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A major branch of combinatorial set theory concerns set mappings, i.e., functions f : X → P (X), where one seeks for a large free subset , that is, a subset Y ⊆ X such that x 6∈ f (y) for any two distinct x, y ∈ Y . P. Erd˝ os [2]

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C O L L O Q U I U M M A T H E M A T I C U M

VOL. LXVIII 1995 FASC. 2

SET MAPPINGS ON GENERALIZED LINEAR CONTINUA

BY

P ´ ETER K O M J ´ A T H (BUDAPEST)

A major branch of combinatorial set theory concerns set mappings, i.e., functions f : X → P (X), where one seeks for a large free subset , that is, a subset Y ⊆ X such that x 6∈ f (y) for any two distinct x, y ∈ Y . P. Erd˝ os [2]

proved that if the ground set is the reals and f (x) is nowhere dense for x ∈ R then there is an infinite free set. Bagemihl [1] extended this by showing that in fact an everywhere dense free set exists. Muthuvel became interested in these questions for R κ , the linear continuum of zero-one sequences of length κ for κ > ω [3, 4]. He showed that under GCH if κ is regular there always exists a free set of cardinal κ [4]. We improve this result by showing that an everywhere dense free set exists.

If κ is a regular cardinal then R κ consists of all nonconstant functions f from κ into 2 = {0, 1} such that there is no last α < κ such that f (α) = 0.

We order R κ by the lexicographic ordering. Subsets of R κ of the form I(g) = {f ∈ R κ : f ⊇ g} where g : γ → 2 for some γ < κ are the intervals.

So what we call intervals are really the nonempty dyadic intervals. Let I be the set of intervals; notice that |I| = 2 = κ under GCH.

A set A ⊆ R κ is everywhere dense if A ∩ I 6= ∅ for every interval I; A is nowhere dense if for every interval I there is a subinterval I 0 ⊆ I such that A ∩ I 0 = ∅; A is of first category if it is the union of κ nowhere dense sets;

otherwise, it is of second category.

First we give a (probably well known) construction of a strong Luzin type set in R κ .

Lemma. (GCH) There is a set A = {r(α) : α < κ + } ⊆ R κ such that A is of second category in every interval and whenever {x α ξ : ξ < µ(α)} are disjoint subsets of κ + with µ(α) < κ then there is an S ⊆ κ + , |S| = κ, and there are intervals J ξ (ξ < µ) such that µ(α) = µ (α ∈ S) and if J ξ 0 ⊆ J ξ are subintervals (ξ < µ) then there is an α ∈ S such that r(x α ξ ) ∈ J ξ 0 for every ξ < µ.

1991 Mathematics Subject Classification: 03E05, 04A20.

Research partially supported by Hungarian Science Research Fund No. T014105.

[193]

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194 P. K O M J ´ A T H

P r o o f. First we remark that the assumption on |S| is inessential as it is obvious from the other assumptions that if there is a good set then there is one of cardinal κ.

By GCH there are κ + objects of the form ((d ξ : ξ < µ), F ) where µ < κ, d ξ are distinct functions γ → 2 for some γ < κ, and F is a function from µ × I such that F (τ, I) is always a subinterval of I. We can, therefore, enumerate them as {((d α ξ : ξ < µ(α)), F α ) : α < κ + }.

To construct r(α) for α < κ + we re-order α as {β(τ ) : τ < κ} and deter- mine successively the digits of r(α), or, in other words, define a descending sequence of intervals {I τ α : τ < κ} and get r(α) as the unique element of the intersection.

At step τ < κ shrink T{I τ α

0

: τ 0 < τ } to an interval b I τ α such that there is at most one d = d β(τ ) ξ for some ξ < µ(β(τ )) such that b I τ α ⊆ I(d).

This is possible as those intervals are disjoint. Then add a final 0 and 1 (this will ensure that r(α) has cofinally many zeroes and ones, i.e., r(α) ∈ R κ ). For τ = 0 let b I 0 α be an arbitrary interval; we only require that every interval should occur κ + times as b I 0 α . Then let I τ α = F β(τ ) (ξ, b I τ α ), where ξ is the above index. Finally, as we have already said, {r(α)} = T{I τ α : τ < κ}.

Assume now that this construction fails to meet the requirements of the Lemma and {x α ξ : ξ < µ(α)} witness this. Then they are disjoint subsets of κ + for α < κ + , µ(α) < κ. Without loss of generality, we can assume that µ(α) = µ for α < κ + . We can also assume that there exists a γ < κ such that r(x α ξ )|γ = d ξ for ξ < µ and these µ functions from γ into 2 are different.

By our indirect assumption there is a function F : µ × I → I such that given I ξ (ξ < µ) there is no α < κ + such that r(x α ξ ) ∈ F (ξ, I ξ ) for every ξ < µ. There is an ε < κ + such that µ(ε) = µ and ((d ξ : ξ < µ), F ) = ((d ε ξ : ξ < µ), F ε ).

Let now α be so large that ε < x α ξ for ξ < µ. When r(x α ξ ) was constructed ε occurred in the τ (ξ)th step for a τ (ξ) < κ. We know that b I x

α ξ

τ (ξ) ⊆ I(d ξ ) by assumption. Then we get the interval I x

α ξ

τ (ξ) = F (ξ, I 0 ) for an I 0 = b I x

α ξ

τ (ξ) . But then {x α ξ : ξ < µ} contradict what was assumed about F .

We need to show that A is of second category in every interval. Assume

not. Then there is an interval I and there are functions F ξ : I → I (for

ξ < κ) such that I ∩ A decomposes as I ∩ A = S{A ξ : ξ < κ}, F ξ (I 0 ) ⊆ I 0

for every I 0 ⊆ I and A ξ ∩ F ξ (I 0 ) = ∅ for ξ < κ. Put I = I(d). Then

(d, F ξ ) occurs in the above enumeration at the α(ξ)th step (say). Select

α > α(ξ) such that b I 0 α = I. Then the successive digits of r(α) are so chosen

that r(α) ∈ I, and for every ξ < κ, r(α) ∈ F ξ (I 0 ) for some I 0 ⊆ I, i.e.,

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GENERALIZED LINEAR CONTINUA

195

r(α) 6∈ A ξ , that is, r(α) 6∈ A, a contradiction.

Theorem. (GCH) If κ is regular and f (x) ⊆ R κ is nowhere dense for x ∈ R κ then there is an everywhere dense free set.

Let A ⊆ R κ be as in the Lemma. Enumerate I as {I µ : µ < κ}. For some µ < κ the sequence {x ξ : ξ < µ} is called bad if f (y) ∩ {r(x ξ ) : ξ < µ} 6= ∅ for all but first category many y in A ∩ I µ .

Claim. There is a δ < κ + such that no set {x ξ : ξ < µ} with x ξ > δ (ξ < µ) is bad.

P r o o f. Otherwise for every δ < κ + there is a bad set above it, so by transfinite recursion we can choose disjoint bad sets {x α ξ : ξ < µ(α)}. The Lemma gives S, µ, and certain intervals J ξ . As {x α ξ : ξ < µ} is bad for α ∈ S and |S| = κ all but first category many elements r(y) of A ∩ I µ

have f (r(y)) ∩ {r(x α ξ ) : ξ < µ(α)} 6= ∅. Let r(y) be such an element.

As f (r(y)) is nowhere dense there exist subintervals J ξ 0 ⊆ J ξ such that J ξ 0 ∩ f (r(y)) = ∅. But then, for some α ∈ S, r(x α ξ ) ∈ J ξ 0 for ξ < µ, and yet f (r(y)) ∩ {r(x α ξ ) : ξ < µ(α)} 6= ∅, a contradiction.

If we now have δ as in the Claim, we can select by transfinite induction the free {r(x ξ ) : ξ < κ}, x ξ > δ, r(x ξ ) ∈ A ∩ I ξ at every step we have a second category set of good extensions.

REFERENCES

[1] F. B a g e m i h l, The existence of an everywhere dense independent set , Michigan Math. J. 20 (1973), 1–2.

[2] P. E r d ˝ o s, Some remarks on set theory III , ibid. 2 (1953–54), 51–57.

[3] K. M u t h u v e l, Nowhere dense set mappings on the generalized linear continua, Or- der 7 (1990), 179–182.

[4] —, Large free set , Colloq. Math. 63 (1992), 107–109.

DEPARTMENT OF COMPUTER SCIENCE R. E ¨ OTV ¨ OS UNIVERSITY

BUDAPEST, M ´ UZEUM KRT. 6–8 1088 HUNGARY

E-mail: KOPE@CS.ELTE.HU

Re¸ cu par la R´ edaction le 7.3.1994

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