = (T, ≺) and φ : C → Br(T ). We show that φ does not have the properties of Definition 4.5.

For some β, T = T

.

Let S = {s ∈ T

: c

∩ T is a subset, mod finite, of a branch of T }. If S = ∅, then by property 14 for all but countably many a ∈ B, a ∩ T 6⊂

φ(a ∩ T ).

Suppose s ∈ S and c

∩ T ⊂

d, where d is a branch of T . Then if s ∈ b

we have a

∩ T ⊂

d. Let S

be the set of T

-minimal elements of S. Since T

is Suslin, S

is countable. So there is s ∈ S

with {b

: a

∈ B and s ∈ b

} uncountable. But then either there are uncountably many α with a

6⊂ φ(a

∩ T ), or φ is not 1-1.

Lemma 4.10. If property 13 holds, then A is anti-Luzin.

P r o o f. Suppose B is an uncountable subset of A. By Lemma 4.7 there are incompatible s, t ∈ T

with C = {b

∈ B : s ∈ b

} and D = {b

∈ B : t ∈ b

} both uncountable. Without loss of generality, we may assume that for some n ≥ sup c

∩ c

if b

∈ C then a

\ c

⊂ n and if b

∈ D then a

\ c

⊂ n. But then S

C ∩ S

D ⊂ n, as desired.

5. Between near-Luzin and strongly near-Luzin

Definition 5.1. An uncountable almost disjoint family A is k-near- Luzin iff for every C

, . . . , C

∈ [A]

, T

i<k

S C

is infinite.

The purpose of this section is to show that these notions are (consis- tently) distinct.

Clearly near-Luzin is 2-near-Luzin and so every Luzin family is 2-near- Luzin, but does not necessarily contain a 3-near-Luzin subfamily, as we will see in Theorem 5.9.

Theorem 5.2. The following is consistent: for each k ∈ [2, ω) there is an

uncountable almost disjoint family A

which is k-near-Luzin, contains no

uncountable Luzin subfamilies, and contains no uncountable (k + 1)-near-

Luzin subfamilies.

The proof proceeds by showing that for every k there is a partial order P

with precaliber ω

forcing A

to exist, and iterating with precaliber ω

. It is easy to see that both k-near-Luzin and “no Luzin subfamilies” are preserved by precaliber ω

forcing. The way we ensure no (k + 1)-near-Luzin subfamilies will also be preserved by precaliber ω

forcing.

In contrast to our earlier constructions, |A

| = 2

for each k. Each A

is again a family of functions, but instead of functions on ω the domains come from a k-linked not (k + 1)-linked family E

with special properties. This family was first constructed by Hajnal; the construction appeared in [JS].

Lemma 5.3. For all k < ω there is a family E

⊂ [ω]

with |E

| = 2

so that:

15. If e

, . . . , e

∈ E

then T

i<k

e

is infinite.

16. If e

, . . . , e

are distinct elements of E

then | T

i≤k

e

| < ω.

17. If X ∈ [E

]

then there are X

, . . . , X

∈ [X]

with  

 \

i≤k

[ X

   < ω.

P r o o f. Let S

= {[

2]

: n < ω}. We will construct E

as a subset of [S

]

; since |S

| = ω, this construction proves the lemma. For f : ω → 2 let e

= S

n<ω

{s ∈ [2

]

: f ¹n ∈ s}. Let E

= {e

: f ∈ 2

}.

We show that property 15 holds: Given f

, . . . , f

distinct, pick m so {f

¹m : i < k} are distinct. But then for each n ≥ m,

{f

¹n : i < k} ∈ \

i<k

e

.

We show that property 16 holds: Given f

, . . . , f

distinct, there is m so that {f

¹m : i ≤ k} are distinct. But then T

i≤k

e

⊂ S

j<m

[2

]

.

We show that property 17 holds: Fix X ∈ [E]

. Let I = {f : e

∈ X}

and let g

, . . . , g

be distinct complete accumulation points of I in the usual topology on 2

. Fix n so that g

¹n, . . . , g

¹n are distinct. Define X

= {e

∈ E : f ¹n = g

¹n}. Then

\

i≤k

[ X

⊂ \

j<n

[2

]

.

Note that by construction properties 15 and 16 are absolute in the fol- lowing sense: Let E = (E

)

, and let M ⊂ N , where M, N are models of enough set theory. Then 15 and 16 hold for E in N .

The next lemma says that property 17 is preserved in some models. In

this and succeeding proofs we will refer to the following easy fact about ccc

forcing:

Fact 5.4. If P is a ccc partial order, P is an uncountable subset of P and ˙ G names the generic filter, then there is p ∈ P such that p ° “P ∩ ˙ G is uncountable”.

Lemma 5.5. Suppose E has the property that if X ∈ [E]

then there are X

, . . . , X

∈ [X]

with

   \

i≤k

[ X

   < ω.

Then E will still have this property in a forcing extension by a precaliber ω

partial order.

P r o o f. Let P have precaliber ω

, and suppose °

X = { ˙x ˙

: α <

ω

} ∈ [E]

. Fix p ∈ P. For each α pick p

≤ p so that for some e

∈ E, p

° ˙x

= e

. Then there is an uncountable centered family {p

: α ∈ I}.

Let Y = {e

: α ∈ I}. By hypothesis there are I

, . . . , I

∈ [I]

and m < ω

with \

i≤k

[ {e

: α ∈ I

} ⊂ m.

Let ˙ G be the generic filter, and define

J ˙

= {α ∈ I

: p

∈ ˙ G}.

List each I

as {α

: γ < ω

}. Let q

≤ p

γ

for all i ≤ k. Let Q = {q

: γ < ω

}. By Fact 5.4 there is p ∈ P with p ° |Q ∩ ˙ G| = ω

. So p ° “ ˙ J

is uncountable” for each i ≤ k, which by a density argument com- pletes the proof.

Lemma 5.6. Suppose E ⊂ ℘(ω) satisfies the following: if X ∈ [E]

there are X

, . . . , X

∈ [X]

with

   \

i≤k

[ X

   < ω.

If A = {f

: e ∈ E} where each f

: e → ω then A has no uncountable (k + 1)-near Luzin subfamilies.

P r o o f. If for some finite m,

\

i≤k

[ X

⊂ m

then \

i≤k

[

e∈X_i

f

⊂ m × ω.

Let Y

∈ [X

]

so ∃σ

∀e ∈ Y

f

¹m = σ

. Then

\

i≤k

[

e∈Yi

f

⊂ \

i

σ

.

This completes the proof.

Proof of Theorem 5.2. Let E

⊂ [ω]

be a family satisfying 15–17 of Lemma 5.3. We have A

= {f

: e ∈ E

}, where f

: e → ω is a function for e ∈ E

. By Lemma 5.6, this assumption guarantees that A

has no (k + 1)-near-Luzin subfamily.

Define P

, a precaliber ω

forcing which adds generic almost disjoint functions f

: e → ω for e ∈ E

, as follows:

• elements of P

have the form p = {σ

: e ∈ E

},

• E

is a finite subset of E

,

• each σ

is a finite function from e to ω.

The order is: p ≤ q iff

• E

⊃ E

,

• for e ∈ E

, σ

⊃ σ

,

• for e 6= e

∈ E

, σ

∩ σ

= σ

∩ σ

.

P

is easily seen to have precaliber ω

. We define p ° ˙ f

(i) = j iff e ∈ E

and σ

(i) = j. By a standard genericity argument, °

[dom ˙ f

= e and if e 6= e

then | ˙ f

∩ ˙ f

| < ω].

Lemma 5.7. °

“ ˙ A

is k-near-Luzin”.

P r o o f. Working in V

, suppose that for each i < k we have an un- countable subset ˙ X

of E

. We want to show that T

i<k

S { ˙ f

: e ∈ ˙ X

} is infinite.

We may assume that the ˙ X

’s are disjoint, and each ˙ X

has a 1-1 enu- meration { ˙e

: α < ω

} . Fix p ∈ P

. For each α there is p

≤ p, for all α, i there is d

so that

p

° ∀i < k ˙e

= d

, and for each α the d

’s are distinct.

We may assume the p

’s are centered. Then, since the enumeration is 1-1, for each i we have d

6= d

if α 6= β.

Pick distinct α

, . . . , α

so that d

6= d

for i 6= j and (by a

∆-system argument) d

6∈ E

. By property 15,  

 \

i<k

d

 = ω.

Let m ∈ T

i<k

d

, m > sup dom σ

for all i and all e ∈ E

. There is q < p

for all i with σ

(m) = 0. So q ° ∃m > n (m, 0) ∈ T

i<k

S { ˙ f

: e ∈ ˙ X

}. A density argument completes the proof.

Lemma 5.8. °

“ ˙ A

has no uncountable Luzin subfamilies”.

P r o o f. Suppose { ˙a

: α < ω

} ⊂ A

and p ° “the enumeration { ˙a

: α < ω

} witnesses that the family is Luzin”. Choose p

≤ p and d

∈ E

with p

° ˙e

= d

∈ E

. We may assume that:

18. The p

’s are centered.

19. {E

: α < ω} is a ∆-system with root E.

20. ∃n ∀α E

= {e

: i < n}.

21. ∀i ∃σ

∀α σ

= σ

. By necessity,

22. ∃m ∀i σ

⊂ m × m.

There is q ≤ p

and k such that q ° ∀i > k ˙a

∩ ˙a

6⊂ m × m. By property 19 there is j ≥ k with E

∩ E

= E.

We define r ≤ q:

• E

= E

∪ E

,

• for e ∈ E

, σ

= σ

,

• for e ∈ E

\ E, σ

= σ

.

Then r ° ˙a

∩ ˙a

= ˙a

∩ ˙a

⊂ m × m, a contradiction. Theorem 5.2 is proved.

Finally, we note that Luzin does not imply 3-near-Luzin:

Theorem 5.9. There is a Luzin almost disjoint family with no uncount- able 3-near-Luzin subfamily.

P r o o f. Let E = E

be as in Lemma 5.3. As in Theorem 5.2, we construct A = {f

: e ∈ E} where dom f

= e for all e, so that A has no uncountable 3-near-Luzin subfamily. Here is how we get Luzin.

Let E = {e

: α < ω

} and f

= f

. Our induction hypothesis at stage α is that for all β < α and all n < ω, {γ < β : f

∩ f

⊂ n × ω} is finite.

This will certainly give us Luzin.

At stage α fix a 1-1 enumeration {β

: n < ω} of α. In the nth step of the construction of f

we ensure that dom f

∩ n = e

∩ n and f

∩ f

6⊂ n × ω, without increasing f

∩ f

for m < n. Since e

∩ e

\ S

{e

: m < n} is

finite, this can be done, and the construction is complete.

References

[AT] U. A b r a h a m and S. T o d o rˇce v i´c, Partition properties of ω

compatible with CH , Fund. Math. 152 (1997), 165–181.

[EF] F. E c k e r t s o n, W. F l e i s s n e r, A. K o r o v i n and R. L e v y, Not realcompact images of not Lindel¨of spaces, Topology Appl. 58 (1994), 115–125.

[HJ] A. H a j n a l and I. J u h ´a s z, Intersection properties of open sets, ibid. 19 (1985), 201–209.

[JN] I. J u h ´a s z, Zs. N a g y, L. S o u k u p and Z. S z e n t m i k l ´o s s y, Intersection properties of open sets, II , in: Papers on General Topology and Applications, Proc. 10th Summer Conference in General Topology and Applications, 1994, E. Coplakova and K. P. Hart (eds.), Ann. New York Acad. Sci. 788, New York Acad. Sci., 1996, 147–159.

[JS] I. J u h ´a s z, L. S o u k u p and Z. S z e n t m i k l ´o s s y, Combinatorial principles from adding Cohen reals, in: Logic Colloquium 95, Haifa, to appear.

[K] S. K o p p e l b e r g, Handbook of Boolean Algebras, Vol. I, North-Holland, Amster- dam, 1987.

[L] N. L u z i n, On parts of the natural series, Izv. Akad. Nauk SSSR Ser. Mat. 11 (1947), 403–410 (in Russian).

[V] B. V e l i c k o v i c, OCA and automorphisms of P(ω)/fin, Topology Appl. 49 (1993), 1–13.

Department of Mathematics University of Kansas Lawrence, Kansas 66045 U.S.A.

E-mail: roitman@math.ukans.edu

Mathematical Institute Hungarian Academy of Sciences POB 127 H-1364 Budapest, Hungary E-mail: soukup@math-inst.hu

Received 24 July 1997