Download Applications of Martin`s Axiom 1. Products of c.c.c. Spaces We

Applications of Martin’s Axiom
1. Products of c.c.c. Spaces
We consider the question of when the product of two c.c.c. topological spaces is
c.c.c.. The next lemma shows the question is the same for partial orders, topological
spaces, and compact Hausdorff spaces. Recall that for any partial order P “ xP, ďy
there is an associated topological space XP on P , where a neighborhood base at
p P P consists of the single open set Np “ tq P P : q ď pu. Clearly this space is not
(except for very trivial partial orders) even T1 .
Lemma 1.1. pZFCq Let κ be an infinite regular cardinal. Then the following are
equivalent.
(1) There are κ-c.c. partial orders P, Q such that P ˆ Q is not κ-c.c.
(2) There κ-c.c. topological spaces X, Y such that the product X ˆ Y is not
κ-c.c.
(3) There are κ-c.c. compact Hasdorff spaces X, Y , such that the product X ˆY
is not κ-c.c.
(4) There are complete Boolean algebras B, C such that their product B ˆ C is
not κ-c.c.
Proof. Note first that for any partial order P , P is κ-c.c. iff XP is κ-c.c. Also
XP ˆQ “ XP ˆ XQ . Suppose first that P , Q are κ-c.c. partial orders, but P ˆ Q is
not κ-c.c. Then XP , XQ are κ-c.c. topological spaces. Since P ˆ Q is not κ-c.c.,
neither is XP ˆQ “ XP ˆ XQ . So (1) implies (2).
To see (2) implies (1), let X, Y be two topological spaces which are κ-c.c., but
X ˆ Y is not κ-c.c. Let P be the collection of non-empty open sets in X ordered
by U ď V iff U Ď V . Likewise define Q. Clearly P , Q are partial orders. P and Q
are κ-c.c. partial orders since a κ antichain in P would be a κ collection of pairwise
disjoint open sets in X. Let tWα uαăκ be a κ family of pairwise disjoint open sets
in X ˆ Y . Without loss of generality we may assume that Wα “ Uα ˆ Vα is a
product of basic open sets. Let pα “ pUα , Vα q P P ˆ Q. The pα form an antichain
in P ˆ Q (since if pU, V q ď pUα , Vα q, pUβ , Vβ q then U Ď Uα X Uβ , V Ď Vα X Vβ , so
U ˆ V Ď Wα X Wβ ). So, P ˆ Q is not κ-c.c.
(4) implies (1) trivially, since Boolean algebras are partial orders. For the other
direction, let P , Q be as in (1). Let B be the regular open algebra of P , and likwise
define C. Thus, B and C are complete Boolean algebras. Since every element of B
(i.e., every regular open subset of XP ) contains a basic open subset of XP , it is
clear that B is κ-c.c., and likewise for Q. For p P P let Bp “ intclpNp q P B and
likewise define Cq P C for q P Q. Let tppα , qα quαăκ be an antichain in P ˆ Q. Let
Bα “ Bpα , and likewise for Cα . Then the pBα , Cα q form an antichain in B ˆ C.
For suppose pB, Cq ď pBα , Cα q, pBβ , Cβ q. We may assume B “ Bp , C “ Cq . So,
Bp Ď Bpα X Bpβ . This says Npα and Npβ are dense below p. Thus, there is an
r P P with r ď pα , pβ . Similarly, qα and qβ are compatible. Hence ppα , qα q and
ppβ , qβ q are compatible, a contradiction.
(3) trivially implies (2). To finish we show that (1) implies (3). Given P and Q,
let B and C again be their completions. Let X 1 be the Stone space of the Boolean
algebra B, that is, X 1 is the collection of ultrafilters on B. X 1 is a topological space
with basic open sets of the form NV “ tU P X 1 : V P Uu for V a regular open set in
1
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XP . X 1 is clearly Hausdorff, and it is a standard fact (see the following exercise)
that X 1 is compact. X 1 is also κ-c.c., for if NVα were pairwisw disjoint open sets in
X 1 , then the Vα would be pairwise disjoint open sets in B, contradicting B being
κ-c.c. Likewise define Y 1 . To see X 1 ˆ Y 1 is not κ-c.c., let ppα , qα q again be a κ
size antichain in P ˆ Q. Let Bα , Cα be as before, and consider the basic open sets
NBα ˆ NCα in X 1 ˆ Y 1 . These are pairwise disjoint since if NBα X NBβ ‰ H then
Bα X Bβ ‰ H, and then (as argued above) pα and pβ are compatible in P (and
likewise for Q).
We showed before that if there is a Suslin tree T , then T gives a partial order
which is c.c.c. (i.e., T with the reverse of the tree ordering) but for which T ˆ T is
not c.c.c. We now show that CH also implies the non-productivity of c.c.c.
Theorem 1.2 (Galvin, Laver). Assume ZFC ` CH. Then there are two c.c.c.
partial orders P , Q whose product P ˆ Q is not c.c.c.
Corollary 1.3. Assuming CH, there are two c.c.c. compact Hausdorff spaces whose
product is not c.c.c.
Proof. Suppose f : pω1 q2 Ñ t0, 1u is a partition of the pairs of countable ordinals.
For i P t0, 1u we define the partial order Pi to consist of all finite p Ď ω1 such that
f 2 p2 “ tiu (that is, Pi consists of finite sets which are homogeneous for color i).
We order Pi by: p ď q iff p Ě q. No matter how we choose f , the product P0 ˆ P1
will not be c.c.c., since the elements pα, αq for α ă ω1 form an antichain. It remains
to show that we can choose f so that P0 and P1 are c.c.c.
Let (using CH) tFα uαăω1 enumerate all ω sequences Fα “ pFα0 , Fα1 , . . . , Fαn , . . . q
of pairwise disjoint finite subsets of ω1 . Suppose we have defined f pα, βq when
maxtα, βu ă γ. We define f pα, γq for α ă γ as follows. Let Sn , n ă ω, enumerate
all objects of the form Sn “ xFαn , Xn , in y such that
(1) αn ă γ,
Ť i P t0, 1u, and Xn is a finite subset of γ.
(2) Xn X k Fαkn “ H and there are infinitely many k P ω such that f 2 pFαkn Y
Xn q “ tiu.
By diagonalizing, there is a sequence tYl ulPω with each Yl of the form Fαkn for some
n “ nplq, k “ kplq, such that the Yl are pairwise disjoint, for each n there are
infinitely many l such that nplq “ n, and f 2 pYl Y Xnplq q “ tinplq qu. For α P Yl ,
define f pα, γq “ inplq . The definition of f gives us the following property.
(*): for any α ă γ, finite X Ď γ, and i P t0, 1u, if suppYFα qq ă γ, X X pYFα q “
H, and there are infiniely many n such that f 2 pFαn Y Xq2 “ tiu, then there are
infinitely many n such that f 2 pFαn Y Xq2 “ tiu and f 2 pFαn Y tγuq2 “ tiu.
To see this works, suppose that tpα uαăω1 was an ω1 -antichain for P0 (the other
case being similar). Thinning to ∆-system, we may assume that pα “ r Y Gα ,
where Gα X Gβ “ H for α ‰ β. To get a contradiction it suffices to find α ‰ β
such that f 2 pGα Y Gβ q “ t0u. Consider the first ω many elements,
Ť and fix α ă ω1
such that Fα “ pG0 , G1 , . . . , Gn , . . . q. Let δ ą α and δ ą sup n Gn . Fix one
of the Gβ with Gβ X δ “ H (which we can do as the Gβ are pairwise disjoint).
Say Gβ “ tγ1 , . . . , γm u. Using (*) m times (at step p we use X “ tγ1 , . . . , γp´1 u,
γ “ γp ) we thin the Gn sequence to an infinite subsequence pG10 , G11 , . . . q such that
f 2 pG1n Y Gβ q2 “ t0u. This contradicts the pα being an antichain.
We now show that MA `
orders is c.c.c.
CH implies that the product of two c.c.c. partial
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Definition 1.4. A partial order P is strongly c.c.c. if whenever A Ď P with |A| “
ω1 , then there is a B Ď A with |B| “ ω1 such that for any finite tp1 , . . . , pn u Ď B
there is a p P P with p ď p1 , . . . , pn .
Lemma 1.5. Assume ZFC`MA` CH. Then every c.c.c. partial order is strongly
c.c.c.
Proof. Let P be c.c.c., and A “ tpα uαăω1 . First we claim that there is a p P A such
that any q ď p is compatible with pα for uncountably many α. For if not, then we
could get qα0 ď pα0 (where α0 “ 0) and α1 such that qα0 is incompatible with all
pβ for β ě α1 . We could then get qα1 ď pα1 and α2 so that qα1 is incompatible
with all pβ for β ě α2 . Continuing we define an ω1 antichain tqαi uiăω1 .
We may assume p0 is such that any q ď p0 is compatible with ω1 many pα .
Let Q be the partial order consisting of all finite subsets u “ tq1 , . . . , qn u Ď P
which contain p0 and which are compatible, that is, there is a q P P with q ď
q1 , . . . , qn . We order Q by u ď v iff u Ě v. We claim that Q is c.c.c. For suppose
tuα uαăω1 were an uncountable antichain in Q. For each α ă ω1 , let rα P P with
rα extending all of the elements of uα . Since P is c.c.c., there are α ‰ β such that
rα k rβ . Let r ď rα , rβ . Then r extends all elements of uα and uβ , so uα Y uβ P Q,
a contradiction. Thus, Q is c.c.c.
For each α ă ω1 , let Dα Ď Q be those u P Q which contain a pβ for some β ą α.
Then Dα is dense in Q. For given u P Q, let q P P extend all of the elements of
u. Since p0 P u, q is compatiblw with ω1 many of the pβ . Let β ą α with pβ k q.
Then u Y tpβ u P Dα .
By MA (and the fact that ω1 ă 2ω from CH) there is a filter G on Q meeting
all of the Dα . Let B “ YG, so B Ď A. Since G meets all of the Dα , |G| “ ω1 .
Since G is a filter,Ť
for any finite tu1 , . . . , un u Ď G, there is a p P G extending all
n
of the elements in i“1 ui . It follws that finite number of elements from B have a
common extension in P .
Lemma 1.6. If P is strongly c.c.c.and Q is c.c.c., then P ˆ Q is c.c.c.
Proof. Suppose ppα , qα q were an uncountable antichain in P ˆQ. Since P is strongly
c.c.c., we may thin the sequence so that for any finite subset u of tpα uαăω1 the
elements of u have a common extension (actually, all we need is that any two are
compatible). Since Q is c.c.c., get now α ă β such that qα k qβ . Then pα k pβ and
qα k qβ , so ppα , qα q k ppβ , qβ q, a contradiction.
Theorem 1.7. Assume ZFC ` MA ` CH. Then the product of two c.c.c. partial
orders is c.c.c. Likewise the product of two c.c.c. topological spaces is c.c.c.
Proof. Immediate from lemmas 1.5 and 1.6.
As a consequence, we have the following topological result.
Theorem 1.8. Assume ZFC`MA`
CH. Then if pXα , τα qαPI are c.c.c. topological
ś
spaces, then their product X “ αPI Xα is also c.c.c.
Proof. Suppose tVα uαăω1 is an uncountable antichain (i.e., pairwise disjoint open
sets) in X. Without loss of generality we many assume the Vα are basic open, say
Vα “ Uβ1α ˆ ¨ ¨ ¨ ˆ Uβnα , where Uβiα is open in Xβi and n depends on α. We call
tβ1α , . . . , βnα u the support of Vα . We may thin the antichain so that the supports
form a ∆-system with say root r “ tβ1 , . . . , βm u. then clearly tWα u also must form
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an antichain, where Wα is the product of the Uβiα for βiα P r. This is a contradiction,
as theorem 1.7 gives that Xβ1 ˆ ¨ ¨ ¨ ˆ Xβm is c.c.c.
2. Almost Disjoint Sets
If x, y Ď ω, we say x and y are almost disjoint if x X y is finite.
Lemma 2.1. There is a 2ω sequence txα uαă2ω of infinite subsets of ω such that if
α ‰ β then xα , xβ are pairwise disjoint.
Proof. Let tyα uαă2ω be distinct infinite subsets of ω. Let f : ω ăω Ñ ω be one-toone. Define xα “ tf pyα X nq : n P ωu Ď ω. Let n P yα ´ yβ , say Then for any
m ą maxtf pyα X 1q, . . . , f pyα X nqu, if m P xα then m R xβ .
Theorem 2.2. Assume ZFC ` MA. Then @κ ă 2ω p2κ “ 2ω q. In particular, 2ω is
regular.
Proof. We use the almost disjoint sets forcing. Fix a sequence txα uαă2ω of infinite,
almost disjoint subsets of ω. Fix κ ă 2ω , and we use this sequence to code subsets
of κ. Let A Ď κ. We claim that there is an x Ď ω such that for all α ă κ if α P A
then x X xα is finite, and if α R A then x X xα is infinite. This clearly suffices to
show that 2κ ď 2ω .
To see the claim, consider the partial order P which consists of all pairs p “ ps, F q
where s P 2ăω and F Ď A is finite. We regard s as the characteristic function of the
subset we are trying to build. If p1 “ ps1 , F 1 q, then we define p1 ď p iff s1 extends s,
F Ď F 1 , and for all n P domps1 q ´ dompsq, if s1 pnq “ 1 then for all α P F we have
xα pnq “ 0.
To see P is c.c.c. simply note that pp, F q is compatible with pp1 , F 1 q if s “ s1 , and
there are only countably many choices for s.
For α P A, let Dα “ tps, F q : xα P F u. Clearly Dα is dense as we may extend
any ps, F q to ps, F Y txα uq. For α ă κ, α R A, and k P ω, let Dα,k “ tps, F q : Dm ě
k pm P xα ^ spmq “ 1qu. To see this is dense, note that xα is almost
Ť disjoint from
all the xβ for β P F . Hence there is am m ě k such that m P xα ´ βPF xβ . Extend
s to s1 where s1 pmq “ 1 and s1 plq “ 0 for all other l P domps1 q ´ dompsq. Then
ps1 , F q ď ps, F q and ps1 , F q P Dα,k .
By MA, let G be Ť
a filter on P meeting all of the Dα for α P A and all of the Dα,k
for α R A. Let x “ ts : DF ps, F q P Gu be the real determined from G. Suppose
first that α P A. Since Dα is dense, let ps, F q P G X Dα . Suppose ps1 , F 1 q P G as
well. Since G is a filter, there is some ps2 , F 2 q P G with ps2 , F 2 q ď ps, F q, ps1 , F 1 q.
Thus, for all n P domps2 q ´ dompsq we have s2 pnq “ 1 Ñ pn R xα q. Hence
x X xα Ď tn : spnq “ 1u, so x is almost disjoint from xα . Suppose now that α R A.
For any k P ω there is an ps, F q P G X Dα,k . Thus there is an m P x X xα with
m ě k. This shows x X xα is infinite.
3. Measure and Category
We first show that MA ` CH implies the ă 2ω additivity of measure and category. First we consider category. Recall a subset of a Polish space is meager if it is
contained in the countable union of closed nowhere dense sets. A set A is comeager
if its complement is meager, equivalently if A contains a countable intersection of
dense open sets. The Baire category theorem (for complete metric spaces) says
that every comeager set is dense, in particular non-empty. It is immediate that
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a countable union of meager sets is meager (equivalently, a countable intersection
of comeager sets is comeager). The following result shows that with MA we may
extend this to ă 2ω size unions.
Theorem 3.1. Assume MA. Then the union of ă 2ω meager sets in a Polish space
is meager.
Proof. Let κ ă 2ω and fix dense
Ş openŞsets Uα , α ă κ. We show that there are
dense open sets Vn such that n Vn Ď αăκ Uα . Fix a base tBn unPω for the Polish
space. Let P consist of all finite sequences p “ xpW0 , F0 q, . . . , pWn , Fn qy
Ş where each
Fi Ď κ is finite, Wi is a finite union of basic open sets, and Wi Ď αPFi Uα . If
1
1
p1 “ xpW02 , F0 q, . . . , pWm
, Fm
qy then we define p1 ď p iff m ě n and for all i ď n we
1
1
have Fi Ě Fi and Wi Ě Wi .
P is c.c.c. since if m “ n and Wi “ Wi1 , then p is compatible with p1 . For each
α ă κ, let Dα be those p such that α P Fi for some i. Clearly each Dα is dense. For
each i and basic open set Bj , let Ei,j be those p such that Wn X Bj ‰ H. Easily
each Ei,j is also dense. From MA, let G be a filter on P meeting all of these dense
sets. Define Vi to be the union of all the Wi such that pWi , Fi q is the ith Ş
coordinate
of a p P G. Since G meets each Ei,j , each Vi is dense open. Let x P i Vi . Fix
α ă κ and we show that x P Wα . Let p P G X Dα , say p “ xpW0 , F0 q, . . . , pWn , Fn qy
1
1
with α P Fi . If q “ xpW01 , F01 q, . . . , pWm
, Fm
qy is also in G, then (if m ě i) Wi1 Ď Uα
as well, since p is compatible with q (any common extension will have ith coordinate
pW, F q where α P F and W Ě Wi Y Wi1 ). Thus, x P Vi Ď Uα .
We now prove the analogous result for measure.
Theorem 3.2. Assume MA. Let µ be a σ-finite Borel measure on a separable
metric space X. Then the union of ă 2ω subsets of X of µ measure 0 has µ
measure 0.
Proof. We may assume that µ is finite, that is, a probability measure. Recall that
any Borel probability measure on a metric space is regular, that is, for every Borel
set B Ď X and every ą 0, there is a closed set F and an open set U with
F Ď B Ď U and µpU ´ F q ă . Recall also that a set A Ď X is defined to have
measure 0 if it is contained in a Borel set of measure 0, and a set is measurable
if is equal to a Borel set modulo a set of measure 0. The measurable sets form a
σ-algebra containing the Borel sets, and µ extends to a countably additive measure
on this algebra.
Fix κ ă 2ω , and suppose Aα , for α ă κ, are measure
Ť zero sets. Fix ą 0, and
we show that there is an open set U containing A “ αăκ Aα of measure ă . Let
P consist of all open sets in X of measure ă . Define p ď q iff p Ě q. We show
that P is c.c.c. Suppose tpα uαăω1 were an uncountable antichain in P. By thinning,
we may assume that for some 1 ă that µppα q ă 1 for all α. For each α there
is a finite union of basic open sets (with respect to a fixed countable base for X)
Uα Ď pα such that µppα ´ Uα q ă p ´ 1 q{2. Fix α ‰ β such that Uα “ Uβ . Then
µppα Y pβ q ď µpUα q ` µppα ´ Uα q ` µppβ ´ Uα q ă .
For each α ă κ, let Dα be those pα containing Aα . Easily each Dα is dense.
By MA, let G be a filter on P meeting all of the Dα . Let V be the union of all
the open sets p P G. Clearly V contains each Aα . We claim that µpV q ď . Since
G is a filter, any finite union of open sets p P G has measure ă . By countable
additivity, any countable union of sets p P G has measure ď . However, in any
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second countable space the union of a family of open sets is given by a union of a
countable subfamily. Hence, µpV q ď .
As a corollary we get a little more:
Corollary 3.3. Assume MA. Let µ be a σ-finite Borel measure on a separable
metric space X. Then the algebra M of measurable sets is closed under κ ă 2ω
unions and intersections for all κ ă 2ω . Also µ is κ-additive for all κ ă 2ω .
Proof. We prove this by induction
on κ ă 2ω . Suppose tAα uαăκŤare given with
Ť
Aα P M. We show that A “ αăκ Aα is in M. Let Bα “ Aα ´ βăα Aβ be the
disjointification of the Aα . By induction, each Bα is in M. From σ-finiteness, only
countably many of the Bα can have non-zero µ measure. The union of the rest has
µ measure 0 by theorem 3.2. Thus, A is the union of countably many sets in M
together with a measure 0 set (which is by definition measurable). Thus A P M.
This show M is ă 2ω -additive. If the Aα are pairwise disjoint, then again only
countably many of them can have non-zero µ measure, and the union
ř of the rest
has measure 0. The countable additivity of µ then gives that µpAq “ µpAα q. 4. Random and Generic Reals
We introduce two new forcings for adding a real, one corresponding to measure
and the other to category, Pm and Pc . The category version turns out to be
isomorphic (actually, have an isomorphic completion) to ordinary Cohen forcing,
so this is not really a new forcing but just a different point of view. In the measure
case, Pm will be a new forcing, which we call the random real forcing.
Let X be a Polish space (e.g., X “ R), and I an ideal on X. For example,
I could be the ideal Im of Lebesgue measure 0 sets, or the ideal Ic of meager
sets. We define an equivalence relation „I of the Borel subsets of X by A „I B iff
A4B P I. For B Ď X a Borel set, we let rBsI , or just rBs if I is understood, denote
the equivalence class of B. We let PI be the partial order whose elements are the
equivalence classes rBs (B a Borel subset of X) with B R I, ordered by rAs ď rBs
iff A ´ B P I. This is easily well-defined and a partial order. In fact, PI is actually
a Boolean algebra under the obvious operations (for example rAs ` rBs “ rA Y Bs,
´rAs “ rX ´ As) provided we add 0 “ rHs back in.
We say I is c.c.c. if the Boolean algebra PI is c.c.c. In other words, there is no
ω1 sequence Aα of Borel subsets of X such that for α ‰ β we have Aα 4Aβ P I.
For a general ideal, this will not be a complete Boolean algebra, but we have the
following.
Lemma 4.1. If I is c.c.c. and countably-additive then PI is a complete Boolean
algebra.
Proof. Consider a collection trAα suř
αăκ u of elements of the Boolean algegra. We
show that the least upper bound rAα s exists. Let B be a maximal collection
subject to being an antichain in PI and for each b P B there is an αŤsuch that
b ď rAα s. Since PI is c.c.c. we may write B “ trBn sunPω . Let B “ n Bn (we
have implicitly
ř chosen representatives for the equivalence classes). We first claim
that rBs “ rBn s. Clearly rBn s ď rBs for each n. On the other hand, if rB 1 s is
1
also such
Ť that rB1n s ď rB s for each n, then by countable additivity of I we get that
B “ n Bn Ď B mod I. Hence rBs is the least upper bound of the rBn s.
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Next we show that rBs is the least upper bound of the rAα s. Sice for each n,
rBn s ď rAα s for some α, clearly any upper bound for the rAα s is an upper bound
for the rBn s. So, it suffices to show that B is an upper bound for the rAα s. Fix α,
and we show that rAα s ď rBs. If not, then Aα ´ B R I. Then rAα ´ Bs cound be
added to the rBn s, contradicting the maximality of B.
Exercise 1. Show that for any Polish space X, the category forcing Pc is c.c.c.
show that for any σ-finite Borel measure µ on a separable metric space X, the
corresponding measure forcing Pm is c.c.c.
Corollary 4.2. Pc and Pm are complete Boolean albegras.