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SH918
SAHARON SHELAH
Abstract. We force 2 to be large and for many pairs in the interval (, 2) astrong version of the polarized partition relations hold. We apply this to prob-lems in general topology. E.g. consistently, every 2 is successor of singularand for every Hausdorff regular space X, hd(X) s(X)+3, hL(X) s(X)+3
and better for s(X) regular, via a half-graph partition relation. For the cases(X) = 0 we get hd(X), hL(X) 2 (we shall get 1 < 20 but in asubsequent work).
Date: June 26, 2010.The author thanks Alice Leonhardt for the beautiful typing. Research supported by the United
States-Israel Binational Science Foundation (Grant No. 2002323). First Typed - 06/Dec/21.
1
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Anotated Content
0 Introduction, pg. 3
1 A Criterion for Strong Polarized Partition Relations, pgs.5-11
[We give sufficient conditions for having strong versions of polarized parti-tion relations after forcing.]
2 The forcing, pgs.12-24
[Assume GCH for simplicity and the parameters p contains < are reg-ular and Reg [, +) and we define Qp which adds Cohen subsetsto but have many kinds of supports, one for each , influencing theorder.]
3 Applying the criterion, pgs. 25-30
[The main result is that (cardinal arithmetic is changed just by making2 = and) using 1 we prove the strong version of polarized partitionrelations hold in many instances.]
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0. Introduction
Out motivation is a problem in general topology and for this we get a consistencyresult in the partition calculus.
In Juhasz-Shelah [JuSh:899] was proved: if ( < )(0 < ) then there is ac.c.c. forcing notion that adds a regular topological space, hereditarily Lindelof ofdensity .
A natural question asked there ([JuSh:899]) is:{0z.1}
Problem 0.1. Assume 1 < 20 . Does there exist (i.e., provably in ZFC) a
hereditary Lindelof regular space of density ?On cardinal invariants in general topology see [Juh80].We prove the consistency of a negative answer, in fact of stronger results by
proving the consistency of strong variants of polarized partition relations (the half-graphs, see below). They are strong enough to resolve the question for hereditarydensity (and Lindelof). Moreover, if = are regular cardinals, > ++, then there is a cardinaland cofinality preserving forcing that makes 2 = and ++ (++, (; ))2
in addition to the main result there 2 []23, see more in [BsSh:287], [Sh:288],[Sh:481], [Sh:546]. The applied notion of forcing (Q, ) is the following: p Q if pis a function from a subset Dom(p) [] into Add(, 1) {} where Add(, 1)denotes the forcing adding a Cohen subset of . p q if Dom(p) Dom(q), p()
q() for Dom(q) and |{ Dom(q) : p() = q()}| < .For n-place simultaneously many polarized partition relation Shelah-Stanley
[ShSt:608] deals with it but there are problems there, so we do not rely on itand also redo it (check!).
Our main result is Theorem 3.10, by it: consistently, G.C.H. fails badly (2
is a successor of a limit cardinal > except when is strong limit singular andthen 2 = +) and hd(X),hL(X) are s(X)+3 for every Hausdorff regular X and
|X| 2(hd(X))+
, w(X) 2(hL(X))+
for any Hausdorff X. (Usually s(X)+2 sufficeso in particular X is hereditary Lindelof X has density 2.
In the present paper we give a generalizaiton of his earlier result, namely, theconsistency of 20 = and ++ (, (; ))2 simultaneously holding for eachregular cardinal such that ++ . This gives a model in which though GCH
fails badly, the hereditary density and the hereditary Lindelof numbers of a T3 spaceX are bounded by s(X)+3 where s(X) stands for spread.
The notion of forcing (P, ) used for the argument is defined as follows. For eachregular cardinal < define the following equivalence relation E on . xEy iffx + = y + . Let [x] denote the equivalence class of x. p P if p is a functionfrom some set Dom(p) into {0, 1} such that |[x] Dom(p)| < holds forevery successor < , x < . p q if p q and for every successor < we have
|{[x] : = Dom(q) [x] = Dom(p) [x]}| < .
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4 SAHARON SHELAH
This notion of forcing (P, ), in a most remarkable way, imitates concurrentlyseveral different posets (Q, ) as defined above. Not surprisingly, in order to show
that (P, ) is cardinal and cofinality preserving, the author uses ideas similar tothose in [Sh:276].
In order to prove the main claim, that is, the partition relation, we introduce anew trick: we find a condition p such that the dense sets we are interested in areall dense below p. It suffices, therefore, to show that forcing with the part belowp gives the required result, and this reduces the problem to showing that a certainnotion of forcing (R, ) forces the sought-for-partition relation where |R| is small(compared to ). As (R,
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1. 1 Strong polarized partition relations
We deal with sufficient conditions on a forcing notion for preserving such par-tition relations. For this, we use an expansion of a forcing notion. Instead of theusual pair (Q, Q), namely, the underlying set and the partial order, we use aquadruple of the form Q = (Q, Q,
prQ , apQ).
The pr stands for pure, and the ap stands for apure. Both are included (aspartial orders) in Q.
Discussion 1.1. We define (below) the notion of (,,)-forcing to give a suffi-cient condition for appropriate cases of Definition 0.2 to hold. We start with thequadruple Q = (Q, Q,
prQ , apQ) such that q Q apQ(q) Q and Q,
prQ
are quasi orders on Q. The idea is that if r apQ(q) then r and q are compatiblein Q, close to r is an a-pure extension of q.
{1c.15}Definition 1.2. 1) We say that Q is a (+, , )-forcing notion when +, are
regular uncountable cardinals, an ordinal and below holds; in writing (+
, , .Let
Y = {q Q : (H(+ ), , cf() using an elaboratedefinition for regulars).
{top}Definition 3.1. Let X be a topological space:
(a) the density of X is:d(X) = min{|S| : S X and S is dense in X}
(b) the hereditary density of X is:hd(X) = sup{ : X has a subspace of density }
(c) hd+(X) =hd(X) = sup{+ : X has a subspace of density }(d) X is not -Lindelof if there is a family {U : < } of open susets of X
whose union is X but w |w| < {U : w} = X
(e) the hereditarily Lindelof number of X is:
hL(X) = hL(X) = sup{ : there are x X and U open(X) for < , such that x U and < x / U}
(f) hL+(X) = sup{+: there are x X,U as above}
(g) the spread of X is s(X) = sup{ : X has a discrete subset with points};s+(X) = s(X) = sup{+ : X has a discrete subspace with points}.
Our starting point was the following problem (0.1) of Juhasz-Shelah [JuSh:899].
Problem 3.2. Assume 1 < < 20 . Does there exist a hereditarily Lindelof
Hausdorff regular space of density ?We answer negatively by a consistency result but then look again at related
problems on hereditary density, Lindelofness and expand spread; our main theoremis 3.10 getting consistency for all cardinals.
We also try to clarify that relationships of this and related partition relations to []22,2, recalling that by [Sh:276], consistently, e.g. 2
0 [1]2n,2 for n < .
So by 3.13 below 20 [1]2n,2 implies 20 (1, (1; 1)n)2 and by 3.14 it
implies < 1 20 ()2n, see on the consistency of this Baumgartner-Hajnal
in [BH73], and Galvin in [Gal75].On cardinal invariants in general topology, in particular, s(X),hd(X),hL(X), see
Juhasz [Juh80]; in particular recall the obvious.
Observation 3.3. For a Hausdorff topological space X:
(a) hL(X) s(X)
(b) hd(X) s(X)
(c) for regularX is hereditarily -Lindelof (i.e. every subspace is -Lindelof)
iff there is x X,U for < as in (e) of 3.1(d) we choose the second statement in (c) as the definition of X is hereditarily
-Lindelof then 3.7, 3.9 holds also for singular.{3c.87}
Conclusion 3.4. Assume =
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() Qp is of cardinality
() Qp is (< )-complete (hence no new sequence of length < is added)
() no cardinal is collapsed, no cofinality is changed
() in VQp we have =
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By the assumption (, (; ))2 at least one of the following cases occurs.
Case 1: There is an increasing sequence : < of ordinals < such that
< < c{, } = 0.This means that < < x / u
2
. But if < < then u2 is an open
neighborhood of x included in u1
which is disjoint to C and x C so
x / u2
.
Lastly, x u2
by the choice of u2 . Together we are done, i.e. (x , u2
) : < is as required.
Case 2: There is a sequence (, ) : < such that:
()1 < < < < <
()2 < c{, } = 1, really suffice.
So
()3 < x u2but now for every < let
()4 y := x2 and u3 := u
22
\c(u22+1).
So
(a) u3 = u22
\c(u22+1) is open (as open minus closed)
(b) y u3.
[Why? Recall y = x2 belongs to u22
(by the choice of u22) and not to u12+1
(as u12+1 is disjoint to C2+1 while x2 C2+1) hence not to c(u22+1
) being
a subset of u12+1 . Together y belongs to u22
\c(u22+1) = u3.]
(c) if < < then y / u3.
[Why? Now y = x2 belongs to u22+1
by ()3 as 2 + 1 < 2 which follows from
< hence y belongs to c(u22+1) hence y / u3 by the definition of u
3.]
(d) if < < then y / u3.
[Why? As u3 u22
and the latter is disjoint to C2 to which x2 = y belongs.]
Together (y, u3) : < exemplifies that we are done.2) Follows trivially. 3.7
{1t.21}Claim 3.9. X has a discrete subspace of size when :
(a) (, (; ))2
(b) X is a Hausdorff moreover a regular (= T3) topological space(c) hL+(X) > , i.e. if is a regular cardinal this means that X is not
hereditarily -Lindelof
Proof. Similar to 3.7. We choose (x, u1) : < such that u1 is an open subset
ofX, x u1 and u1 {x : (, )} = . We can choose them as hL
+(X) > .We then choose an open neighborhood u2 of x such that c(u
2) u
1. We then
define c : []2 {0, 1} as follows
() if < then c{, } = 1 iff x u2.
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We continue as in the proof of 3.7, but now, in Case 2
()3
< x u2
and let
()4 y := x2 , u3 := u
22
\c(u22+1).
3.9
Now we come to our main result.{1t.41}
Theorem 3.10. The Main TheoremIt is consistent (using no large cardinals) that:
() () 2 is + if is strong limit singular and always 2
is the successor of a singular cardinal
() for every we have < 2 2 = 2
() hd(X) hL(X) s(X) for any limit cardinal and Hausdorff regular (= T3) topological space X
() hd(X) s(X)+3 and hL(X) s(X)+3 for any Hausdorff regular(= T3) topological space
() in () we can replace s(X)+3 by s(X)+2 except when 2s(X) = s(X)+
() in particular, if X is a (Hausdorff regular topological space which is)Lindelof or of countable density or just s(X) = 0 thenhd(X) + hL(X) 2
() if X is a Hausdorff space6 then |X| < 2(hd(X)+)
() if X is a Hausdorff space then w(X) 2(hL(X)+)
() if 2 > + then ++ (, (; ))2 for < +.{1t.42}
Remark 3.11. In the Theorem 3.10 above:1) If we use less sharp results in 1,2,3 we should above just use (hd(X))+n() forlarge enough n().2) We may like to improve clause () to < 2hd(X). If below we choose +1 stronglyinaccessible (so we need to assume V |= there are unboundedly many stronginaccessible cardinals and clause () is changed), nothing is lost, we have +1 =+1 then we can add
()+ for any Hausdorff space X, |X| < 2hd(X) except (possibly) when hd(X) isstrong limit singular.
3) Similarly for clause ().
4) Probably using large cardinal we can eliminate also the exceptional case in ( )+
;it seemed that a similar situation is the one in Cummings-Shelah [CuSh:541], butwe have not looked into this.5) We may wonder whether in clause () we can replace 2 by 1 and similarly forother cardinals, we have started proving the relevant consistency, see [Sh:F884].
Proof. We can assume V satisfies G.C.H. We choose (, ) : an ordinal suchthat:
6is interesting because usually 2 = 2(+), see clause ()
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(a) 0 = 0 = 0,
(b) < cf(+1) < +1
(c) +1 is the first regular +1,
(d) for limit we have is the first regular cardinal := { : < }.
Now let p = (, +1, , ) where , are defined by = Reg [, +1], = : ,
= , so are chosen as in 3.4.
So p : an ordinal is a class. We define an Easton support iteration P,Q :
Ord so {P : Ord} is a class forcing, choosing the P-name Q such that
P Q = Qp, i.e. Q
is defined as in Definition 2.5 for the parameter p.
As in VP section two is applicable for p so in VP+1 , the conclusions of 3.4, 3.5
hold and 2 = +1 so cardinal arithmetic should be clear, in particular, clause() holds. Of course, forcing with P/P+1 does not change those conclusions as
it is +1-complete.In VP we have enough cases of + (, (; ))2, i.e. clause () by 2.13. So,first, if = s(X) belongs to [, +1) we have +2 (; (; ))2 and hd(X),hL(X) +2.
Second, if = s(X) belongs to no such interval then + = , = > cf()for some hence recalling =
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we get a contradiction by cf() > ) say 0, 1 are -incomparable and withoutloss of generality 0 sup{ : < }
(b) 0 < (c) U is > (d) 1 .
So Case (c)1 of Definition 0.2(2) holds. So we are done. 3.13
We can remark also{1t.37}
Claim 3.14. Assume = 2 of cardinality > =
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4. Private Appendix
Moved from pg.2, Anotated Content:
4 Simultaneous Partitions and General Topology, pgs. 29-36
[We deduce the consistency of 2 is a successor of singular and manypartition relations hold between and 2 for every . We deduce consis-tency results on hereditary density/Lindelof and the spread of T3 topolog-ical spaces (and so for Boolean Algebras).]
5. Polarized partitions in Sh:918
To [Sh:918, 2.12], we can add{d2}
Claim 5.1. Assume
(a) \{}(b) = min(\+) but = <
(c) = : < n with + non-decreasing for simplicity
(d) = : < n
(e) < and + (), see below.
Then in VQp we have ().{d3}
Discussion 5.2. This also gives an alternative proof of a version of the maintheorem, losing by replacing +2 by +n(), n() < large enough but winning byhaving n-place partition relations for n < n() or so.
{d4}Definition 5.3. Let n < , = : < n, = : < n and be given with
, > 0 and e an equivalence relation on n such that , respect e, i.e., ek = k = k then e (), i.e., e ()
(1)n means: if c :
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(f) if s n(S) let tp(s) = {(i, ): for some i < n we have si {} }
(g) we call s standard when (i1, 1), (i2, 2) tp(s) [1 < 2 i1 < i2
1 2] and [(i1, ), (i2, )(h) t is a subset of {tp(s) : s n(S) for some n is standard}
(i) S,t = {s n()(S) : tp(s) = t} for t t, S,n = {s
n(S) : s isstandard}
(j) i f n > then tn(()) = {(, ) : < n} wherei then tn(()) =
{tn(()) : n n}; so below we may write n instead oft and i instead of{1}.
1) We say 1e ()t when: if cn :
n(S) for n < then there is U such that
() U = U : < ()() U has order type () sup(U) : < () respects e
()1 if t t then cn(t){s n(t)(S) : k < g(s) sk = (i, ) Ui and
tp(s) = t} is constant.
2) We say 0e ()t when : if cn :
n(S) for n < then thre is U such thatclause (), (), () above hold and
()2 ift t and m = |t|, n < and si n(S) and s S,t and ss
n+m(S)for = 1, 2 and k < g(s) sk = (i, ) Ui and similarly for s1, s2then cn+m(ss1) = cn+m(ss2).
3) We say 3e ()
t
when: if f = fs : s S, fs a condition in Cohen(), =sup(), then we can find U satisfying clauses (), (), () above and
()3 for each t t the functions {fs : s SU and tp(s) = t} are pairwisecompatible.
4) We say 4e ()t when: if F = Fs : s S each Fs a maximal antichain of
Cohen(), , = sup() and cn : {(s, f) : s n(S), f Fs} then we can
find U satisfying clauses (), (), () above and
()4 we can find fs : s SU such that fs Fs
fs : s SU,i are pairwise7 compatible
if n, s1, s2 are as in part one thencn((s1, fs1)) =
cn((s2, fs2).
5) Let + ()n be ()t when = 3 (or is it u) and we use n{(n)}.
{d8}Claim 5.6. 1) If =
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Proof. See [Sh:95].
Proof. Proof of 5.1We start as in [Sh:918, 2.12] but
1 q Qp c
:
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References
[BH73] J. Baumgartner and A. Hajnal, A proof (involving Martins axiom) of a partition relation,
Polska Akademia Nauk. Fundamenta Mathematicae 78(3) (1973), 193203.[EH78] Paul Erdos and Andras Hajnal, Embedding theorems for graphs establishing negative par-
tition relations, Period.Math.Hungar. 9 (1978), 205230.[EHMR84] Paul Erdos, Andras Ha jnal, A. Mate, and Richard Rado, Combinatorial set theory:
Partition Relations for Cardinals, Studies in Logic and the Foundation of Math., vol. 106, NorthHolland Publ. Co, Amsterdam, 1984.
[Gal75] F. Galvin, On a partition theorem of Baumgartner and Hajnal, Infinite and Finite Sets(Colloq., Keszthely, 1973; dedicated to P. Erdos on his 60th Birthday) (Amsterdam), Colloq.Math. Soc. Janos Bolyai, vol. 10, North-Holland, 1975, pp. 711729.
[Juh80] Istvan Juhasz, Cardinal functions in topologyten years later. Second edition, Mathe-matical Centre Tracts, vol. 123, Mathematisch Centrum, Amsterdam, 1980.
[Sh:95] Saharon Shelah, Canonization theorems and applications, The Journal of Symbolic Logic46 (1981), 345353.
[Sh:276] , Was Sierpinski right? I, Israel Journal of Mathematics 62 (1988), 355380.[BsSh:287] Andreas Blass and Saharon Shelah, Near coherence of filters. III. A simplified consis-
tency proof, Notre Dame Journal of Formal Logic 30 (1989), 530538.[Sh:288] Saharon Shelah, Strong Partition Relations Below the Power Set: Consistency, Was
Sierpinski Right, II?, Proceedings of the Conference on Set Theory and its Applications in honorof A.Hajnal and V.T.Sos, Budapest, 1/91, Colloquia Mathematica Societatis Janos Bolyai. Sets,Graphs, and Numbers, vol. 60, 1991, math.LO/9201244, pp. 637668.
[Sh:481] , Was Sierpinski right? III Can continuumc.c. times c.c.c. be continuumc.c.?,Annals of Pure and Applied Logic 78 (1996), 259269, math.LO/9509226.
[CuSh:541] James Cummings and Saharon Shelah, Cardinal invariants above the continuum, An-nals of Pure and Applied Logic 75 (1995), 251268, math.LO/9509228.
[Sh:546] Saharon Shelah, Was Sierpinski right? IV, Journal of Symbolic Logic 65 (2000), 10311054, math.LO/9712282.
[ShSt:608] Saharon Shelah and Lee Stanley, Forcing Many Positive Polarized Partition RelationsBetween a Cardinal and its Powerset, Journal of Symbolic Logic 66 (2001), 13591370.
[Sh:F884] Saharon Shelah, Partition relations andT3-spaces complimentary to Sh918.[JuSh:899] Istvan Juhasz and Saharon Shelah, Lindelof spaces of singular density, Studia Scien-
tiarum Mathematicarum Hungarica accepted, math.LO/0702295.[Sh:918] Saharon Shelah, Many partition relations below density, Israel Journal of Mathematicsaccepted, 0902.0440.
Einstein Institute of Mathematics, Edmond J. Safra Campus, Givat Ram, The He-
brew University of Jerusalem, Jerusalem, 91904, Israel, and, Department of Mathe-
matics, Hill Center - Busch Campus, Rutgers, The State University of New Jersey, 110
Frelinghuysen Road, Piscataway, NJ 08854-8019 USA
E-mail address: [email protected]: http://shelah.logic.at