T H E J O U R N A L OF SYMBOLIC L O G I C
Volume 44, N u m b e r 3, Sept. 1979
ALMOST DISJOINT SETS AND MARTIN'S AXIOM
MICHAEL L.
WAGE1
Abstract. We present a number of results involving almost disjoint sets and Martin*s
axiom. Included is an example, due to K. Kunen, of a c.c.c. partial order without property
K whose product with every c.c.c. partial order is c.c.c.
Two sets are called almost disjoint if their intersection is finite. Our first theorem
is a basic Martin's axiom result on almost disjoint sets. The case |A'j = co is well
known and is a corollary of the theorem on p. 154 of [MS].
m
THEOREM 1 (MA). Suppose co < £ < 2 and {Ta: a < £} is a collection of countable subsets of a set X such that \ Ta f] T$ | < cofor all a # B. Then there exists a
set M c Xsuch that | M \ = | X \ and \ M f] Ta \ < cofor all a < f
PROOF. Fix X and {Ta: a < £} as in the hypothesis and let K = | X\ . For
convenience we assume that X = /r = {a: a < /r}. The theorem is clearly true if
K < co or £ < /:, so we assume that w < K < | . Let P = {(F / ) : f c {F<Ja<i-,
I F | < © , / c «, | / | < ©}. For ( F / ) , (G, g) e P define ( F , / ) < (G, #) if and only
ifF=> G , / = > g , a n d ( / - g ) fl ( Q G ) = 0 Before we can apply Martin's axiom we must show that (F, < ) satisfies the
countable chain condition (abbreviated c.c.c). Aiming for a contradiction, assume
that {(Fa, fa): a < coi} is a pairwise incompatible subset of P. In the present situation this means that for each a, 8 < coi with a # 3 we have either ((J Fa) H
(/*\/«) * 0 or ( ( J F J ) n (fa\ffi)*0% ^ e J-system lemma [Ju, Appendix 2],
there exists an / a n d uncountable B c w\ such t h a t / a f| fe = / f o r all a, 8e P.
Note that for a, p e P, we have (F a , /„) and (F^, /^) incompatible if and only if
(F„,/ a — / ) and (F$, / 3 — / ) are incompatible. Hence, without loss of generality,
we will assume fa 0 ^ = 0 for all a, 3 e u>\, a ^ p. Since each / a is finite, we
can also assume that | fa | = \f$ | = jfor all a, j3 6 coi- Another application of the
_j-system lemma yields a set F and uncountable set P c coi such that Fa f] F^ = F
for all a, ,3 e B,a*3. Let F ' = {a e P : / a f] (1J F ) = 0 } - S i n c e t h e A's are disjoint and (jFis countable, | B' j = coi- But now for a, 8 e B' it is true that (F a , / J
and (F^, /^) are incompatible if and only if (Fa — F, fa) and (F$ — F, /^) are
incompatible. Thus, without loss of generality we assume that F 0 fl F^ = 0 for
all a, S e coi- Now for each n e co it follows that (\JF„) f] fa = 0 for all but
countably many aeco\. Thus for all but countably many a e coi we have ((jF a ) f)
/„ ?*= 0 for all «eco. Since \f„ \ = / for all rceco, this implies that | (\jFr) f]
(U^"«) I = ^ f° r some 7-, oecoi, 7"#o. This is a contradiction since F r fl F5 = 0
Received August 1, 1975; revised September 1, 1978.
'This research was partially supported by the Institute for Medicine and Mathematics, Athens,
Ohio, and the National Science Foundation, Grant ;MCS77-07731.
313
© 1979 Association for Symbolic Logic
OO22-48I2/79/4403-OOO3/S02.50
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314
MICHAEL L. WAGE
and the F's were assumed to be almost disjoint. Therefore, (P, < ) is c.c.c.
Define Ua = {(F,f) :TaeF} for each a < £, tfW K- = {(F,f) :f [)(($ +wi)
- 3 ) # 0 } for each p < K. Each £/a is dense in (P, < ) since it is always true that
( F U {7;},/) < (F,/). Given any (F.f)ePand
-3 < K. there exists a j e((p + wi)
- ,3) such that r 4 (jF, so that ( F , / U M ) ^ (*"»/)• s i n c e ( ^ / U M ) e ^ , this
shows that each V^ is dense in (P, < ). Now Martin's axiom guarantees the existence of a filter g? c P such that £/„ fl 2? # 0 for all a < 6 and V$ f] S? # 0
for each /3 < /r. We claim that M = (J {/: ( F , / ) e S? for some F } is the desired set
of cardinality K. Given any a < £ there is a (G, g) e ^ such that Ta e G. Since
# is pairwise compatible, this yields
I Ta n ^ I < I g ! 1 M | = | g I < <u.
For each jS < «, K^ f) S? # 0 . so M f| ((£ + &n) — p) # 0 also. Since /c was
assumed infinite, this implies \M\ = K and the proof is complete.
•
Note that R. Laver has given an argument similar to the above in [L, p. 1035].
The following theorem is a stronger version of Theorem 1. We do not give the proof
since Theorem 1 is all that is needed in the rest of this note and the proof of Theorem 2 is similar to that of Theorem 1. To prove the c.c.c. in Theorem 2, one uses
the fact that, under MACNJ),
("•WT
'
which is a special case of the lemma on p. 1030 of [L].
THEOREM 2 (MA(/c)). Let stf and & be families of sets such that:
(1)1 J ^ I < rcand\gg\ < K,
(2)\X\ <
wforeachXesJ,
(3) if S/Q is an uncountable subset of si, then | f]ss/o I is finite, and
(4) ifX e 3$ and 3? is a finite subset of si, then X <£ \)3F.
Then there is a set S a \j<% such that S fl X is finite for each X e s-J and S fl Y
=£ 0 for each F e ^ .
Consider the following statements concerning a partial order P.
(a) Every subset of P of cardinality o)X is the union of a> filters. (A set X <= P is
called a. filter if every finite subset of X is compatible.)
(b) Every uncountable subset of P contains an uncountable filter.
(c) Every uncountable subset of P contains an uncountable set of pairwise compatible elements.
(d) If s is any sequence of elements of P of length u>\, then 5 has a subsequence
s' of length a>\ such that, for some n < co, every subsequence of s' of length o> + n
contains a pair of compatible elements.
(e) P x O is c.c.c. for every c.c.c. Q.
(f) P is c.c.c.
Then (a) -» (b) -» (c) -» (d) -> (e) -+ (f). Only the fourth implication is nontrivial.
It follows from the following lemma, which is probably well-known folklore. The
proof uses the partition relation o>i -* (a^, w + l) 2 , a special case of [ER, Corollary l , p . 459].
LEMMA 1. If O is a c.c.c. partially ordered set and n < w, then every sequence of
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ALMOST DISJOINT SETS AND MARTIN'S AXIOM
315
elements of Q of length co\ has a pairwise compatible subsequence of length co + n.
Assuming MA(coi), each of the above implications can be reversed [Ju, Theorem 5.3, p. 60]; assuming the continuum hypothesis, the first, second, third and
fifth implications cannot be reversed. (What about the fourth?) A counterexample
to the reversal of the second implication is the collection of all compact subsets of
the reals that have positive Lebesgue measure, ordered by inclusion, see [KT].
Counterexamples to the fifth were given by Laver and Galvin, [G]. The following
(unpublished) theorem of Hajnal implies that, assuming CH, the first implication cannot be reversed.
S
THEOREM 3 (HAJNAL). There is a partially ordered set P of cardinality 2 ° such
that:
(1) Every uncountable subset of P contains an uncountable filter, and
(2) P is not the union of Ho par ise compatible sets.
Ken Kunen pointed out thai me method of proof of Theorem 1 can be used to
show that (d) •** (c). We need a lemma before presenting Kunen's example.
Let £ be arbitrary and let {Ta: a < £} be a collection of countable subsets of a
set X with pairwise finite intersection. Let P be the partial order defined from this
collection as in the proof of Theorem 1.
LEMMA 2. P satisfies condition (d) above.
PROOF. Let S = (,(Fa, fa): a < a;i> be a sequence of elements of P of length w\.
As in the proof of Theorem 1, we can assume t h a t / a f] f$ = Fa f] F$ = 0 and
\fa I = l/fli = ' for each a, ,3 e a ) , , « # p\
Itt
S' = <(FaJa): Vp < a(\jFp)
f]fa * 0>-
Note that S' is uncountable, for if not, by the pressing down lemma [Ju, Theorem
A1.2] there would be a, /3 such that
OjFp) D /„ # 0
for uncountably many a. This would contradict the facts that [JFa is countable
and/ a f| fa> = 0 f ° r all a, a' e cob a j= a . Let n = / + 1 ( = \fa \ + 1). We must
show that an arbitrary subsequence <(G„, ga): a e w + n) of 5" of length co + n
contains a pair of compatible elements. If no two elements of <(Ga, ga): a e co + »y
are compatible, then g. fj ( I j G ^ ) # 0 for each j < co and k < n. Since each
gj had cardinality /" ( = n — 1) and there are n different G^*, this implies that
d j G ^ t ) H (UG^y) is infinite for some distinct k, k' < n. This contradiction
proves the lemma.
•
A partial order that satisfies condition (c) above is said to have property K
(for Knaster—see [KT]).
THEOREM 4 (KUNEN). (CH) There is a c.c.c. partial order P such that
(i) P satisfies condition (d) above, but
(ii) P does not have property K.
PROOF. P will be an instance, under CH, of the partial order in Lemma 2. X will
be R and $ will be co\. Let Xa (a < oil) enumerate R and Ya (a < co\) enumerate
the collection of all countably infinite subsets of R. For each a < co\, choose Ta =
{ i ; : n e co} so that the sequence (1%: n e m ) converges to xa and V/3 < a [xa is a
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316
MICHAEL L. WAGE
limit point of Y^ -> T f| Y$ # 0 ] . Since <?2 : H e o>> converges to x a , r„ f| 7^ is
finite for each a, j3 < coi, a # /3. By Lemma 2, P satisfies condition (d). To prove
that P does not have property K, set Pa - ({Ta}, {xa}) for each a < o)\. Let S be
an arbitrary uncountable subset w\. Fix /3 so that Y^ c {x a : a e S } c c l ^ ) and
fix a e S, a > /3 such that xa i Y$. Then Ta f) Y$ # 0 , so there exists a 7- e S with
x r € Ta. But now /?„ and pT are incompatible. D
Note that Kunen's Theorem 4 is intimately related to the following theorem of
Hajnal (which is essentially the case a — 0 of Theorem 1 in [H, p. 278]). A function
/ : X -* 0>(X) is called a set mapping if V„v e X, x 4f(x). A subset M <= X is called
free with respect to a set mapping,/, if Vx, y € M(x £f(y) A j>£/(x)).
THEOREM 5 (HAJNAL). (CH) 77zere « a set mapping/: a>i -> ^(coO -rac/* ?/zar
( l ) / ( a ) c a/o/ - a// a € <wi,
(2) / ( a ) |~) /(/3) is finite whenever a ^ / 3 , cnrf
(3) every subset ofcoi which is free with respect to f is countable.
Kunen's Theorem 4 can be derived from Hajnal's Theorem 5 as follows. L e t /
be a set mapping satisfying the conditions of Hajnal's theorem and let P be the
family of all finite subsets of w1 which are free with respect t o / . Partially order P
by reverse inclusion. Then P is a counterexample to (d) -> (c). The uncountable
subset {{a}: a e a j j has no uncountable pairwise compatible subset, yet an easy
^-system argument shows that P satisfies (d). Next let P be as in the proof of
Kunen's Theorem 4. Sttpa = ({Ta}, {xa}) for each a e co\. Fix a function <f>: w\ -*
wi such that /3 < a -*-0(/3) < 6(a) and x^a) i TW). Define/(a) ={/3 < a: xW) e
T^fa)} = {|3 < a: JPCKS) is incompatible with /*<$(#}. This / trivially satisfies conditions (1) and (2) of Hajnal's Theorem 5 and satisfies (3) because the set {pa:
a < coi} has no uncountable pairwise compatible subsequence.
Kunen's original proof of Theorem 4 was model theoretic. His proof that P
satisfies (i) was extremely short: the proof that P is c.c.c. given after Theorem 1
can be carried out in any generic extension of the universe V°, so if Q is c.c.c.
by [J, Lemma 86], so is P x Q. We can transform Theorem 4 into a theorem on
topological spaces by considering the Stone space of the Boolean algebra above.
THEOREM 6 (KUNEN). (CH) There is a c.c.c. topological space X such that
(i') X x Y is c.c.c. for each c.c.c. space Y, but
(ii') X does not have property K, i.e. there exist open subsets of X, Ua, (a < o>i)
such that
- . 3 5 c «,[ I S\ = a>! A Va, j8 e S Ua C\ U?*
0].
We next prove a corollary of Theorem 1 that has been used in [DRW] to construct a Moore space with a weakly uniform base that has no point countable base.
Theorem 7 (MA + —1 CH). There exists a collection, 3tf, of subsets of o)2, and a
partition, {jfa: a e 0)2}, ofj>!f such that:
(1) H fl K is finite for each
H,Ks^,H^K,
(2) if M <r Q)z with \ M | = o>j, then there exists an He 3^ with \H [\ M \ = co\,
and
(3) if M c o>2 with \ M\ = a)2 and a e 02, then there exists H e 3^Ca with
\Hf\M\=a>1.
PROOF. Enumerate all subsets of a>2 of order type cm as < 5 r : y e 2<">. We will
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ALMOST DISJOINT SETS AND MARTIN'S AXIOM
317
construct sets Ha for each a < 2<° by induction. Suppose we have denned Ha for
all a < p < 2W such that
(a) | Ha C: H51 < a for all a, 5 < 3, a ¥= 5,
(b) for each y < 3, there exists an a < 3 such that | ST f] Ha \ = coi, and
(c) # a c 5 a for all a < 3.
If there exists a < ,3 with | Ha f) &}] = a>i, then let # ? = 0 . If | #„ p; •% | < a>i,
then Theorem 1 applies (with Ta = Ha f, Sp, £ = /3, and X = S^) to yield a set
Af c S 3 with | Af | = wi and | Ha f] M \ < w for all a < 3. Define /fg = M.
By [U], we can partition {a e a;2: cf(a) = <o\} into a>2 stationary pieces, {Aa :
a e (oz). For a < a>2 define .Wa = {//s: sup Hp e Aa}. Let 3V = \J {jfa: a e cu2}.
Since the ,4a's are disjoint, the J^ a 's do indeed partition j ^ . Moreover, by construction, (1) and (2) are satisfied. To prove (3), assume that M c w2 w ^ h I Af I = <^2
and a e a>2- Since Aa is stationary, there exists some element X e cl(Af) f] y^. Thus
Af contains a set S, of order type o>i, with sup S = A. By (2), there is an /f e ^
with | /f f] S | = cui- But now sup H = X also, so by definition of yfa, H sj^a.
Since of course | / / f> M | = on also, this completes the proof.
•
F. Galvin has noticed that the existence of a set jf as in the theorem above
implies the negation of the continuum hypothesis. To see this, assume both the
continuum hypothesis and the existence of such an Jtf. By induction we construct
a set M c co2 such that | M | = co2, yet for all H e tf, \ H f] M | < a>. Let m0 = 0.
Assume that we have defined ma for all a < fl < o,2 s u c n t n a t l # (1 {»V a < /3}[
< w for all Heji?. Let
S={W6#:!//nK:a<|3}|=4
Since the //'s are almost disjoint, and j3 < a>2, CH implies that | S | < a,]. Hence
we can choose m$ e co2 such that m^ > sup H for each H eS. Since now | i / f|
{mtt: a < ,3} | = a> implies H f] {ma: 3 < a < a>2} = 0 , M = {«v- a < <D2}
is the desired set.
In Theorem 8 we strengthen Theorem 7 under the additional assumption 2<" =
o>2. Note that Baumgartner has constructed a family of countable sets with properties analogous to those below, see [EH, pp. 278-279].
THEOREM 8 (MA + 2°> = o>2). There exists a collection, jf, of subsets of a>2, and
a partition, \^Ca: a < o>2}, of #f such that:
(1) tp(//) = <ui for each Hejf,
(2) H fl K is finite for each H, Kej^,H
# K,
( 3 ) ; / M c o)2, e/f/ier JW <= (J-Jf /or jome countable <g cz JP, or else 3N c M
(] N | = <wi A Va < a,2 lHzjraH
c AT).
PROOF. Let {/{,.: 7- < a>2} enumerate all subsets of <y2 of cardinality wj and let
0 : a,2 -• a,2 x o)2 be a bijection. Recursively choose a collection, {Hf. jS < co2},
of almost disjoint sets of order type OJ\ such that for each 3 < (v2, H$ c R$^
(where <pi(j$) denotes the first coordinate of ^(/3)) if possible (i.e. if 3 / / c
Rr(\ H\ = a>2AVV<p\Haf\H\<
a,)). Set j>T = {/^: i/^ c R^(s)} and, for
each a < <D2, #Ca = {//^ e Jf: 0(3) = (7-, a) for some f } .
Clearly (1) and (2) are satisfied. To prove (3), assume that Af is a subset of w2
that is not contained in any countable number of elements of JF. It is not hard to
show that there exists a 7- < o>2 such that RT <=• M and Rr is not contained in any
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318
MICHAEL L. WAGE
countable number of elements of ^f. Then Va < a^S-K <= -Rr(l -R I = wi A V77 <
0-1(r> a ) I -R, (I JR I ^ <M), and thus Theorem 1 (applied to i? and {/?, f] R : rj <
(p~l(T, cc)}) implies that
V a < (x>z I H^- i ( f i a ) I =a)\.
The theorem follows.
D
I would like to thank the referee for many helpful suggestions and for informing me of Hajnal's theorems and the strengthening of Theorems
1 and 7.
ACKNOWLEDGEMENT.
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YALE UNIVERSITY
NEW HAVEN, CONNECTICUT 06520
Current address: Harvard Medical School, Harvard-MIT Division of Health Sciences and Technology, 17-512 M I T , Cambridge, MA 02139
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