Comparing the closed almost disjointness and dominating numbers
Abstract.
We prove that if there is a dominating family of size , then there is are many compact subsets of whose union is a maximal almost disjoint family of functions that is also maximal with respect to infinite partial functions.
Key words and phrases:
maximal almost disjoint family, dominating family2010 Mathematics Subject Classification:
03E35, 03E65, 03E17, 03E051. Introduction
Recall that two infinite subsets and of are almost disjoint or a.d. if is finite. A family of infinite subsets of is said to be almost disjoint or a.d. in if its members are pairwise almost disjoint. A Maximal Almost Disjoint family, or MAD family in is an infinite a.d. family in that is not properly contained in a larger a.d. family.
Two functions and in are said to be almost disjoint or a.d. if they agree in only finitely many places. We say that a family is a.d. in if its members are pairwise a.d., and we say that an a.d. family is MAD in if . Identifying functions with their graphs, every a.d. family in is also an a.d. family in ; however, it is never MAD in because any function is a.d. from the vertical columns of . MAD families in that become MAD in when the vertical columns of are thrown in were considered by Van Douwen.
We say that is an infinite partial function if it is a function from some infinite set to . An a.d. family is said to be Van Douwen if for any infinite partial function there is such that . is Van Douwen iff is a MAD family in , where is the th vertical column of . The first author showed in [3] that Van Douwen MAD families always exist.
Recall that is the least size of an unbounded family in , is the least size of a dominating family in , and is the least size of a MAD family in . It is well known that . Whether could consistently be larger than was an open question for a long time, until Shelah achieved a breakthrough in [4] by producing a model where and . However, it is not known whether can be larger than when ; this is one of the few major remaining open problems in the theory of cardinal invariants posed during the earliest days of the subject (see [5] and [2]). In this note we take a small step towards resolving this question by showing that if , then there is a MAD family in which is the union of compact subsets of . More precisely, we will establish the following:
Theorem 1.
Assume . Then there exist compact subsets of whose union is a Van Douwen MAD family.
The cardinal invariant was recently introduced and studied by Brendle and Khomskii [1] in connection with the possible descriptive complexities of MAD families in certain forcing extensions of .
Definition 2.
is the least such that there are closed subsets of whose union is a MAD family in .
Obviously, . Brendle and Khomskii showed in [1] that behaves differently from by producing a model where . They asked whether implies that . As , our result in this paper provides a partial positive answer to their question.
2. The construction
Assume in this section. We will build many compact subsets of whose union is a Van Douwen MAD family. To this end, we will construct a sequence of finitely branching subtrees of such that has the required properties. Henceforth, will mean is a subtree of .
Definition 3.
Let . Let and . For any ordinal and define to mean
Note that if and , then , and that for a limit ordinal , if , then . Also, for any , if and , then . Moreover, if and if and are such that , , and , then there is such that . Therefore, if there is with , and and there is some ordinal such that , then is some and some ordinal such that , thus allowing us to construct an infinite, strictly descending sequence of ordinals. So if with , then for any and any ordinal , . On the other hand, suppose that with . Then there is with and such that , , and , allowing us to construct with such that .
Definition 4.
Suppose , , and . Assume that is a.d. from each . Then define by .
Note the following features of this definition


for all with , if there exists such that and , then .
On the other hand, notice that if there is a function such that () and () hold when is replaced with , then must be a.d. from .
Definition 5.
is said to be an interval partition if , where , and . For , denotes the interval .
Given two interval partitions and , we say that dominates and write if .
It is well known that is also the size of the smallest family of interval partitions dominating any interval partition. So fix a sequence of interval partitions such that


for any interval partition , there exists such that .
Fix an scale such that . For each , define and by induction on as follows. If is a successor, then is any onto function, and . If is a limit, then let enumerate . Now, define and such that


.
Observe that such an must be a surjection. For each , put .
Now fix and assume that has been defined for each such that each is finitely branching and is an a.d. family in . Let enumerate , possibly with repetitions. For a tree and , denotes , and denotes . We will define a sequence of natural numbers and determine by induction on . . Assume that and are given. Suppose also that we are given a sequence of natural numbers such that


.
Let denote the member of that is right most with respect to the lexicographical ordering on . Suppose we are also given , an injection. Here is the set of all pairs such that

there are , and numbers such that


for each ,

such that


There is such that (if , this means that ). For each , such that () and () hold when is replaced there with , is replaced with , with , and with .
Assume that for each , we are also given , which we will call the active node at stage . Note that , and so . For each , let . For, , say if either or and is to the left of in the lexicographic ordering on . Let be the minimal member of . will be active at stage . The meaning of this is that none of the other nodes in will be allowed to branch at stage . Choose greater than all for such that . Let be the set of all pairs such that

there exist and a natural number such that


for each ,

such that


There is such that . For each , such that () and () are satisfied when is replaced with , is replaced with , with , and with .
Note that is always finite. Now, the construction splits into two cases.
Case I: . Put . Let be as in (7) above, and let be as in (8). Let
Here is as in (9), and is as in (10) with respect to . Now choose large enough so that and so that it is possible to pick and such that the following conditions are satisfied.

for each , , and for each ,

for each , if , then there exists such that . For each , there exists such that . For , if , then .

for each and , . For , if , then .

for each , , and , . For each , , and , if , then and .
Define as follows. For any , . For any , . This finishes case 1.
Case II: . For each , let . Let witness (7) for and let witness (8) for . Let be the set of all such that there is no so that
Here and witness (9) and (10) respectively with respect to . Choose large enough so that and so that it is possible to choose , , and , subsets of , satisfying the following conditions.

. For each , . For each , .

is the right most branch of . For each , there exists such that . For each , if , then there is so that . For each , if , then .

For each and , . For each and for each , , and for each , . For each , if , then for all , .

For each , , , and , and . For each , , and , if , then and .
For each , define . For each , set . This completes the construction. We now check that it is as required.
Lemma 6.
For each , there are infinitely many such that .
Proof.
For each put . It is clear from the construction that . If the lemma fails, then there are and with the property that for infinitely many , there is a such that and . Let be the left most node in with this property. Choose and such that , , and there is no such that and is to the left of . Note that . So is to the left of , and is not to the left of , whence . But then there is some where was active, a contradiction. ∎
Note that Lemma 6 implies that for any , there is a unique minimal extension of which is active.
Lemma 7.
is finitely branching and is a.d. in .
Proof.
It is clear from the construction that is finitely branching. Fix , with . Let . It is clear from the construction that .
Next, fix . Suppose . Let and , and suppose for a contradiction that . So there are infinitely many such that . For any , this can only happen if and for some . Put . Note that in this case . For such , let be as in (10) with respect to . So for infinitely many such , . But then for infinitely many such , , producing an infinite strictly descending sequence of ordinals. ∎
Lemma 8.
For each and , there are and such that .
Proof.
Suppose for a contradiction that there are and such that is a.d. from , for each . Let be a countable elementary submodel containing everything relevant. Put . For each , let denote , and note that and are members of . Let . Find such that for , . Since , it follows from (4) that for all but finitely many , for all , . Now, find such that and . Note that for any , () and () are satisfied when is replaced there with , is replaced with , with , and with . But now, it follows from the construction that there is such that for infinitely many , there is such that . ∎
3. Remarks and Questions
The construction in this paper is very specific to ; indeed, it is possible to show that is not always an upper bound for . A modification of the methods of Section 4 of [4] shows that if is a measurable cardinal and if , then there is a c.c.c. poset such that , and forces that and .
As mentioned in Section 1, we see the result in this paper as providing a weak positive answer to the following basic question, which has remained open for long.
Question 9.
If , then is ?
There are also several open questions about upper and lower bounds for .
Question 10 (Brendle and Khomskii [1]).
If , then is ?
Question 11.
Is ?
Regarding Question 10, it is proved in Brendle and Khomskii [1] that if is any ground model satisfying and is any poset forcing that all splitting families in remain splitting families in , then also forces that . This result suggests that Question 10 should have a positive answer, and showing this would be an improvement of the result in this paper.
References
 [1] J. Brendle and Y. Khomskii, perfect MAD families, In Preparation.
 [2] E. Pearl (ed.), Open problems in topology. II, Elsevier B. V., Amsterdam, 2007.
 [3] D. Raghavan, There is a Van Douwen MAD family, Trans. Amer. Math. Soc. 362 (2010), no. 11, 5879–5891.
 [4] S. Shelah, Two cardinal invariants of the continuum and FS linearly ordered iterated forcing, Acta Math. 192 (2004), no. 2, 187–223.
 [5] E. K. van Douwen, The integers and topology, Handbook of settheoretic topology, NorthHolland, Amsterdam, 1984, pp. 111–167.