Saharon Shelah and Boaz Tsaban- On a Problem of Juhasz and Van Mill

8/3/2019 Saharon Shelah and Boaz Tsaban- On a Problem of Juhasz and Van Mill

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ON A PROBLEM OF JUHASZ AND VAN MILL

SAHARON SHELAH AND BOAZ TSABAN

Abstract. A 27 years old and still open problem of Juhasz andvan Mill asks whether there exists a cardinal such that everyregular dense in itself countably compact space has a dense initself subset of cardinality at most . We give a negative answerfor the analogous question where regular is weakened to Hausdorff,and coutnably compact is strengthened to sequentially compact.

1. Introduction

Every compact Hausdorff dense in itself topological space containsa countable subset which is dense in itself [1]. A space is countablycompact if each countably infinite subset of that space has a clusterpoint. Tkacenko [3] asked whether every regular countably compactdense in itself topological space contains a countable subset which isdense in itself. This was answered in the negative by Juhasz and vanMill [1], who concluded their paper with the following.

Problem 1 (Juhasz and van Mill [1]). Does there exist a cardinal such that every regular dense in itself countably compact topologicalspace has a dense in itself subset of cardinality at most ?

Despite several efforts made in the past, the problem is still open.We will prove the following. A topological space is scattered if each ofits nonempty subsets has an isolated point.

Theorem 2. Assume that 0 = . There exists a Hausdorff space Ssuch that:

(1) |S| = ;(2) S is dense in itself;(3) Each subspace of S of cardinality less than the cofinality of is

scattered; and(4) S is sequentially compact.

The cardinals satisfying the assumption of the theorem are those ofthe form = 0. Note that in this case, cf() > 0. By the HausdorffFormula, if 0 = , then we have that

(+)0 = max{+, 0} = +,1


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2 SAHARON SHELAH AND BOAZ TSABAN

so for each cardinal we can take = (0)+ (which is regular andgreater than ) to obtain a negative answer to Problem 1 when stated

for Hausdorff spaces.Of course, the original Problem 1 is still interesting. We hope that

this paper will revive the interest in this problem.

2. Proof of Theorem 2

Definition 3. For a cardinal , let denote the collection of finitesequences of elements of , and define a topology on with subbaseconsisting of the -neighborhoods

[] = {} { : and (||) },

of each , where < .

Lemma 4.

(1) || = ;(2) is dense in itself;(3) For each Y with |Y| < cf():

(a) There exists < such that the -neighborhoods [], Y, are pairwise disjoint.

(b) Y is closed and discrete.(4) Every two disjoint subsets of of cardinality smaller thancf()

can be separated by disjoint open subsets of .

Proof. (1) is immediate. To prove (2), note that each

is theaccumulation point of { : < }.

(3a) We claim that for all distinct , , there exists , < such that [], [], = . Indeed, if and are incomparable withregards to , then we can take , = 0. Otherwise assume that .Then we can take , = (||) + 1.

Assume that |Y| < cf(). Define = sup{, : , Y}. As|Y| < cf(), < and is as required.

(3b) Assume that |Y| < cf(). By (3a), Y is discrete. To see thatY is closed, assume that y is in the closure of Y, and apply (3a) toY {y} to see that y Y.

(4) follow from (3a), either.

The proof of the forthcoming Proposition 6 uses weak bases. A weakbase of a topological space X is a family of the form

xX B(x) such

that:

(1) For each x X:(a) x B for all B B(x);(b) For all B1, B2 B(x), B1B2 contains an element ofB(x);


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ON A PROBLEM OF JUHASZ AND VAN MILL 3

(2) For each U X, U is open in X if, and only if, for each x U,U contains an element of B(x).

A space X with a weak base

xX B(x) is weakly locally-countable/Haus-dorff/regular if, respectively:

(1) For each x X, there is a countable element in B(x);(2) For all x, y X, there are disjoint Bx B(x), By B(y);(3) X is weakly Hausdorff and all elements of

xX B(x) are closed.

Lemma 5 (Nyikos-Vaughan [2]). For each weakly locally-countablespace X:

(1) X is locally countable; and(2) If X is weakly regular, then it is Hausdorff.

Proof. For the readers convenience, we reproduce the proofs.(1) Let x X. Set U0 = {x} and inductively, for each y Un choose

a countable Dy B(y), and set Un+1 =

yUnDy. Then

n Un is a

countable open neighborhood of x.(2) Let x, y X be distinct. By (1), there is a countable open

neighborhood U of x, y. Enumerate U = {xn : n N}. Let U0 ={x}, V0 = {y}. By induction on n, assume that Un, Vn are disjointand closed. If xn Un, pick B B(xn) such that B U \ Vn andlet Un+1 = Un B, Vn+1 = Vn. Otherwise, pick B B(xn) such thatB U \ Un and let Un+1 = Un, Vn+1 = Vn B.

The setsU

=

nUn

, V=

nVn are disjoint open neighborhoods ofx, y, respectively.

Proposition 6. Assume that 0 = . There exists a Hausdorff spaceX such that:

(1) |X| = ;(2) X is locally countable;(3) X is scattered;(4) There are isolated points in X; and(5) X is sequentially compact.

Proof. This is proved as in Nyikos and Vaughans Construction of

Xu [2, Page 313]. In our case, we should replace c with (and use thepremise 0 = ), need not worry about the ultrafilter u, and verifythat there are isolated points in X. We give a full direct proof.

The topology is constructed on the space X = .Enumerate []0 = {D : < }, such that D for each

. We construct, inductively on with , topologies on, starting with = and taking the discrete topology on . (Foreach n, we set B(n) = {n}.)


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When is a limit ordinal: Let be the topology having


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induction hypothesis, B is closed in (, ), and therefore \ B ,and contains a cofinite subset of D. Thus, \ B = (\ B) {} .

This completes the proof that X is Hausdorff.Each nonempty subset A of X has an isolated point: Take =

min A. Then + 1 +1 , and + 1 is disjoint from A \ {}.X has many isolated points: Let < be such that D = . For

each > , {} is a countable subset of which is not a subsetof . Thus, these sets are considered in (successor) stages later than + 1, where they already have a cluster point. Consequently, in eachof these many stages, a new isolated point is added.

X is countably compact: Let A be a countable subset of . Let besuch that D = A. Then either A has a cluster point in (, ) and thusin X (since is open in X), or else A converges to (in (+ 1, +1)and therefore in X).

X is sequentially compact: This can be proved by quoting basic facts.Indeed, X is countably compact and has a countable weak base at eachpoint. Thus, X is sequential, and therefore sequentially compact.

Alternatively, we can argue directly. Let A be a countable subset of. Let be the minimal cluster point of A. As is not isolated in X,D converges to . In (, ), A is closed, and therefore U = \A isopen in X. IfD\ U is finite, then U {} is an open neighborhood of, and (U {}) A {}, in contradiction to being a cluster pointofA. Thus, D\ U = D A is infinite, and is therefore a subsequence

of A converging to . Proof of Theorem 2. Assume that 0 = , and let X be as in Lemma6. We may assume that X = . We will define the requiredtopology on S = X .

Let I be the set of isolated points of X, and let {Ax : x I} bemaximal almost disjoint in []0. The open sets in the topology of Sare those of the form

U V,

where U is open in X, V is open in , and for each x I U, Ax \ Vis finite. (Note that for U = , we get that each open set in is also

open in S.)Clearly, |S| = .S is Hausdorff: Consider any distinct x0, x1 X and any distinct

y0, y1 . We will find disjoint open sets G0, G1 such that xi, yi Gi,

i = 0, 1. X is Hausdorff and locally countable. Take disjoint countableopen subsets U0, U1 of X containing x0, x1, respectively. EnumerateI (U0 U1) = {xn : n N}. For each n, define Axn = Axn \

k


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disjoint open sets V0, V1 in such that

({yi}

xIUi

Ax) \ {y1i} Vi,

i = 0, 1. The sets Gi = Ui Vi, i = 0, 1, are as required.S is dense in itself: is dense in itself, hence (by the definition of

open sets in S) each y is an accumulation point in S. Let x X.If x is an accumulation point in X, then x is an accumulation point inS. Otherwise, x is in the closure ofAx: Indeed, the sets {x} V whereV is open in and Ax \ V is finite form a neighborhood base at x,and therefore the elements of Ax converge to x.

Each Z S with |Z| < cf() is scattered:

Case 1: Z X. As X is scattered, we can find x Z isolated in X.Then {x} is a neighborhood ofx in S and is disjoint from Z\ {x}.

Case 2: Z = . |Z | < cf() = |I|. Thus, Z isdiscrete in . As each open set in is also open in S, we are done.

S is sequentially compact: Assume that Z [S]0. If Z X isinfinite, then there are x X and a subsequence of Z X convergingto x. It is clear that the same subsequence will also converge to x inthe topology of S. And if not, then Z is infinite. As {Ax : x I}is maximal almost disjoint in []0 , there is x I such that AxZ

is infinite. Being a subsequence of Ax, it converges to x.

This completes the proof of Theorem 2.

Acknowledgments. We thank Istvan Juhasz for his substantial sim-plification of our proof, and Peter Nyikos and Lyubomyr Zdomskyy fortheir useful suggestions.

Our research is partially supported by: The Israel Science Founda-tion (first author); Koshland Center for Basic research (second author).

This is publication 921 of the first author.

References

[1] I. Juhasz and J. van Mill, Countably compact spaces all countable subsets ofwhich are scattered, Commentationes Mathematicae Universitats Carolinae 22(1981), 851855.

[2] P. Nyikos and J. Vaughan, The Scarborough-Stone problem for Hausdorffspaces, Topology and its Applications 44 (1992), 309316.

[3] M. G. Tkacenko, On compacta representable as countable unions of left sepa-rated subspaces, I, Commentationes Mathematicae Universitats Carolinae 20(1978), 741749.


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Institute of Mathematics, Hebrew University of Jerusalem, Givat

Ram, 91904 Jerusalem, Israel, and Mathematics Department, Rutgers

University, New Brunswick, NJ 08903, U.S.A.E-mail address: [email protected]

Department of Mathematics, Bar-Ilan University, Ramat-Gan 52900,

Israel; and Department of Mathematics, Weizmann Institute of Sci-

ence, Rehovot 76100, Israel.

E-mail address: [email protected]: http://www.cs.biu.ac.il/~tsaban

Saharon Shelah and Boaz Tsaban- On a Problem of Juhasz and Van Mill

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