On regular but not completely regular spaces
aa r X i v : . [ m a t h . GN ] J a n ON REGULAR BUT NOT COMPLETELY REGULARSPACES
PIOTR KALEMBA AND SZYMON PLEWIK
Abstract.
We present how to obtain non-comparable regular butnot completely regular spaces. We analyze a generalization of Mys-ior’s example, extracting its underlying purely set-theoretic frame-work. This enables us to build simple counterexamples, using theNiemytzki plane, the Songefrey plane or Lusin gaps. Introduction
Our discussion focuses around a question:
How can a completelyregular space be extended by a point to only a regular space?
BeforeA. Mysior’s example, such a construction seemed quite complicated,compare [10] and [3]. R. Engelking included a description of Mysior’sexample in the Polish edition of his book [4, p. 55-56]. In [2] thereis considered a modification of Mysior’s example which requires noalgebraic structure on the space. We present a purely set-theoreticapproach which enables us to obtain non-comparable examples, suchspaces are X ( ω, λ ) and X ( λ , κ ) , see Section 2. This approach is astep towards a procedure to rearrange some completely regular spacesonto only regular ones. One can find a somewhat similar idea in [6],compare "the Jones’ counterexample machine" in [2, p. 317]. Thestarting point of our discussion are the cases of completely regularspaces which are not normal. For example, subspaces of the Niemytzkiplane are examined in [1] or [11], some Ψ -spaces are studied in [5],also the Songefrey plane is commentated in [11]. The key idea of ourconstruction of counterexamples looks roughly as follows. Start from acompletely regular space X , which is not normal. In fact, we need that X contains countable many pairwise disjoint closed subsets which, evenafter removal from each of them a small subset, cannot be separated Mathematics Subject Classification.
Primary: 54D10; Secondary: 54G20.
Key words and phrases.
Niemytzki plane; Songefrey plane; Lusin gap. y open sets. By numbering these closed sets as ∆ X ( k ) and assumingthat the collections of small sets form proper ideals I X ( k ) , we shouldcheck that the property ( ∗ ) is fulfilled. Copies of X are numbered byintegers and then the k -th copy is glued along the set ∆ X ( k ) to the ( k − -copy, moreover copies of sets ∆ X ( m ) , for k = n = k − , areremoved from the k -th copy. As a result we get the completely regularspace Y X , which has a one-point extension to the regular space whichis not completely regular.In fact, given a completely regular space X, which we do not knowwhether it has one-point extension to the space which is only regular,we can build a space Y X which has such an extension. A somewhatsimilar method was presented in [6]. For this reason, we look for waysof comparing such spaces. Following the concept of topological ranks,compare [7, p. 112] or [12, p. 24], which was developed in Polish Schoolof Mathematics, we say that spaces X and Y have non-comparable regularity ranks , whenever X and Y are regular but not completelyregular and there does not exist a regular but not completely regularspace Z such that Z is homeomorphic to a subspace of X and Z ishomeomorphic to a subspace of Y .2. On Mysior’s example
We modify the approach carried out in [2], which consists in ageneralization of Mysior’s example, compare [10]. Despite the factthat our arguments resemble those used in [2], we believe that thispresentation is a bit simpler and enables us to construct some non-homeomorphic examples, for example spaces X ( λ, κ ) . Let κ be anuncountable cardinal and { A ( k ) : k ∈ Z } be a countable infinite par-tition of κ into pairwise disjoint subsets of the cardinality κ , where Z stands for the integers. Denote the diagonal of the Cartesian product κ by ∆ = { ( x, x ) : x ∈ κ } and put ∆( k ) = ∆ ∩ A ( k ) .Fix an infinite cardinal number λ < κ and proper λ + -complete ideals I ( k, λ ) on the sets A ( k ) . In particular, we assume that singletons arein I ( k, λ ) , hence H ∈ I ( k, λ ) for any H ⊆ A ( k ) such that | H | λ .Consider a topology T on X = κ generated by the basis consisting ofall singletons { a } , whenever a ∈ κ \ ∆ , and all the sets { ( x, x ) } ∪ ( { x } × ( A ( k − \ G )) ∪ (( A ( k + 1) \ F ) × { x } ) , here x ∈ A ( k ) and | G | < λ and F ∈ I ( k + 1 , λ ) . We denote not-singleton basic sets by Γ( x, G, F ) . Lemma 1.
Assume that H ⊆ ∆( k ) ∩ V, where V is an open set in X .If the set { x ∈ A ( k ) : ( x, x ) ∈ H } does not belong to the ideal I ( k, λ ) ,then the difference ∆( k − \ cl X ( V ) has the cardinality less than λ .Proof. Suppose that a set { ( b α , b α ) : α < λ } ⊆ ∆( k − of the car-dinality λ is disjoint from cl X ( V ) . For each α < λ , fix a basic set Γ( b α , G α , F α ) disjoint from V , where F α ∈ I ( k, λ ) . The ideal I ( k, λ ) is λ + -complete and the set { x ∈ A ( k ) : ( x, x ) ∈ H } does not belong tothis ideal. So, there exists a point ( x, x ) ∈ H such that x ∈ A ( k ) \ [ { F α : α < λ } . Therefore ( x, b α ) ∈ ( A ( k ) \ F α ) × { b α } ⊆ Γ( b α , G α , F α ) for every α < λ . Fix a basic set Γ( x, G x , F x ) ⊆ V and α < λ such that b α ∈ A ( k − \ G x . We get ( x, b α ) ∈ { x } × ( A ( k − \ G x ) ⊆ V , acontradiction. (cid:3) Corollary 2.
The space X is completely regular, but not normal.Proof. The base consists of closed-open sets and one-points subsets of X are closed. So X , being a zero-dimensional T space, is completelyregular. Subsets ∆( k + 1) and ∆( k ) are closed and disjoint. By Lemma1, if a set V ⊆ X is open and ∆( k + 1) ⊆ V , then cl X ( V ) ∩ ∆( k ) = ∅ ,which implies that X is not a normal space. (cid:3) Proposition 3.
Assume that the cardinal λ has an uncountable cofi-nality. If f : X → R is a continuous real valued function, then for anypoint x ∈ κ there exists a basic set Γ( x, G x , F x ) such that the function f is constant on it.Proof. Without loss of generality, we can assume that f ( x, x ) = 0 .For each n > , fix a base set Γ( x, G n , F n ) ⊆ f − (( − n , n )) . Then put G x = ∪{ G n : n > } and F x = ∪{ F n : n > } . Since λ has anuncountable cofinality, we get that the set Γ( x, G x , F x ) belongs to thebase. Obviously, if ( a, b ) ∈ ∩{ Γ( x, G n , F n ) : n > } = Γ( x, G x , F x ) , then f ( a, b ) = 0 . (cid:3) hen λ has the countable cofinality, then the above proof also works,but then the set G x = ∪{ G n : n > } may have the cardinality λ , andtherefore Γ( x, G x , F x ) does not necessarily belong to the base, and alsoit could be not open. Furthermore, any continuous real valued functionmust also be constant onto other large subsets of X. Lemma 4.
Let k ∈ Z . If f : X → R is a continuous real valuedfunction, then for any ε > there exists a real number a such that f [∆( k − ⊆ [ a, a + 3 · ε ] for all but less than λ many points ( x, x ) ∈ ∆( k − .Proof. Fix a real number b and ε > . The ideal I ( k, λ ) is λ + -complete,so we can choose an integer q ∈ Z such that the subset { x ∈ A ( k ) : f ( x, x ) ∈ [ b + q · ε, b + ( q + 1) · ε ] } does not belong to I ( k, λ ) . Use Lemma 1, putting a = b + ( q − · ε and H = f − ([ a + ε, a + 2 · ε ]) ∩ ∆( k ) and V = f − (( a, a + 3 · ε )) . Since cl X ( V ) ⊆ f − ([ a, a + 3 · ε ]) , the proof is completed. (cid:3) Corollary 5. If f : X → R is a continuous real valued function, thenfor any k ∈ Z there exists a real number a k such that f ( x, x ) = a k forall but λ many points ( x, x ) ∈ ∆( k ) . Moreover, if λ has uncountablecofinality, then f ( x, x ) = a k for all but less than λ many x ∈ A ( k ) .Proof. Apply Lemma 4, substituting consecutively n for ε , for n > ,and k + 1 for k . (cid:3) Theorem 6. If f : X → R is a continuous real valued function, thenthere exists a ∈ R such that f ( x, x ) = a for all but λ many x ∈ κ .Moreover, when λ has an uncountable cofinality, then f ( x, x ) = a forall but less than λ many x ∈ κ .Proof. We shall to prove that the numbers a k which appear in Corollary5 are equal. To do this, suppose that a k = a k − for some k ∈ Z . Choosedisjoint open intervals J and I such that a k ∈ J and a k − ∈ I . ApplyLemma 1, taking H = { ( x, x ) ∈ ∆( k ) : f ( x, x ) = a k } and V = f − ( J ) .Since cl X ( V ) ∩ f − ( I ) = ∅ , we get f ( x, x ) = a k − for all but less than λ many x ∈ A ( k − , a contradiction. (cid:3) Knowing infinite cardinal numbers λ < κ and proper λ + -completeideals I ( k, λ ) on sets A ( k ) , one can extend the space X by one ortwo points so as to get a regular space which is not completely regular. his is a standard construction, compare [6], [10] and [2] or [4, Example1.5.9], so we will describe it briefly. Fix points + ∞ and −∞ that donot belong to X . On the set X ∗ = X ∪ {−∞ , + ∞} we introduce thefollowing topology. Let open sets in X be open in X ∗ , too. But thesets V + m = { + ∞} ∪ [ { A ( n ) × κ : n > m } form a base at the point + ∞ and the sets V − m = {−∞} ∪ [ { A ( n ) × κ : n m } \ ∆( m ) form a base at the point −∞ . Thus we have ∆( m ) = cl X ∗ ( V + m ) ∩ cl X ∗ ( V − m ) = cl X ( V + m ∩ X ) ∩ cl X ( V − m ∩ X ) , which gives that the space X ∗ is regular and not completely regular.Indeed, consider a closed subset D ⊆ X ∗ and a point p ∈ X ∗ \ D .When p ∈ X , then p has a closed-open neighborhood in X ∗ whichis disjoint with D . When p = + ∞ , then consider the basic set V + m which is disjoint with D and check cl X ∗ ( V + m +1 ) ⊆ V + m . Analogously,when p = −∞ , then consider the basic set V − m which is disjoint with D and check cl X ∗ ( V − m − ) ⊆ V − m . By Theorem 6, no continuous realvalued function separates an arbitrary closed set ∆( k ) from a point p ∈ { + ∞ , −∞} . Hence the space X ∗ is not completely regular. Thesame holds for subspaces X ∗ \ { + ∞} and X ∗ \ {−∞} . Moreover, if f : X ∗ → R is a continuous function, then f (+ ∞ ) = f ( −∞ ) . Now for convenience, the above defined space X is denoted X ( λ, κ ) ,whenever the ideals I ( λ, κ ) consist of sets of the cardinality less than λ . Assuming ω < λ < λ < κ we get two (non-comparable) non-homeomorphic spaces X ( ω, λ ) and X ( λ , κ ) , since the first one hasthe cardinality λ . But a subspace of X ( λ , κ ) of the cardinality λ is discrete and its closure in ( X ( ω, λ )) ∗ , being zero-dimensional, iscompletely regular. In other words, spaces ( X ( ω, λ )) ∗ and ( X ( λ , κ )) ∗ have non-comparable regularity ranks.3. General approach
The analysis conducted above can be generalized using some knowncounterexamples. We apply such a generalization to the Niemytzkiplane, cf. [4, p. 34] or [11, pp. 100 - 102], the Songefrey’s half-opensquare topology, cf. [11, pp. 103 - 105] and special Isbell-Mrówkaspaces (which are also known as Ψ -spaces). iven a space X and a closed and discrete subset ∆ X ⊆ X , assumethat ∆ X can be partitioned onto pairwise disjoint subsets ∆ X ( k ) . Foreach k ∈ Z , let I X ( k ) be a proper ideal on ∆ X ( k ) . Suppose that thefollowing property is fulfilled:( ∗ ). If a set V ⊆ X is open and the set ∆ X ( k ) \ V belongs to I X ( k ) ,then the set ∆ X ( k − \ cl X ( V ) belongs to I X ( k − . Then it is possible to give a general scheme of a construction of a com-pletely regular space Y = Y X , which has one-point extension to aregular space which is not completely regular and two-point extensionto a regular space such that no continuous real valued function sepa-rates the extra points, whereas removing a single point we get a regularspace which is not completely regular. To get this we put x k = ( ( k, x ) , when x ∈ X \ ∆ X ; { ( k, x ) , ( k + 1 , x ) } , when x ∈ ∆ X ( k ) . And then put Y X = { x k : x ∈ X and k ∈ Z } . Endow Y X with thetopology as follows. If k ∈ Z and V ⊆ X \ ∆ X is an open subset of X ,then the set { x k : x ∈ V } is open in Y X . Thus we define neighborhoodsof the point x k where x / ∈ ∆ X . To define neighborhoods of the point x k , where x ∈ ∆ X , we use the formula: If k ∈ Z and V ⊆ X is an opensubset, then the set { x k : x ∈ V } ∪ { x k +1 : x ∈ V \ ∆ X } is open in Y X . To get a version of ( ∗ ) , we put the following: ∆ Y ( k ) = { x k : x ∈ ∆ X ( k ) } ; ∆ Y = S { ∆ X ( k ) : k ∈ Z } ; Let I Y ( k ) be a properideal which consists of sets { x k : x ∈ A } for A ∈ I X ( k ) ; Y k = { y k : y ∈ X \ ∆ X } . So, if k ∈ Z , then ∆ Y ( k ) = cl Y ( { y k : y ∈ X \ ∆ X } ) ∩ cl Y ( { y k +1 : y ∈ X \ ∆ X } ) . As we can see, the properties of the space Y X can be automaticallyrewritten from the relevant properties of X , so we leave details to thereader. Proposition 7.
Assume that a space X satisfied ( ∗ ) and the space Y is as above. If a set V ⊆ Y is open and the set ∆ Y ( k ) \ V belongs to I Y ( k ) , then the set ∆ Y ( k − \ cl Y ( V ) belongs to I Y ( k − . (cid:3) Proposition 8.
If a space X is completely regular, then the space Y is completely regular, too. (cid:3) ow, fix points + ∞ and −∞ that do not belong to Y . On the set Y ∗ = Y ∪ {−∞ , + ∞} we introduce the following topology. Let opensets in Y be open in Y ∗ , too. But the sets V + m = { + ∞} ∪ [ { Y n : n > m } ∪ [ { ∆ Y ( n ) : n > m } form a base at the point + ∞ and the sets V − m = {−∞} ∪ [ { Y n : n m } ∪ [ { ∆ Y ( n ) : n < m } form a base at the point −∞ . Thus we have ∆ Y ( m ) ⊆ cl Y ∗ ( V + m ) ∩ cl Y ∗ ( V − m ) = ∆ Y ( m ) ∪ Y m , which implies the following. Theorem 9. If f : Y ∗ → R is a continuous real valued function, then f (+ ∞ ) = f ( −∞ ) .Proof. Suppose f : Y ∗ → R is a continuous function such that f (+ ∞ ) = 1 and f ( −∞ ) = 0 . Fix a decreasing sequence { ǫ n } whichconverges to . Thus f − (( ǫ n , ⊆ cl Y ∗ ( f − (( ǫ n , ⊆ f − ([ ǫ n , ⊆ f − (( ǫ n +1 , . By Proposition 7, if K m ∈ I Y ( m ) and ∆ Y ( m ) \ K m ⊆ f − (( ǫ n , , then f − (( ǫ n +1 , ⊇ ∆ Y ( m − \ K m − , for some K m − ∈ I Y ( m − . Since there exists m ∈ Z such that + ∞ ∈ V + m ⊆ f − (( ǫ , , inductively, we get ∆ Y \ [ { K n : n ∈ Z } ⊆ f − ([ 12 , , which implies that each V − n contains a point y ∈ Y such that f ( y ) > . Hence f ( −∞ ) > , a contradiction. (cid:3) Application of the Niemytzki plane.
Recall that theNiemytzki plane P = { ( a, b ) ∈ R × R : 0 b } is the closed half-plane which is endowed with the topology generated by open discsdisjoint with the real axis ∆ P = { ( x,
0) : x ∈ R } and all sets of theform { a } ∪ D where D ⊆ P is an open disc which is tangent to ∆ P atthe point a ∈ ∆ P . Choose pairwise disjoint subsets ∆ P ( k ) ⊆ ∆ P , where k ∈ Z , such that each set ∆ P ( k ) meets every dense G δ subset of the realaxis. To do that is enough to slightly modify the classic constructionof a Bernstein set. Namely, fix an enumeration { A α : α < c } of alldense G δ subsets of the real axis. Defining inductively at step α choose (1-1)-numerated subset { p αk : k ∈ Z } ⊆ A α \ { p βk : k ∈ Z and β < α } .Then, for each k ∈ Z , put ∆ P ( k ) = { p αk : α < c } . Let us assume that if F ⊆ R × R , then the topology on F inducedfrom the Euclidean topology will be called the natural topology on F . Aset, which is a countable union of nowhere dense subsets in the naturaltopology on F , will be called a set of first category in F . Our proofof the following lemma is a modification of known reasoning justifyingthat P is not a normal space, compare [11, pp. 101 -102]. Lemma 10.
Let a set F ⊆ ∆ P be a dense subset in the natural topologyon the real axis ∆ P . If a set V is open in P and F ⊆ V , then the set ∆ P \ cl P ( V ) is of first category in ∆ P .Proof. To each point a ∈ ∆ P \ cl P ( V ) there corresponds a disc D a ⊆ P \ cl P ( V ) of radius r a tangent to ∆ P at the point a . Put S n = { a ∈ ∆ P \ cl P ( V ) : r a > n } and use density of F to check that each S n is nowhere dense in thenatural topology on ∆ P . So S { S n : n > } = ∆ P \ cl P ( V ) . (cid:3) The space Y P is completely regular. The subspaces Y P ∪ {−∞} , Y P ∪{ + ∞} and and the space Y ∗ P are regular. Moreover, if f : Y ∗ P → R is a continuous real valued function, then f (+ ∞ ) = f ( −∞ ) .3.2. Application of the Songefrey plane, i.e. application of theSongenfrey’s half-open square topology.Recall that the Songefrey plane S = { ( a, b ) : a ∈ R and b ∈ R } isthe plane endowed with the topology generated by rectangles of theform [ a, b ) × [ c, d ) . Let ∆ S = { ( x, − x ) : x ∈ R } . Since ∆ S with thetopology induced from the Euclidean topology is homeomorphic withthe real line, we can choose pairwise disjoint subsets ∆ S ( k ) ⊆ ∆ S suchthat each set ∆ S ( k ) meets every dense G δ subset of ∆ S . The followinglemma can be proved be the second category argument used previouslyin the proof Lemma 10, so we omit it, compare also [11, pp. 103 -104]. Lemma 11.
Let a set F ⊆ ∆ S be a dense subset in the topology on ∆ S which is inherited from the Euclidean topology. If a set V is open in S and F ⊆ V , then the set ∆ S \ cl S ( V ) is of first category in ∆ S . (cid:3) gain, the space Y S is completely regular. The subspaces Y S ∪{−∞} , Y S ∪ { + ∞} and and the space Y ∗ S are regular. Moreover, if f : Y ∗ S → R is a continuous real valued function, then f (+ ∞ ) = f ( −∞ ) .3.3. Applications of some Ψ -spaces. Let us recall some notionsneeded to define a Lusin gap, compare [8]. A family of sets is called almost disjoint , whenever any two members of it have the finite in-tersection. A set C separates two families, whenever each member ofthe first family is almost contained in C , i.e. B \ C is finite for any B ∈ Q , and each member of the other family is almost disjoint with C .An uncountable family L , which consists of almost disjoint and infinitesubsets of ω , is called Lusin-gap, whenever no two its uncountable anddisjoint subfamilies can be separated by a subset of ω . Adapting con-cepts discussed in [9] or [5], to a Lusin-gap L , let Ψ( L ) = L ∪ ω . Atopology on Ψ( L ) is generated as follows. Any subset of ω is open, alsofor each point A ∈ L the sets { A } ∪ A \ F , where F is finite, are open. Proposition 12. If L is a Lusin-gap and S { ∆ L ( k ) : k ∈ Z } = L ,then the space Ψ( L ) satisfies the property ( ∗ ) , whenever sets ∆ L ( k ) areuncountable and pairwise disjoint and each ideal I L ( k ) consists of allcountable subsets of ∆ L ( k ) .Proof. Consider uncountable and disjoint families A , B ⊆ L . Suppose
A ⊆ V and B ⊆ W, where open sets V and W are disjoint. Let C = [ { A ⊆ ω : { A } ∪ A is almost contained in V } . The set C separates families A and B , which contradicts that L is aLusin-gap. Setting A = ∆ L ( k ) and B = ∆ L ( k − , we are done. (cid:3) The space Y L is completely regular. Again by Theorem 9, we getthe following. The subspaces Y L ∪ {−∞} , Y L ∪ { + ∞} and and thespace Y ∗L are regular. Moreover, if f : Y ∗L → R is a continuous realvalued function, then f (+ ∞ ) = f ( −∞ ) .4. Comment
In [6], F. B. Jones formulated the following problem:
Does a non-completely regular space always contain a substructure similar to thatpossessed by Y ? Jones’ space Y is constructed by gluing (sewing) count-ably many disjoint copies of a suitable space X . This method fixes two ubsets of X and consists in sewing alternately copies of either of them.On the other hand, our method consists in gluing different sets at eachstep. The problem of Jones may be understood as an incentive tostudy the structural diversity of regular spaces, which are not com-pletely regular. Even though the meaning of "a substructure similarto that possessed by Y " seems vague, we think that an appropriatecriterion for the aforementioned diversity is a slightly modified conceptof a topological rank, compare [7] or [12]. We have introduced regular-ity ranks, but our counterexamples are only a preliminary step to thestudy of diversity of regular spaces. References [1] D. Chodounský,
Non-normality and relative normality of Niemytzki plane .Acta Univ. Carolin. Math. Phys. 48 (2007), no. 2, 37-41.[2] K. Ch. Ciesielski and J. Wojciechowski,
Cardinality of regular spaces admittingonly constant continuous functions . Topology Proc. 47 (2016), 313-329.[3] R. Engelking,
General topology . Mathematical Monographs, Vol. 60, PWN—Polish Scientific Publishers, Warsaw, (1977).[4] R. Engelking,
Topologia ogólna I , Państwowe Wydawnictwo Naukowe,Warszawa (1989).[5] F. Hernández-Hernández and M. Hrušák,
Q-sets and normality of Ψ -spaces .Spring Topology and Dynamical Systems Conference. Topology Proc. 29(2005), no. 1, 155-165.[6] F. B. Jones, Hereditarily separable, non-completely regular spaces , Proceed-ings of the Blacksburg Virginia Topological Conference, March (1973).[7] K. Kuratowski,
Topology-Volume I . Transl. by J. Jaworowski, Academic Press,New York-London; Pañstwowe Wydawnictwo Naukowe Polish Scientific Pub-lishers, Warsaw (1966).[8] N. Luzin,
On subsets of the series of natural numbers , Isv. Akad. Nauk. SSSRSer. Mat. 11 (1947), 403-411.[9] S. Mrówka,
On completely regular spaces , Fund. Math. 41 (1954), 105-106.[10] A. Mysior,
A regular space which is not completely regular . Proc. Amer. Math.Soc. 81 (1981), no. 4, 652-653.[11] L.A. Steen and J.A.jun. Seebach,
Counterexamples in topology . New York etc.:Holt, Rinehart and Winston, Inc., XIII, (1970).[12] W. Sierpiński,
Introduction to General Topology . Lectures in Mathematics atthe University of Toronto. The University of Toronto Press (1934). iotr Kalemba, Institute of Mathematics, University of Silesia, ul.Bankowa 14, 40-007 Katowice E-mail address : [email protected] Szymon Plewik, Institute of Mathematics, University of Silesia, ul.Bankowa 14, 40-007 Katowice
E-mail address : [email protected]@math.us.edu.pl