Peter Komjath and Saharon Shelah- Two consistency results on set mappings

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Two consistency results on set mappings

Peter Komjath and Saharon Shelah

Abstract. It is consistent that there is a set mapping from the four-tuplesof n into the finite subsets with no free subsets of size tn for some naturalnumber tn. For any n < it is consistent that there is a set mapping fromthe pairs of n into the finite subsets with no infinite free sets. For anyn < it is consistent that there is a set mapping from the pairs of n inton with no uncountable free sets.

In this paper we consider some problems on set mappings, that is, for ourcurrent purposes, functions of the type f : []k [] then there is a free set of cardinal (remember, k = 1 is assumed).Several cases were soon proved, for example S. Piccard solved the case when is regular ([9]), but only in 1950 was the full conjecture established byPaul Erdos ([1]) with the assumption of GCH, and ten years later withoutthis assumption, by A. Hajnal ([5]). In the fifties Erdos and Hajnal startedthe research on the case k > 1 following the observation of Kuratowski andSierpinski (see [4]) that for set mappings on []k there always exists a freeset of cardinal k + 1 iff +k.

In ZFC alone, Hajnal and Mate extended the Kuratowski-Sierpinskiresults by showing ([6]) that if k = 2 and +2 then there are arbitrarily

large finite free sets, and Hajnal proved (see [3]) that a similar result holdsfor k = 3, +3. One of the problems emphasized in [3] is if the resultcan be extended to k = 4, +4. In Theorem 1 we show that it is not

Research of the first author was partially supported by the HungarianNational Science Research Grant No. T 016391. Research of second the au-thor was partially supported by the United States-Israel Binational ScienceFoundation. Publication number 645.

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the case; for every natural number n there exists a natural number tn suchthat for any given regular it is consistent that there is a set mappingf : [+n]4 [+n]


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The case n = 0 is obvious, since we can take f({x0, x1, x2, x3}) =(x1, x2).

Assume that V is a model of set theory satisfying the Theorem for n,and for + in place of . That is, for , 2 = + holds, and there

is a set mapping F : []4[] satisfying F({x0, x1, x2, x3}) (x1, x2)with no free subset of cardinality tn. We are going to force with a notion offorcing (P, ) in which the conditions will be some pairs of the form (s, g)with s []


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We have to show that q P, that is, there is no 7-element F-closed,g-free subset of s. Assume that B is such a set. As p, p are conditions,B a b, B a b . There are, therefore, 0 B b, 1 B b .

An easy calculation shows that no matter what position 0, 1 occupy

in B, there is a five-tuple y0 < y1 < y2 < y3 < y4 in B such that 0,1 {y0, y1, y3, y4}. (This is the point where the choice of 7 plays role.)We get, therefore, that g({y0, y1, y3, y4}) = F({y0, y1, y3, y4}) s y2 so Bcannot be F-closed and g-free.

Let G P be a generic subset of P. Set S =

{s : (s, g) G} andf =

{g : (s, g) G}. Clearly, f is a set mapping of the required type on

the set S.

Claim 3. There is a p P forcing that |S| = .

Proof of Claim. Otherwise, 1 forces that S is bounded in , and as (P, )is < -c.c., it forces a bound, say < . But as

{},

S, we get

a contradiction.

Claim 4. In V[G], f has no free subset of cardinality tn+1.

Proof of Claim. Assume that A S is a free subset of cardinality tn+1.Color the five-tuples ofA as follows. If{x0, x1, x2, x3, x4} [A]

5, x0 < x1 i(x1, x2) > whichis impossible.

Assume now that < , GCH holds up to and including and there

is a set mapping F : []2 [] with no infinite free sets and withF

{x, y}

x for x < y < . We are going to define a < -closed partialordering (P, ) which adds a set mapping f : [S]2 [S]


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r, r can be given. To show this, it suffices to prove, that if u is a g-free,F-closed set, which is new, that is, has points in b, as well as in b , thenit cannot end extend an old secured set (one in p or in p). Assumethat x0 < x1 < x2 < are the elements of u. If xi b, xj b , and i,

j = 0, then x0 F

{xi, xj}

, so x0 g

{xi, xj}

by the definition of g andso our set is not g-free. We get, therefore, that x0 is the only element ofu b (say). The possibility that both x1 and x2 are in a is ruled out by ourabove condition that b F

{x1, x2}

= . This means that {x0, x1, x2} is

a new set, so u is indeed not an end extension of an old secured set as weassumed that secured sets have at least three elements.

If G P is a generic subset, then define S =

{s : (s,g,r) G},f =

{g : (s,g,r) G}, R =

{r : (s,g,r) G}, .

Claim 2. |S| = .

Proof of Claim. As in the corresponding proof in Theorem 1.

Claim 3. F has no infinite free set in V[G].

Proof of Claim. This is a well-known fact. It follows from the rankcharacterization of the nonexistence of free sets.

Claim 4. f has no infinite free set.

Proof of Claim. Assume that x0 < x1 < form an infinite f-free set. ByRamseys theorem we can assume that either for every triplet i < j < k < ,xi F(xj , xk) holds or for every triplet i < j < k < , xi / F(xj, xk)holds. The latter is impossible by Claim 3. Therefore {x0, x1, . . .} is f-free,F-closed, but then R

{x0, x1, x2}

> R

{x0, x1, x2, x3}

> , which is

impossible.

An easy application of Theorem 2 solves another problem of [6].

Corollary 3. For every n < it is consistent that there exists a setmapping f : [n]2 [n]

1 with no uncountable free set.

Proof. Applying Theorem 2 assume that F : [n]2 [n]

0 is a setmapping with no infinite free sets so that F({x, y}) x for all x < y < n.Define the notion of forcing as follows, (s, g) P iff s [n]


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Proof of Claim. Straightforward.

Claim 2. (P, ) is c.c.c.

Proof of Claim. Assume that p P for < 1. By the usual thinning

out procedure we can assume that p = (s s, g) where s F(x, y) = holds for x, y s, and the functions g|[s]

2 are identical. Now any two p-sare compatible.

If G P is a generic set, put f =

{g : (s, g) G}.

Claim 3. f has no uncountable free set.

Proof of Claim. Assume that p X is an uncountable free set. Thereare, for < 1, conditions p p and ordinals with p X.

Again, we can assume, that p = (s s, g), s, and the functionsg [s]

2 are identical. As F has no infinite free sets (no uncountablesuffices) there are ordinals 0, 1, 2 < 1 such that 0 F

1 , 2

. We

can now extend p to a condition p = (s, g) where

s = s s0 s1 s2 ,

g extends g0 , g1 , g2 and g

{1 , 2}

= 0 .

Now Corollary 3 follows from the claims above.

References

[1] P. Erdos: Some remarks on set theory, Proceedings of the AmericanMathematical Society, 1(1950), 127141.

[2] P. Erdos, A. Hajnal: On the structure of set mappings, Acta Mathe-matica Acad. Sci.Hung., 9(1958), 111131.

[3] P. Erdos, A. Hajnal, A. Mate, R. Rado: Combinatorial Set Theory:

Partition Relations for Cardinals, North-Holland, Akademiai Kiado,1984.[4] G. Grunwald: Egy halmazelmeleti tetelrol, Mathematikai es Fizikai

Lapok, 44(1937), 5153.[5] A. Hajnal: Proof of a conjecture of S. Ruziewicz, Fund. Math. 50

(1961), 123-128.[6] A. Hajnal, A. Mate: Set mappings, partitions, and chromatic numbers,

in: Logic Colloquium 73, Bristol, North-Holland, 1975, 347379.

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[7] P. Komjath: A set mapping with no infinite free subsets, Journal ofSymbolic Logic 56 (1991), 304306.

[8] K. Kuratowski: Sur une characterization des alephs, Fundamenta Ma-thematicae, 38(1951), 1417.

[9] S. Piccard: Sur un probleme de M. Ruziewicz de la theorie des relations,Fundamenta Mathematicae, 29(1937), 59.

[10] S. Ruziewicz: Une generalisation dun theoreme de M. Sierpinski, Publ.Math. de lUniversite de Belgrade 5 (1936), 23-27.

Peter KomjathDepartment of Computer ScienceEotvos UniversityBudapest, Muzeum krt. 681088, Hungarye-mail: [email protected]

S. ShelahInstitute of Mathematicsthe Hebrew UniversityJerusalem, Israele-mail: [email protected]

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Peter Komjath and Saharon Shelah- Two consistency results on set mappings

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