On regular separable countably compact R -rigid spaces
aa r X i v : . [ m a t h . GN ] J u l ON REGULAR SEPARABLE COUNTABLY COMPACT R -RIGID SPACES SERHII BARDYLA AND LYUBOMYR ZDOMSKYY
Abstract.
We call a regular topological space X to be R -rigid if any continuous real-valuedfunction on X is constant. In this paper we construct a number of consistent examples of count-ably compact R -rigid spaces with additional properties like separability and first countability.This way we answer several questions of Tzannes, Banakh, Ravsky, and the first author, as wellas get a consistent R -rigid counterexample to a classical problem of Nyikos. Introduction R -rigid spaces (or, more generally, spaces on which every continuous function into a given space Y is constant) are of particular interest in general topology, see, e.g., [1, 6, 7, 8, 12, 14, 15, 24,25, 26, 27]. For a long time all known examples of regular R -rigid spaces were far from beingcountably compact. In [26] Tzannes constructed a Hausdorff countably compact R -rigid space T such that no pair of distinct points of T have disjoint closed neighborhoods, i.e., this space is farfrom being regular. This motivated Tzannes to ask the following two questions: Problem 1 ([23, Problem C65]) . Does there exist a regular (first countable, separable) countablycompact space on which every continuous real-valued function is constant?
Problem 2 ([23, Problem C66]) . Does there exist, for every Hausdorff space R , a regular (firstcountable, separable) countably compact space on which every continuous function into R is con-stant? Recently, the first author jointly with Osipov obtained the following affirmative answer to bothquestions of Tzannes without the properties mentioned in brackets.
Theorem 1.1 ([4, Theorem 3]) . For each cardinal κ ≥ ω there exists a regular infinite space R κ such that the closure in R κ of any X ∈ [ R κ ] ≤ κ is compact, and for any T space Y of pseudochar-acter ψ ( Y ) ≤ κ each continuous map f : R κ → Y is constant. However, for any κ ≥ ω the space R κ is neither separable nor first-countable. Since R -rigidspaces cannot be Tychonoff (and hence are non-compact), the ultimate version of Problem 1may be thought of as a strong form of the following notoriously difficult old problem of Nyikos,see [19, 20] for the history thereof and related results. Problem 3 ([20, Problem 1]) . Does ZFC imply the existence of a regular separable, first countable,countably compact, non-compact space?
In [2] Banakh, Ravsky and the first author posed the following problem, which is a weaker formof Nyikos’ Problem 3.
Problem 4 ([2, Question 4.9]) . Does there exist in ZFC an example of a separable regular se-quentially compact space which is not Tychonoff ?
In this paper we construct a ZFC example of a regular separable totally countably compact R -rigid space, thus answering Problem 1 for separable spaces in the affirmative, see Theorem 3.3.On the other hand, it is easy to see that the answer to Problem 2 for separable spaces is negative, Mathematics Subject Classification.
Primary 54C30. Secondary 54D35, 03E17.
Key words and phrases.
Countably compact space, separable space, constant function, R -rigid space.The authors were partially supported by the Austrian Science Fund FWF Grant I 3709-N35. The second authoralso thanks the Austrian Science Fund FWF (Grant I 2374-N35) for generous support for this research. see Remark 4.5. In Theorem 4.2 we prove that consistently, for each cardinal κ < c there exists aregular separable totally countably compact space admitting only constant continuous functionsinto any T space of pseudocharacter ≤ κ , see also Proposition 4.3. In addition we construct aconsistent example of a regular separable first-countable countably compact R -rigid space, thusproviding a consistent counterexample to Problem 3 with additional strong topological properties,see Theorem 5.2. Finally, by using the non-preservation of the normality by products of “nice”spaces, in Theorem 5.1 we provide an affirmative answer to Problem 4 outright in ZFC.The notions used but not defined in this paper are standard and can be found in [9, 10, 18].2. van Douwen’s extension For any distinct points a, b of a topological space X by Const( X ) a,b we denote the class of T spaces such that f ( a ) = f ( b ) for any continuous mapping f : X → Y ∈ Const( X ) a,b .In this section we slightly modify the construction of van Douwen [8]. In particular, for anydistinct non-isolated points a, b of a separable regular space X we shall construct a space D a,b ( X )such that for any Y ∈ Const( X ) a,b , each continuous function h : D a,b ( X ) → Y is constant.Moreover, the space D a,b ( X ) remains regular and separable.Let X be a regular separable space and a, b ∈ X be non-isolated points. Put B ( a ) = { U ⊂ X | U is an open neighborhood of a and b / ∈ U } and B ( b ) = { U ⊂ X | U is an open neighborhood of b and a / ∈ U } . Let Z be the Tychonoff product X × ω where ω is endowed with the discrete topology. For any x ∈ X and n ∈ ω by x n we denote the point ( x, n ). Analogously, for any B ⊂ X and n ∈ ω theset B ×{ n } is denoted by B n . Let D be a dense countable subset of X such that D ⊂ X \ { a, b } .Put A = ∪ n ∈ ω D n . Fix any bijection f : A → ω such that f ( x n ) = n for each n ∈ ω and x n ∈ D n .On the set Z consider the smallest equivalence relation ∼ which satisfies the following conditions: • x n ∼ a f ( x n ) for any n ∈ ω and x ∈ D ; • b n ∼ b m for any n, m ∈ ω .Let D a,b ( X ) be the quotient space Z/ ∼ . By π we denote the natural projection of Z onto D a,b ( X ). Put b ∗ = π ( b n ) ∈ D a,b ( X ). Observe that the definition of the element b ∗ does notdepend on the choice of n ∈ ω . A subset B ⊂ Z is called saturated if π − ( π ( B )) = B . Lemma 2.1.
Let X be a regular separable space. Then the space D a,b ( X ) is regular, separableand any continuous map g : D a,b ( X ) → Y ∈ Const( X ) a,b is constant.Proof. Observe that the countable set π ( A ) is dense in D a,b ( X ) witnessing that the space D a,b ( X )is separable. Since each equivalence class of the relation ∼ is closed in Z the space D a,b ( X ) is T .Fix any q ∈ D a,b ( X ) \ { b ∗ } and an open neighborhood U of q . Let { x n } = π − ( q ) \ { a k | k ∈ ω } .By the definition of the quotient topology, to prove the regularity of D a,b ( X ) at the point q it issufficient to find an open saturated subset B ⊂ Z and a closed saturated subset C ⊂ Z such that x n ∈ B ⊂ C ⊂ π − ( U ) (note that if x n ∈ B , then π − ( q ) ⊂ B ). The subsets B and C will beconstructed inductively. Since X n is a clopen subspace of the regular space Z and x n / ∈ { a n , b n } there exists an open subset V ⊂ X n such that a n , b n / ∈ V and x n ∈ V ⊂ V ⊂ π − ( U ). Put B (0) = C (0) = ∅ , B (1) = V and C (1) = V . Assume that we already constructed subsets B ( n ) and C ( n ) for every n ≤ k . Put B ∗ ( k ) = ( B ( k ) \ B ( k − ) ∩ A and C ∗ ( k ) = ( C ( k ) \ C ( k − ) ∩ A . Fixany function t : C ∗ ( k ) → B ( a ) such that t ( y ) f ( y ) ⊂ π − ( U ) for each y ∈ C ∗ ( k ) . Since π − ( U ) isan open saturated subset and the space Z is regular such a function exists. Put s = t ↾ B ∗ ( k ) . Let B ( k +1) = ∪{ s ( y ) f ( y ) | y ∈ B ∗ ( k ) } ∪ B ( k ) and C ( k +1) = ∪{ t ( y ) f ( y ) | y ∈ C ∗ ( k ) } ∪ C ( k ) . Finally, put B = ∪ n ∈ ω B ( n ) and C = ∪ n ∈ ω C ( n ) . At this point it is straightforward to check that the set B ( C , resp.) is an open (closed, resp.) saturated subset of Z such that π − ( q ) ⊂ B ⊂ C ⊂ π − ( U ).Hence the space D a,b ( X ) is regular at any point q = b ∗ .Fix any open neighborhood U of b ∗ in D a,b ( X ). By the regularity of Z , for each n ∈ ω thereexists an open neighborhood V n ⊂ X n of b n such that V n ⊂ V n ⊂ π − ( U ). Put B (0) = C (0) = ∅ B (1) = ∪ n ∈ ω V n and C (1) = ∪ n ∈ ω V n . Observe that the family { V n } n ∈ ω is locally finite which yields N REGULAR SEPARABLE COUNTABLY COMPACT R -RIGID SPACES 3 that the set C (1) is closed. Next, repeating the previous inductive construction we can find a desiredsets B and C such that B is open saturated, C is closed saturated and π − ( b ∗ ) ⊂ B ⊂ C ⊂ π − ( U ).Hence the space D a,b ( X ) is regular.Fix any continuous map g : D a,b ( X ) → Y ∈ Const( X ) a,b . Note that for any n ∈ ω , x ∈ X n and an open neighborhood U ⊂ X n of x there exists an open saturated subset W ⊂ Z such that U ⊂ W and W ∩ X n = U . Thus the map h : X → π ( X n ), h ( x ) = π ( x n ) defines a homeomorphismbetween X and the subspace π ( X n ) of D a,b ( X ). Hence for any n ∈ ω , g ( π ( a n )) = g ( π ( b n )) by thechoice of Y . The definition of the equivalence relation ∼ implies that g ( π ( a n )) = g ( b ∗ ) for any n ∈ ω . Since the set { π ( a n ) | n ∈ ω } = π ( A ) is dense in D a,b ( X ) and Y is a T space, the map g is constant on the whole D a,b ( X ). (cid:3) The following lemmas will be useful in the next sections.
Lemma 2.2.
Let X be a regular separable countably compact space. Then a subset C of D a,b ( X ) is closed and discrete iff | C ∩ π ( X n ) | < ω for each n ∈ ω .Proof. Let C be a closed discrete subset of D a,b ( X ). Recall that the subspace π ( X n ) ⊂ D a,b ( X ) ishomeomorphic to X . Since X is countably compact and C is closed and discrete, the set C ∩ π ( X n )is finite for each n ∈ ω .Let C be any subset of D a,b ( X ) such that | C ∩ π ( X n ) | < ω for each n ∈ ω . Observe that forany x ∈ C the set U = Z \ π − ( C \ { x } ) is an open saturated subset of Z which contains π − ( x ).Thus π ( U ) is an open neighborhood of x in D a,b ( X ) such that π ( U ) ∩ C = { x } . Hence the set C is discrete.It is straightforward to check that the set π − ( C ) is closed in Z . By the definition of thequotient topology the set C is closed in D a,b ( X ). (cid:3) Corollary 2.3.
Let X be a regular separable countably compact space. Then for each countablefamily F = { C ( n ) } n ∈ ω of countable closed discrete subsets of D a,b ( X ) there exists a countableclosed discrete subset C ⊂ D a,b ( X ) such that C ( n ) ⊂ ∗ C for each n ∈ ω .Proof. Fix any countable family F = { C ( n ) } n ∈ ω of countable closed discrete subsets of D a,b ( X ).By Lemma 2.2, the set C ( n ) ∩ π ( X k ) is finite for any n, k ∈ ω . Put C ∗ ( n ) = C ( n ) \ ( ∪ k ≤ n π ( X k )), n ∈ ω . Let C = ∪ n ∈ ω C ∗ ( n ) . Observe that | C ∩ π ( X n ) | < ω for each n ∈ ω . Lemma 2.2 impliesthat C is closed and discrete in D a,b ( X ). Since C ( n ) ⊂ ∗ C ∗ ( n ) , we have that C ( n ) ⊂ ∗ C for each n ∈ ω . (cid:3) Following [3], a subset P of a space X is called strongly discrete if each point x ∈ P has aneighborhood O x ⊂ X such that the family { O x | x ∈ P } is disjoint and locally finite in X . Aspace X is said to have property D if any countable closed discrete subset of X is strongly discrete. Lemma 2.4.
Let X be a regular separable countably compact space. Then D a,b ( X ) has property D .Proof. Fix any countable closed discrete subset P of D a,b ( X ). With no loss of generality we canassume that b ∗ / ∈ P . Lemma 2.2 implies that for each n ∈ ω the set π − ( P ) ∩ X n is finite. Itfollows that π − ( P ) is a closed discrete subset of Z . For each p ∈ P there exists x ( p ) ∈ X and n ( p ) ∈ ω such that { x ( p ) n ( p ) } = π − ( p ) \ { a k | k ∈ ω } . Fix an open subset V ( p ) ⊂ X n ( p ) whichsatisfies the following conditions:( i ) x ( p ) n ( p ) ∈ V ( p );( ii ) V ( p ) ∩ V ( q ) = ∅ if p = q ;( iii ) a n ( p ) / ∈ V ( p );( iv ) b n ( p ) / ∈ V ( p ).Next we shall inductively construct open saturated pairwise disjoint subsets W ( p ) ⊂ Z , p ∈ P such that π − ( p ) ⊂ W ( p ) and the family { π ( W ( p )) | p ∈ P } is locally finite. For each p ∈ P put W ( p ) (0) = ∅ and W ( p ) (1) = V ( p ). Assume that we already constructed subsets W ( p ) ( n ) for each n ≤ k and p ∈ P . Put W ( p ) ∗ ( k ) = ( W ( p ) ( k ) \ W ( p ) ( k − ) ∩ A and fix any function g : W ( p ) ∗ ( k ) → B ( a ) which satisfies the following condition: S. BARDYLA AND L. ZDOMSKYY ( v ) g ( z ) f ( z ) ∩ V ( q ) = ∅ for every z ∈ W ( p ) ∗ ( k ) and q ∈ P .Such a function exists by the choice of V ( p ), p ∈ P (see condition ( iii ) above). Let W ( p ) ( k +1) = ∪{ g ( z ) f ( z ) | z ∈ W ( p ) ∗ ( k ) } ∪ W ( p ) ( k ) . Finally, for each p ∈ P put W ( p ) = ∪ n ∈ ω W ( p ) ( n ) . It is easyto see that subsets W ( p ), p ∈ P are open and saturated. Since the map f is bijective the sets W ( p ), p ∈ P are pairwise disjoint.It remains to show that the family { π ( W ( p )) | p ∈ P } is locally finite. For this fix any n ∈ ω and x n ∈ X n \ { a n } . Next we shall inductively construct an open saturated subset U ⊂ Z whichcontains x n and intersects only finitely many members of the family { W ( p ) | p ∈ P } . If x n ∈ W ( p )for some p ∈ P , then put U = W ( p ). If x n / ∈ ∪{ W ( p ) | p ∈ P } there are two cases to consider:1. x n = b n . Put U (0) = ∅ and U (1) = X n \ ( π − ( P ) ∪{ a n } ). Assume that we already constructed U ( n ) for each n ≤ k . Let U ∗ ( k ) = ( U ( k ) \ U ( k − ) ∩ A . Consider any function g : U ∗ ( k ) → B ( a ) suchthat g ( z ) f ( z ) ∩ V ( p ) = ∅ for any p ∈ P and z ∈ U ∗ ( k ) . The existence of such a function is providedby the bijectivity of f together with the choice of the sets V ( p ) (see condition ( iii ) above). Put U ( k +1) = ∪{ g ( y ) f ( y ) | y ∈ U ∗ ( k ) } ∪ U ( k ) . Finally, let U = ∪ n ∈ ω U ( n ) . Observe that there exist onlyfinitely many p ∈ P such that W ( p ) ∩ X n = ∅ . It is easy to check that U is an open saturatedsubset of Z which contains x n and intersects only finitely many elements of the set { W ( p ) | p ∈ P } .Namely, U ∩ W ( p ) = ∅ iff W ( p ) ∩ X n = ∅ .2. x n = b n . The choice of the sets V ( p ) (see condition ( iv ) above) and the definition of the sets W ( p ) imply that for each k ∈ ω there exists an open neighborhood V k of b k such that ∪{ W ( p ) | p ∈ P } ∩ V k = ∅ . Put U (0) = ∅ , U (1) = ∪ k ∈ ω V k and repeat the previous inductive constructionto obtain an open saturated subset U ⊂ Z such that { b k | k ∈ ω } ⊂ U and U ∩ W ( p ) = ∅ for all p ∈ P .Hence for any x ∈ Z there exists an open saturated set U such that x ∈ U and π ( U ) ∩ π ( W ( p )) = ∅ for all but finitely many p ∈ P witnessing that the family { π ( W ( p )) | p ∈ P } is locally finite. (cid:3) Separable regular totally countably compact R -rigid spaces in ZFC A space X is called totally countably compact if each infinite set in X contains an infinite subsetwith compact closure in X . A family F of closed subsets of a topological space X is called a closedultrafilter if it satisfies the following conditions: • ∅ / ∈ F ; • A ∩ B ∈ F for any A, B ∈ F ; • a closed set F ⊂ X belongs to F if F ∩ A = ∅ for every A ∈ F .For any A ⊂ X put h A i = {F ∈ W ( X ) | there exists F ∈ F such that F ⊂ A } . Following [10, § W ( X ) of a T space X consists of closed ultrafilters on X and carries the topology generated by the base consisting of the sets h U i where U runs throughopen subsets of X .By [10, Theorem 3.6.21], the Wallman extension W ( X ) is T and compact. Notice that a T space X is normal if and only if W ( X ) is Hausdorff [10, 3.6.22]. Consider the map j X : X → W ( X )assigning to each point x ∈ X the principal closed ultrafilter consisting of all closed sets F ⊂ X containing the point x . It is clear that the image j X ( X ) is dense in W ( X ). Since the space X is T , Theorem 3.6.21 from [10] provides that the map j X : X → W ( X ) is a topological embedding.So, X can be identified with the subspace j X ( X ) of W ( X ). Before formulating the main result ofthis section we need a few auxiliaries lemmas. Lemma 3.1 ([4, Lemma 1]) . Let
F ∈ W ( X ) and A ⊂ X . Then F ∈ cl W ( X ) ( A ) if and only if cl X ( A ) ∈ F . For a topological space X by cW ( X ) we denote the following subspace of the Wallman extension W ( X ): cW ( X ) = ∪{ cl W ( X ) ( A ) | A is a countable discrete closed subset of X } . N REGULAR SEPARABLE COUNTABLY COMPACT R -RIGID SPACES 5 Lemma 3.2.
Let X be a regular space which has property D and for each countable family F = { C ( n ) } n ∈ ω of countable closed discrete subsets of X there exists a countable closed discrete subset C ⊂ X such that C ( n ) ⊂ ∗ C for each n ∈ ω . Then the space cW ( X ) is regular and countablycompact.Proof. Since the Wallman extension W ( X ) is T , the space cW ( X ) is T as well. To prove theregularity of cW ( X ) we will show that W ( X ) is regular at any point of cW ( X ). Fix any x ∈ X andopen neighborhood h U i of x in W ( X ). By the regularity of X , there exists an open neighborhood V ⊂ X of x such that cl X ( V ) ⊂ U . Lemma 3.1 provides that h V i ⊂ h cl X ( V ) i = cl W ( X ) ( h V i ) ⊂h U i . Hence the space W ( X ) is regular at any point x ∈ X .Consider any closed ultrafilter F ∈ cW ( X ) \ X and fix any open neighborhood h U i of F in W ( X ). Take any F ∈ F such that F ⊂ U . By the definition of cW ( X ), there exists a countableclosed discrete subset G ⊂ X such that F ∈ cl W ( X ) ( G ). Lemma 3.1 implies that G ∈ F . Put H = F ∩ G and note that H is closed and discrete, H ⊂ U and H ∈ F . Since the space X satisfies property D , the subset H is strongly discrete. Hence there exists a locally finite family { W ( y ) | y ∈ H } of pairwise disjoint open subsets of X such that y ∈ W ( y ) for each y ∈ H . Also,with no loss of generality we can assume that ∪{ W ( y ) | y ∈ H } ⊂ U . By the regularity of X , forevery y ∈ H there exists an open subset V ( y ) of X such that y ∈ V ( Y ) ⊂ cl X ( V ( y )) ⊂ W ( Y ).Since the family { W ( y ) | y ∈ H } is locally finite the set ∪{ cl X ( V ( y )) | y ∈ H } is closed in X .Then Lemma 3.1 implies the following: F ∈ h∪{ V ( y ) | y ∈ H }i ⊂ h∪{ cl X ( V ( y )) | y ∈ H }i = cl W ( X ) ( h∪{ V ( y ) | y ∈ H }i ) ⊂ h U i . Hence the space W ( X ) is regular at any point of cW ( X ) witnessing that cW ( X ) is regular.It remains to show that cW ( X ) is countably compact. Assuming the contrary, let A be aninfinite closed discrete subset of cW ( X ). The definition of the space cW ( X ) implies that the set A ∩ X is finite. Consider any countable infinite subset D = {F ( n ) | n ∈ ω } ⊂ A ∩ ( cW ( X ) \ X ).For each n ∈ ω there exists a countable closed discrete subset C ( n ) ⊂ X such that C ( n ) ∈ F ( n ) .By the assumption, there exists a countable closed discrete subset C ⊂ X such that C ( n ) ⊂ ∗ C .Note that C ∈ F ( n ) for each n ∈ ω . Then {F ( n ) | n ∈ ω } is contained in a compact subsetcl cW ( X ) ( C ) = cl W ( X ) ( C ) which implies a contradiction. Hence the space cW ( X ) is countablycompact. (cid:3) Now we are able to formulate the main result of this section which gives an affirmative answerto Problem 1 for separable spaces.
Theorem 3.3.
Each separable regular non-normal (totally) countably compact space X can beembedded into a regular separable (totally) countably compact R -rigid space R ( X ) .Proof. To see that separable regular non-normal totally countably compact spaces exist, fix anyultrafilter p ∈ ω ∗ such that the subspace β ( ω ) \ { p } is non-normal. The existence in ZFC of suchan ultrafilter was proved by Blaszczyk and Szyma´nski in [5]. We claim that β ( ω ) \ { p } is separable,regular, non-normal and totally countably compact. The first three conditions are obvious. Fixany countable subset A ⊂ β ( ω ) \ { p } . Since β ( ω ) contains only trivial convergent sequences thereexists an open neighborhood U of p in β ( ω ) such that the set B = A \ U is infinite. Then B is acompact subset of β ( ω ) \ { p } witnessing that the space β ( ω ) \ { p } is totally countably compact.Let X be any separable regular non-normal (totally) countably compact space. We shall makethree steps to construct a desired space R ( X ). Step 1 . Using Jones machine [16], we shall construct a separable regular (totally) countablycompact space J ( X ) which is not functionally Hausdorff. Let A and B be two disjoint closedsubsets of X which cannot be separated by disjoint open subsets. Let Z be a discrete spaceof integers and a, b be distinct points which do not belong to X × Z . By R we denote the set X × Z ∪ { a, b } endowed with the topology τ which satisfies the following conditions: • the Tychonoff product X × Z is open in R ; • if a ∈ U ∈ τ , then there exists n ∈ N such that { ( x, − k ) | x ∈ X and k > n } ⊂ U ; • if b ∈ U ∈ τ , then there exists n ∈ N such that { ( x, k ) | x ∈ X and k > n } ⊂ U . S. BARDYLA AND L. ZDOMSKYY
One can check that the space R is separable, regular and (totally) countably compact.On the space R consider the smallest equivalence relation ∼ such that( x, n ) ∼ ( x, n + 1) and ( y, n ) ∼ ( y, n − n ∈ Z , x ∈ A and y ∈ B .Let J ( X ) be the quotient space R/ ∼ . Being a continuous image of the separable (totally)countably compact space R the space J ( X ) is separable and (totally) countably compact. Itis straightforward to check that J ( X ) is regular and contains a homeomorphic copy of X (forinstance, one can consider the subspace π ( X × { n } ) ⊂ J ( X ) for any integer n ). We claim that f ( a ) = f ( b ) for each continuous map f : J ( X ) → R . Indeed, if f ( a ) = f ( b ), then fix a sequence { U ( n ) | n ∈ ω } of open neighborhoods of f ( a ) such that f ( b ) / ∈ ∪ n ∈ ω U ( n ) and U ( n ) ⊂ U ( n +1) for each n ∈ ω . Let k = max { n ∈ Z | π ( A ×{ n } ) ⊂ f − ( U ) } . Observe that B = f − ( U ) ∩ π ( B ×{ k + 1 } ) is a nonempty subset of f − ( U (1) ) witnessing that f − ( U (1) ) ∩ π ( X × ( k + 1)) = ∅ .It is easy to check that the closed sets B and π ( A × ( k + 2)) cannot be separated by disjointopen subsets in J ( X ). Thus A = f − ( U ) ∩ π ( A ×{ k + 2 } ) is a nonempty subset of f − ( U (2) )and therefore f − ( U (2) ) ∩ π ( X × ( k + 2)) = ∅ . By the induction on n ∈ ω it can be shown that π ( X ×{− k + n } ) ∩ f − ( U ( n ) ) = ∅ for each n ∈ ω . Since f ( b ) / ∈ ∪ n ∈ ω U ( n ) there exists an openneighborhood V of f ( b ) such that ∪ n ∈ ω U ( n ) ∩ V = ∅ . By the definition of the topology on J ( X ),there exists m ∈ ω such that π ( X ×{ m } ) ⊂ f − ( V ). However, f − ( V ) ∩ f − ( U ( k + m ) ) = ∅ whichimplies a contradiction. Hence f ( a ) = f ( b ) for each continuous real-valued function on J ( X ) or,equivalently, R ∈ Const a,b ( J ( X )). Step 2 . Consider the space D a,b ( J ( X )). By Lemma 2.1, the space D a,b ( J ( X )) is separable, re-gular and admits only constant continuous real-valued functions. Also, recall that D a,b ( J ( X )) con-tains a homeomorphic copy of J ( X ) and, as a consequence, of X . However, the space D a,b ( J ( X ))fails to be countably compact. Step 3 . Consider the space R ( X ) = cW ( D a,b ( J ( X ))). By Lemma 2.4 and Corollary 2.3, thespace D a,b ( J ( X )) satisfies conditions of Lemma 3.2. Hence the space R ( X ) is regular separablecountably compact and contains D a,b ( J ( X )) as a dense subspace which implies that R ( X ) is R -rigid.It remains to check that the space R ( X ) is totally countably compact whenever X is totallycountably compact. Recall that in this case J ( X ) is totally countably compact. Fix any countableinfinite subset D ⊂ R ( X ). There are two cases to consider:1. The set D ∩ D a,b ( J ( X )) is infinite. If there exists n ∈ ω such that the set M = π ( J ( X ) n ) ∩ D is infinite, then by the total countable compactness of J ( X ) (recall that π ( J ( X ) n ) is homeomorphicto J ( X )) there exists an infinite subset N ⊂ M such that N is compact in R ( X ). If for each n ∈ ω the set D ∩ π ( J ( X ) n ) is finite, then Lemma 2.2 implies that the set D is closed and discrete in D a,b ( J ( X )). Hence the set cl R ( X ) ( D ) = cl W ( D a,b ( J ( X ))) ( C ) is compact.2. The set D ∩ D a,b ( J ( X )) is finite. Let T = {F ( n ) | n ∈ ω } = D ∩ ( R ( X ) \ D a,b ( J ( X ))). Foreach n ∈ ω there exists a closed discrete subset C ( n ) ⊂ D a,b ( J ( X )) such that C ( n ) ∈ F ( n ) . ByCorollary 2.3, there exists a countable closed discrete subset C ⊂ D a,b ( J ( X )) such that C ( n ) ⊂ ∗ C .Note that C ∈ F ( n ) for each n ∈ ω . Then {F ( n ) | n ∈ ω } is contained in a compact subsetcl R ( X ) ( C ) = cl W ( D a,b ( J ( X ))) ( C ). Hence the space R ( X ) is totally countably compact. (cid:3) Consistent generalization
For a cardinal θ a subset T = { T α } α ∈ θ ⊂ [ ω ] ω is called a tower if for any α < β < θ , T α ⊂ ∗ T β and | T β \ T α | = ω . A tower T = { T α } α ∈ θ is called regular if the cardinal θ is regular. A tower T is called maximal if for any infinite subset I of ω there exists T α ∈ T such that the set I ∩ T α isinfinite.In this section, we provide a consistent generalization of Theorem 3.3. For this purpose we willessentially use an example of Franklin and Rajagopalan constructed in [11] (see also Example 7.1in [9]).For any ordinals α, β by [ α, β ] we denote the set of all ordinals γ such that α ≤ γ ≤ β . Further, ifsome ordinal α is considered as a topological space, then α is assumed to carry the order topology. N REGULAR SEPARABLE COUNTABLY COMPACT R -RIGID SPACES 7 For any distinct cardinals α, β , by T α,β we denote the punctured Tychonoff ( α, β )-plank, i.e., thesubspace ( α + 1) × ( β + 1) \ { ( α, β ) } of the Tychonoff product ( α + 1) × ( β + 1). Lemma 4.1 ([4, Lemma 3]) . Let κ be a cardinal and Y be a T space of pseudo-character ψ ( Y ) ≤ κ . Then for any regular cardinals λ, ξ such that κ + < λ < ξ and for each continuous map f : T λ,ξ → Y there exist y ∈ Y , α ∈ λ and µ ∈ ξ such that [ α, λ ] × [ µ, ξ ] \ { ( λ, ξ ) } ⊂ f − ( y ) . The following theorem is the main result of this section.
Theorem 4.2.
Let κ be a cardinal such that there exist two maximal towers T , T satisfying κ + < | T | < | T | . Then there exists a separable totally countably compact space Q such that forany T space Y of pseudo-character ψ ( Y ) ≤ κ each continuous map f : Q → Y is constant.Proof. Fix an infinite cardinal κ and two maximal towers T = { T α | α ∈ t } and T = { T α | α ∈ t } such that κ + < t < t . For every i ∈ { , } , consider the space Y i = T i ∪ ω which istopologized as follows. Points of ω are isolated and a basic open neighborhood of T ∈ T i has theform B ( S, T, F ) = { P ∈ T i | S ⊂ ∗ P ⊆ ∗ T } ∪ (( T \ S ) \ F ) , where S ∈ T i ∪ {∅} satisfies S ⊂ ∗ T and F is a finite subset of ω .Repeating arguments used in [9, Example 7.1] one can check that the space Y i is sequentiallycompact, separable and locally compact for every i ∈ { , } . For every i ∈ { , } choose any point ∞ i / ∈ Y i and let X i = {∞ i } ∪ Y i be the one-point compactification of the locally compact space Y i . Consider the space Π = ( X × X ) \ { ( ∞ , ∞ ) } . It is easy to see that the space Π is regularseparable and sequentially compact. Observe that Π is not normal, because the closed subsets A = {∞ }× T and B = T ×{∞ } cannot be separated by disjoint open sets. Also, note thatthe sets A and B are homeomorphic to the cardinals t and t , respectively. Thus the subspace T × T ∪ A ∪ B ⊂ Π is homeomorphic to the Tychonoff plank T t , t .Next, using Jones machine, we shall construct the space J (Π) similarly as the space J ( X ) wasconstructed in the proof of Theorem 3.3. More precisely, let Z be a discrete set of integers and a, b distinct points which do not belong to Π × Z . By R we denote the set Π × Z ∪ { a, b } endowedwith the topology τ which satisfies the following conditions: • the Tychonoff product Π × Z is open in R ; • if a ∈ U ∈ τ , then there exists n ∈ N such that { ( x, − k ) | x ∈ Π and k > n } ⊂ U ; • if b ∈ U ∈ τ , then there exists n ∈ N such that { ( x, k ) | x ∈ Π and k > n } ⊂ U .On the space R consider the smallest equivalence relation ∼ such that( x, n ) ∼ ( x, n + 1) and ( y, n ) ∼ ( y, n − n ∈ Z , x ∈ A and y ∈ B .Let J (Π) be the quotient space R/ ∼ . It is easy to check that J (Π) is separable regular andsequentially compact space. Fix any T space Y with pseudo-character ≤ κ and a continuous map f : Π → Y . We claim that f ( a ) = f ( b ). Consider the map f ◦ π : R → Y . Recall that for each n ∈ Z the subspace ( T × T ∪ A ∪ B ) ×{ n } is homeomorphic to T t , t . Thus, Lemma 4.1 impliesthat for each n ∈ Z there exist y n ∈ Y and ordinals α n ∈ t , β n ∈ t such that { ( x, ∞ , n ) | x ∈ T \ { T ξ | ξ ≤ α n }} ∪ { ( ∞ , y, n ) | y ∈ T \ { T ξ | ξ ≤ β n }} ⊂ π − ( f − ( y n )) . Put α = sup { α n | n ∈ Z } . Since π (( ∞ , x, n )) = π (( ∞ , x, n + 1)) and π (( y, ∞ , n )) = π (( y, ∞ , n − n ∈ Z , x ∈ T and y ∈ T we obtain that y n − = y n = y n +1 , n ∈ Z .Hence there exists a unique y ∈ Y such that D = π ( { ( x, ∞ , n ) | n ∈ Z and x ∈ T \ { T ξ | ξ ≤ α }} ) ⊂ f − ( y ) . Observe that { a, b } ⊂ D ⊂ f − ( y ) = f − ( y ). Hence any T space Y of pseudo-character ≤ κ belongs to Const a,b ( J (Π)).Finally, consider the space D a,b ( J (Π)) and put Q = cW ( D a,b ( J (Π))). Similarly as in the proofof Theorem 3.3 it can be showed that the space Q is regular, separable, totally countably compactand admits only constant continuous mappings into any T space of pseudocharacter ≤ κ . (cid:3) S. BARDYLA AND L. ZDOMSKYY
Next we show that the assumption of Theorem 4.2 is consistent with ZFC and may hold for many κ < c simultaneously. This fact may be thought of as a folklore, e.g., in private communicationwe have been informed by V. Fischer and J. Schilchan that they have come across the same ideas.Nonetheless we give a somewhat detailed proof thereof using the Mathias forcing M F associated toa free filter F on ω . M F consists of pairs h s, F i such that s ∈ [ ω ] <ω , F ∈ F , and max s < min F .A condition h s, F i is stronger than h s ′ , F ′ i if F ⊂ F ′ , s is an end-extension of s ′ , and s \ s ′ ⊂ F ′ .This poset introduces the generic subset X ∈ [ ω ] ω such that X ⊂ ∗ F for all F ∈ F , namely X = S { s : ∃ F ∈ F ( h s, F i ∈ G ) } , where G is M F generic. Proposition 4.3.
Let V be a model of GCH and κ a cardinal of uncountable cofinality. Thenthere is a ccc poset P such that in V P , ω = κ and for every regular cardinal λ ≤ κ there is a tower { T λξ : ξ < λ } .Proof. P will be a two step iteration P ∗ ˙ P , where P = F in ( κ × ω,
2) is the standard Cohenposet adding κ -many Cohen reals. Let Λ be the set of all regular cardinals λ ≤ κ . ˙ P is a( P -name for) the finitely supported product Q λ ∈ Λ Q λ , where Q λ = Q λλ in the finite supportiteration h Q λα , ˙ R λβ : β < λ, α ≤ λ i defined as follows. Suppose that we have already constructed h Q λα , ˙ R λβ : β < ξ, α ≤ ξ i for some ξ < λ , giving rise to a sequence h ˙ A λβ : β < ξ i of Q λξ -names forinfinite subsets of ω such that 1 Q λξ forces h ˙ A λβ : β < ξ i to be ⊂ ∗ -decreasing. Then ˙ R λξ is the Q λξ -name for the Mathias forcing M ˙ F λξ , where ˙ F λξ is the name for the filter generated by h ˙ A λβ : β < ξ i ,and ˙ A λξ is the M ˙ F λξ -generic real. This completes our definition of P . In V P , every Q λ is σ -centeredbeing an iteration of σ -centered posets of length at most continuum, see [13]. Thus P is ccc in V P being a finite support product of σ -centered posets, and hence P is ccc as well. A standardcounting of nice names for reals shows that 2 ω = κ in V P .Let G be P -generic filter and A λβ := ( ˙ A λβ ) G for all λ ∈ Λ and β < λ . Thus in V [ G ], h A λβ : β < λ i has the property that A λβ ⊂ ∗ A λξ for all ξ < β < λ . Let h λβ be the enumerating function of A λβ , i.e.,the unique increasing bijection from ω onto A λβ . Since for every λ ∈ Λ the poset P can be writtenin the form P ∗ ( Q ν ∈ Λ \{ λ } ˙ Q ν × ˙ Q λ ), we conclude that the family { h λβ : β < λ } is unboundedin V [ G ], and hence the family { A λβ : β < λ } has no infinite pseudointersection. It follows that { ω \ A λβ : β < λ } is a tower of length λ in V [ G ]. (cid:3) Theorem 4.2 motivates the following
Problem 5.
Is it consistent that there exists a cardinal κ such that κ < c ≤ κ ++ and a separable(totally) countably compact regular space Q such that for any T space Y with ψ ( Y ) ≤ κ , anycontinuous map f : Q → Y is constant? Remark 4.4.
Theorem 4.2 cannot be generalized to the case κ = c . Indeed, let Y be a regularseparable space. Then ψ ( X ) ≤ ω = c by inequality 2.7, page 15 of [17], and thus the identitymap id : Y → Y is continuous and non-constant. Remark 4.5.
The answer to Problem 2 in the case of separable spaces is negative. For thisobserve that the cardinality of any regular separable space is bounded by the cardinal 2 c . Thusthere exists a set S consisting of separable regular spaces such that any separable regular space ishomeomorphic to some element of S . Consider the Tychonoff product Π S . Obviously the spaceΠ S is regular and for each regular separable space X there exists a non-constant (even injective)continuous map f : X → Π S .5. Around the problem of Nyikos
First we answer in the affirmative Problem 4, the latter being variation of Nyikos’ problemobtained by weakening the disjunction of first-countability and countable compactness to thesequential compactness.
Theorem 5.1.
There exists a regular separable sequentially compact non-functionally Hausdorffspace S within ZFC. N REGULAR SEPARABLE COUNTABLY COMPACT R -RIGID SPACES 9 Proof.
Fix any maximal tower T = { T α | α ∈ κ } on ω and let Y be the space of Franklin andRajagopalan which corresponds to the tower T (see spaces Y , Y in the proof of Theorem 4.2or Example 7.1 from [9]). Recall that Y is separable normal locally compact and sequentiallycompact. By Y ∗ we denote the one point compactification of Y . Observe that the Tychonoffproduct of Z = Y × Y ∗ is separable regular and sequentially compact. Claim.
The space Z is not normal.Proof. Recall that the space Y contains a closed homeomorphic copy of the cardinal κ and thespace Y ∗ contains a closed homeomorphic copy of the ordinal κ + 1. Therefore the space Z contains a closed homeomorphic copy of the Tychonoff product K = κ × ( κ + 1). It remains toshow that that the space K is not normal. To derive a contradiction, assume that K is normal.Consider the closed disjoint subsets A = { ( α, α ) | α ∈ κ } and B = { ( α, κ ) | α ∈ κ } of K . Since K is pseudocompact, Glicksberg’s Theorem (see [10, Exercise 3.12.20(c)]) implies that β ( K ) = β ( κ ) × β ( κ + 1) = ( κ + 1) × ( κ + 1). Since the space K is normal, Corollary 3.6.4 from [10] providesthat cl β ( K ) ( A ) ∩ cl β ( K ) ( B ) = ∅ . However, it is easy to see that ( κ, κ ) ∈ cl β ( K ) ( A ) ∩ cl β ( K ) ( B )which implies a contradiction. (cid:3) Finally, using Jones machine (similarly as in the proof of Theorem 3.3 and Theorem 4.2) weconstruct the space S = J ( Z ) which is regular separable sequentially compact but not functionallyHausdorff. (cid:3) The rest of this section is devoted to the following theorem whose proof will be divided intoseveral steps and rely onto the example of a non-normal regular separable first-countable countablycompact space constructed by Nyikos and Vaughan in [21] under t = ω . We shall need the set-theoretic assumption “ ω = t < b = c and there exists a simple P c -point” which we denote by ♥ . Theorem 5.2. ( ♥ ) There exists a regular separable first-countable R -rigid countably compactspace. Theorem 5.2 implies that the affirmative answer to the strongest version of Problem 1 is con-sistent, as well as gives consistent counterexamples to the Nyikos’ problem which are R -rigid.Recall that an ultrafilter F on ω is called a simple P κ -point , where κ is a regular cardinal, ifthere exists a basis { F α : α < κ } of F which is a tower of length κ . The following fact showsthat ♥ is consistent with ZFC and seems to be a kind of folklore. We are grateful to A. Dow forsharing with us the proof sketched below. Proposition 5.3. ♥ is consistent.Proof. Let V be a model of GCH and P be a poset adding ω Cohen subsets of ω with countablesupports. Thus 2 ω = ω and 2 ω = ω holds in V P . In V P , let P be a finite support iteration oflength ω of ccc posets of size ω producing a simple P ω -point and forcing in addition b = c = ω .For this it is enough that we take cofinally often Hechler forcing as well as the Mathias forcing forultrafilters which are made always larger by including into the next ones the Mathias generics forthe previous ones.Since 2 ω = ω > c = ω in V P ∗ ˙ P , we conclude that p = ω in this model because 2 κ = c for any infinite κ < p . The latter statement is also well-known and follows from the existenceof a countable dense subset Q of 2 κ as well as the fact that each point x of 2 κ is a limit of asequence in Q convergent to x because 2 κ has character κ < p (see [9, p.130]), which gives that2 κ ≤ | Q | ω = c . (cid:3) Observe that the extension D a,b ( X ) does not preserve first-countability. Therefore we shall usea suitable modification of this extension defined below. Let X , a , b , B ( a ), B ( b ), Z , D , D n , A , b ∗ , f : A → ω , ∼ and π be such as in Section 2. In addition, we shall assume that X is first-countable.Next we introduce two kinds of sets in Z which will be useful in the definition of the topology on Z/ ∼ which is weaker than that of D a,b ( X ). • For any n ∈ ω and subsets U ⊂ X n \ { a n , b n } , V ⊂ X \ { b } such that int( V ) ∈ B ( a ) we shallinductively construct the subset [ U ] V ⊂ Z as follows: let U (0) = ∅ , U (1) = U and assume that for some n ∈ ω the sets U ( k ) , k ≤ n are already constructed. Put U ∗ ( n ) = ( U ( n ) \ U ( n − ) ∩ A and U ( n +1) = ∪{ V k | k ∈ f ( U ∗ ( n ) ) } ∪ U ( n ) . Finally, let [ U ] V = ∪ n ∈ ω U ( n ) . • For any subsets U ⊂ X \ { a } and V ⊂ X \ { b } such that int( U ) ∈ B ( b ) and int( V ) ∈ B ( a )we shall inductively construct the subset [ U b ] V ⊂ Z as follows: let U (0) = ∅ , U (1) = ∪ n ∈ ω U n and assume that for some n ∈ ω the sets U ( k ) , k ≤ n are already constructed. Put U ∗ ( n ) =( U ( n ) \ U ( n − ) ∩ A and U ( n +1) = ∪{ V k | k ∈ f ( U ∗ ( n ) ) } ∪ U ( n ) . Finally, let [ U b ] V = ∪ n ∈ ω U ( n ) . • Let E a,b ( X ) be the quotient set Z/ ∼ endowed with the topology τ generated by the base B ( τ ) = { π ([ U n ] V ) | n ∈ ω, U is an open subset of X \ { a, b } and V ∈ B ( a ) }∪∪ { π ([ U b ] V | U ∈ B ( b ) and V ∈ B ( a ) } . Observe that the topology τ is weaker than the quotient topology on Z/ ∼ , that is the identitymap id : D a,b ( X ) → E a,b ( X ) is continuous. This immediately yields that E a,b ( X ) is separableand any continuous map g : E a,b ( X ) → Y ∈ Const( X ) a,b is constant. Lemma 5.4.
The space E a,b ( X ) is first-countable.Proof. Let { A ( n ) | n ∈ ω } ⊂ B ( a ) and { B ( n ) | n ∈ ω } ⊂ B ( b ) be open neighborhood bases at thepoints a ∈ X and b ∈ X , respectively. It is straightforward to check that the family { π ([ B ( n ) b ] A ( n ) ) | n ∈ ω } forms an open neighborhood base at the point b ∗ ∈ E a,b ( X ). Fix any q ∈ E a,b ( X ) \ { b ∗ } and notice that there exist x ∈ X \ { a, b } and n ∈ ω such that π − ( q ) \ { a n | n ∈ ω } = { x n } . Let { U ( k ) | k ∈ ω } be an open neighborhood base at the point x ∈ X such that U ( k ) ⊂ X \ { a, b } forany k ∈ ω . It is easy to check that the family { π ([ U ( k ) n ] A ( k ) ) | k ∈ ω } forms an open neighborhoodbase at the point q ∈ E a,b ( X ). Hence the space E a,b ( X ) is first-countable. (cid:3) It remains to verify that E a,b ( X ) is regular. For this we need the following auxiliary lemma: Lemma 5.5.
For any n ∈ ω and subsets W ∈ B ( b ) , V ∈ B ( a ) , and U ⊂ X \ { a, b } such that a cl X ( U ) the following inclusions hold: ( i ) cl E a,b ( X ) ( π ([ U n ] V )) ⊂ π ([cl X ( U ) n ] cl X ( V ) ) ; ( ii ) cl E a,b ( X ) ( π ([ W b ] V )) ⊂ π ([cl X ( W ) b ] cl X ( V ) ) . Proof.
Fix any n ∈ ω and subsets W ∈ B ( b ), V ∈ B ( a ), U ⊂ X \ { a, b } such that a cl X ( U ).Let q ∈ E a,b ( X ) be such that π − ( q ) ⊂ Z \ [cl X ( U )] cl X ( V ) . Observe that the set [cl X ( U ) n ] cl X ( V ) is closed in Z . Two cases are possible:1. q = b ∗ . Let { x m } = π − ( q ) \ { a k | k ∈ ω } . Fix an open subset T m ⊂ X m \ ([cl X ( U ) n ] cl X ( V ) ∪{ a m , b m } ) which contains x m , and an O ∈ B ( a ) such that O ∩ U = ∅ . It can be checked that π ([ T m ] O ) is an open neighborhood of q in E a,b ( X ) which is disjoint from the set π ([ U n ] V ), whichyields q / ∈ cl E a,b ( X ) ( π ([ U ] V )).2. q = b ∗ . Fix any T ∈ B ( b ) such that T ⊂ X \ cl X ( V ) and T ∩ U = ∅ . Let O ∈ B ( a ) be suchthat O ∩ U = ∅ . A routine verifications show that [ T b ] O ∩ [ U n ] V = ∅ . Therefore, the set π ([ T b ] O )is an open neighborhood of b ∗ in E a,b ( X ) which is disjoint with the set π ([ U n ] V ), thus witnessing b ∗ / ∈ cl E a,b ( X ) ( π ([ U n ] V )). This proves the inclusion ( i ).The second inclusion can be proved analogously. (cid:3) Lemma 5.6.
The space E a,b ( X ) is regular.Proof. Fix any q ∈ E a,b ( X ) \ { b ∗ } , and let π − ( q ) \ { a k | k ∈ ω } = { x n } . Observe that { q } = ∩{ π ([ U n ] V ) | V ∈ B ( a ) , x ∈ U ⊂ X \ { a, b } and U is open in X } and { b ∗ } = ∩{ π ([ U b ] V ) | V ∈ B ( a ) , U ∈ B ( b ) } . Hence E a,b ( X ) is a T space.Let π ([ U n ] V ) be any basic open neighborhood of the point q . Recall that U is an open neighbor-hood of x in X . By the regularity of X , there exists an open neighborhood W ⊂ X of x such thatcl X ( W ) ⊂ U . Using one more time the regularity of X find any F ∈ B ( a ) such that cl X ( F ) ⊂ V .Lemma 2.4 implies that q ∈ π ([ W n ] F ) ⊂ cl E a,b ( X ) ( π ([ W n ] F )) ⊂ π ([cl X ( W ) n ] cl X ( F ) ) ⊂ π ([ U ] V ) . N REGULAR SEPARABLE COUNTABLY COMPACT R -RIGID SPACES 11 Fix any open neighborhood π ([ U b ] V ) of b ∗ . By the regularity of X , there exist W ∈ B ( b ) and F ∈ B ( a ) such that cl X ( W ) ⊂ U and cl X ( F ) ⊂ V . Lemma 2.4 provides thatcl E a,b ( X ) ( π ([ W b ] F )) ⊂ π ([cl X ( W ) b ] cl X ( F ) ) ⊂ π ([ U b ] V ) . Hence the space E a,b ( X ) is regular. (cid:3) Let ♥ be the assertion “ b = c and there exists a simple P c -point”. The following propositionwill be crucial in our construction of regular separable first-countable R -rigid countably compactspace. Proposition 5.7. ( ♥ ) Let X be a regular first-countable space of cardinality κ < c . Then thereexists a regular first-countable countably compact space O ( X ) which contains X as a dense sub-space.Proof. Let X be a regular first-countable space such that | X | < c . If X is countably compact, thenput O ( X ) = X . Otherwise, let D = { A ∈ [ X ] ω | A is closed and discrete } and fix any bijection h : D ∪ [ c ] ω → c such that h ( a ) ≥ sup( a ) for any a ∈ [ c ] ω . It is easy to see that such a bijectionexists. Next, using ideas of Ostaszewski [22], for every α ≤ c we shall inductively construct atopology τ α on X ∪ α by defining an open neighborhood base B α ( x ) at each point x ∈ X ∪ α .For convenience, by Y α we denote the space ( X ∪ α, τ α ). At the end, we will show that the space O ( X ) = Y c has the desired properties.Let Y = X and for each x ∈ X fix an open neighborhood base B ( x ) = { U n ( x ) | n ∈ ω } at thepoint x such that U ( x ) = X and U n +1 ( x ) ⊂ U n ( x ). Assume that for each α < ξ the spaces Y α arealready constructed by defining for each x ∈ Y α the family B α ( x ) = { U αn ( x ) | n ∈ ω } which formsan open neighborhood base at x in Y α . Moreover, additionally suppose that the spaces Y α are T , U α ( x ) = Y α , cl Y α ( U αn +1 ( x )) ⊂ U αn ( x ) and U αn ( x ) = U βn ( x ) ∩ Y α for any x ∈ Y α and α < β < ξ .Then we have three cases to consider:1) ξ = η + 1 for some η ∈ c and h − ( η ) already has an accumulation point in Y η ;2) ξ = η + 1 for some η ∈ c and h − ( η ) does not have an accumulation point in Y η ;3) ξ is a limit ordinal.1) Let B ξ ( x ) = B η ( x ) for each x ∈ Y η . The open neighborhood base B ξ ( η ) at the point η ∈ Y ξ is defined as follows: U ξ ( η ) = Y ξ and U ξn ( η ) = { η } for each n > h − ( η ) = { z n } n ∈ ω . Fix any simple P c -point p which exists by the assumption. For any y ∈ Y η fix m ( y ) ∈ ω such that F y = { n ∈ ω | z n ∈ U ηm ( y ) ( y ) \ U ηm ( y )+1 ( y ) } ∈ p . Since the set h − ( η ) is closed the integer m ( y ) exists for any y ∈ Y η . Since | Y η | < c there exists F η ∈ p suchthat F η ⊂ ∗ F y for any y ∈ Y η . Let d η = { z n | n ∈ F η } . Note that d η ⊂ ∗ U ηm ( y ) ( y ) \ U ηm ( y )+1 ( y ) ⊂ U ηm ( y ) ( y ) \ cl Y η ( U ηm ( y )+2 ( y )) for any y ∈ Y η . Since the set U ηm ( y ) ( y ) \ cl Y η ( U ηm ( y )+2 ( y )) is open thereexists a function f y ∈ ω ω such that cl Y η ( U ηf y ( n ) ( z n )) ⊂ U ηm ( y ) ( y ) \ cl Y η ( U ηm ( y )+2 ( y )) for all butfinitely many n ∈ F η . Let f η ∈ ω ω be a function such that f η ≥ ∗ f y for each y ∈ Y η . Such an f η exists because | Y η | < b . Next we define the open neighborhood base at the point η ∈ Y ξ : Put U ξ ( η ) = Y ξ and U ξk ( η ) = ∪{ U ηf η ( n )+ k ( z n ) | n ∈ F η \ k } ∪ { η } for all k ≥
1. For each y ∈ Y η and n ∈ ω let U ξn ( y ) = U ηn ( y ) if n ≥ m ( y ) + 1 and U ξn ( y ) = U ηn ( y ) ∪ { η } if n ≤ m ( y ). Obviously, thespace Y ξ is T . We claim that the family { U ηf η ( n ) ( z n ) | n ∈ F η } is locally finite in Y η . Recall thatfor each y ∈ Y η , cl Y η ( U ηf y ( n ) ( z n )) ⊂ U ηm ( y ) ( y ) \ cl Y η ( U ηm ( y )+2 ( y )) for all but finitely many n ∈ F η .Since f η ≥ ∗ f y for any y ∈ Y η we obtain that cl Y η ( U ηf η ( n ) ( z n )) ⊂ U ηm ( y ) ( y ) \ cl Y η ( U ηm ( y )+2 ( y )) forall but finitely many n ∈ F η . Hence for any y ∈ Y η the set U ηm ( y )+2 ( y ) intersects only finitelymany elements of the family { U ηf η ( n ) ( z n ) | n ∈ F η } witnessing that this family is locally finite. Fixany k ∈ ω . Since U ηf η ( n )+ k ( z n ) ⊂ U ηf η ( n ) ( z n ) we obtain that the family { U ηf η ( n )+ k ( z n ) | n ∈ F η } is locally finite as well. This proves that Y ξ is Hausdorff: For every y ∈ Y η the intersection U ξm ( y )+2 ( y ) ∩ U ξk ( η ) = U ηm ( y )+2 ( y ) ∩ U ξk ( η ) is empty provided that k is larger than all of the(finitely many) elements n ∈ F η such that U ηf y ( n ) ( z n ) ∩ U ξm ( y )+2 ( y ) = ∅ . Next, we shall prove that cl Y ξ ( U ξn +1 ( y )) ⊂ U ξn ( y ) for each y ∈ Y ξ . First consider the case y ∈ Y η .If n ≥ m ( y ) + 1, thencl Y ξ ( U ξn +1 ( y )) = cl Y ξ ( U ηn +1 ( y )) = cl Y η ( U ηn +1 ( y )) ⊂ U ηn ( y ) ⊂ U ξn ( y )because in this case n + 1 ≥ m ( y ) + 2 and thus η , the only element of Y ξ \ Y η , has a neighbourhooddisjoint from U ηn +1 ( y ), as we have established above. If n ≤ m ( y ), then we havecl Y ξ ( U ξn +1 ( y )) ⊂ cl Y ξ ( U ηn +1 ( y ) ∪ { η } ) = cl Y ξ ( U ηn +1 ( y )) ∪ { η } = cl Y η ( U ηn +1 ( y )) ∪ { η } ⊂⊂ U ηn ( y ) ∪ { η } = U ξn ( y ) , the second equality is again a consequence of the fact that { η } = Y ξ \ Y η . Now let us consider thecase y = η . For each k ∈ ω we havecl Y ξ ( U ξk +1 ( η )) = cl Y ξ (cid:0) [ { U ηf η ( n )+ k +1 ( z n ) | n ∈ F η \ ( k + 1) } ∪ { η } (cid:1) == cl Y ξ (cid:0) [ { U ηf η ( n )+ k +1 ( z n ) | n ∈ F η \ ( k + 1) } (cid:1) ∪ { η } == cl Y η (cid:0) [ { U ηf η ( n )+ k +1 ( z n ) | n ∈ F η \ ( k + 1) } (cid:1) ∪ { η } == [ { cl Y η (cid:0) U ηf η ( n )+ k +1 ( z n ) (cid:1) | n ∈ F η \ ( k + 1) } ∪ { η } ⊂⊂ [ { U ηf η ( n )+ k ( z n ) | n ∈ F η \ ( k + 1) } ∪ { η } ⊂ U ξk ( η ) . The latest equality is a consequence of the local finiteness of the family { U ηf η ( n )+ k +1 ( z n ) | n ∈ F η } in Y η .3) For each y ∈ Y ξ put U ξn ( y ) = ∪ α ( y ) <γ<ξ U γn ( y ), where α ( y ) = min { α < ξ : y ∈ Y α } . The nextclaim implies that the space Y ξ is Hausdorff. Claim.
For any γ < ξ , n, m ∈ ω and distinct points y , y ∈ Y γ , if U γn ( y ) ∩ U γm ( y ) = ∅ , then U ξn ( y ) ∩ U ξm ( y ) = ∅ .Proof. To derive a contradiction, assume that U ξn ( y ) ∩ U ξm ( y ) = ∅ . It is easy to see that U ξn ( y ) ∩ U ξm ( y ) ⊂ [ γ, ξ ). Let δ = min U ξn ( y ξ ) ∩ U ξm ( y ). It follows that U δn ( y ) ∩ U δm ( y ) = ∅ and U δ +1 n ( y ) ∩ U δ +1 m ( y ) = { δ } . Then the set d δ is closed and discrete in Y δ . By the definition of U δ +1 n ( y ) (see case2 above), δ ∈ U δ +1 n ( y ) ∩ U δ +1 m ( y ) iff d δ ⊂ ∗ U δn ( y ) ∩ U δm ( y ) witnessing that U δn ( y ) ∩ U δm ( y ) = ∅ .A contradiction. (cid:3) It remains to check that cl Y ξ ( U ξn +1 ( y )) ⊂ U ξn ( y ) for each y ∈ Y ξ . Fix any y ∈ Y ξ and n ∈ ω .By the assumption, cl Y γ U γn +1 ( y ) ⊂ U γn ( y ) for each γ < ξ . To derive a contradiction, assume thatthere exists z ∈ Y ξ such that z ∈ cl Y ξ U ξn +1 ( y ) \ U ξn ( y ). Fix any δ < ξ such that z ∈ Y δ . Weclaim that z ∈ cl Y δ U δn +1 ( y ). Indeed, if z / ∈ cl Y δ ( U δn +1 ( y )), then there exists m ∈ ω such that U δm ( z ) ∩ U δn +1 ( y ) = ∅ . The Claim proved above implies that U ξm ( z ) ∩ U ξn +1 ( y ) = ∅ witnessing that z / ∈ cl Y ξ ( U ξn +1 ( y )). The obtained contradiction implies that z ∈ cl Y δ ( U δn +1 ( y )) ⊂ U δn ( y ) ⊂ U ξn ( y ),which contradicts the choice of z . Hence cl Y ξ ( U ξn +1 ( y )) ⊂ U ξn ( y ) for each y ∈ Y ξ .Let O ( X ) = Y c . By the construction, the space O ( X ) is regular and first-countable. Let A beany countable subset of O ( X ). If the set B = A ∩ c is infinite, then consider h ( B ) ∈ c . By theconstruction, either B has an accumulation point in Y h ( B ) or h ( B ) is an accumulation point of B in Y h ( B )+1 . In both cases B has an accumulation point in O ( X ). If A ⊂ ∗ X , then either ithas an accumulation point in X , or A is closed and discrete in X . In the latter case either A hasan accumulation point in Y h ( A ) or h ( A ) is an accumulation point of A in Y h ( A )+1 . Thus O ( X ) iscountably compact. (cid:3) We are in a position now to present the
Proof of Theorem 5.2.
Let X be regular non-normal separable first-countable countably compactspace constructed by Nyikos and Vaughan under t = ω in [21]. Note that | X | = ω . ApplyingJones machine to X we obtain regular separable first-countable countably compact space J ( X ) ofcardinality ω containing two points a, b which cannot be separated by any real-valued continuous N REGULAR SEPARABLE COUNTABLY COMPACT R -RIGID SPACES 13 map. Then consider the space E a,b ( J ( X )) which is regular separable first-countable R -rigid and | E a,b ( J ( X )) | = ω . Proposition 5.7 implies that the space O ( E a,b ( J ( X ))) is regular first-countableand countably compact. Moreover, the space O ( E a,b ( J ( X ))) is separable and R -rigid, because itcontains a dense copy of E a,b ( J ( X )). (cid:3) Remark 5.8.
Example 2 from [3] shows that there exists a regular first-countable space of car-dinality d which cannot be embedded into Urysohn countably compact spaces. Also, it can beproved that for any mad family A on ω the Mr´owka-Isbell space Ψ( A ) is Tychonoff first-countableand cannot be embedded into any Hausdorff countably compact space of character < b . HenceProposition 5.7 cannot be proved within ZFC. In particular its conclusion cannot hold in anymodel of min { d , a } < c .Theorems 5.1 and 5.2 motivate the following Problem 6.
Does there exist a ZFC example of separable sequentially compact regular R -rigidspace? By d ( X ), w ( X ), χ ( X ), ψ ( X ), s ( X ), e ( X ), and c ( X ) is denoted the density, weight, character,pseudo-character, spread, extent, and cellularity of the space X , respectively. We refer the readerto [10] for their definitions. We end up with a remark which shows how the extensions D a,b ( X )and E a,b ( X ) can be generalized for a non-separable space X and reveals some of their properties.Its proof is fairly standard and is therefore left to the reader. Remark 5.9.
Let a, b ∈ X , B ( a ), and B ( b ) be such as in Section 2. Let Z be the Tychonoff product X × d ( X ) where d ( X ) is endowed with the discrete topology. For any x ∈ X and α ∈ d ( X ) by x α we denote the point ( x, α ). Analogously, for any B ⊂ X and α ∈ d ( X ) the set B ×{ α } isdenoted by B α . Let D be a dense subset of X such that D ⊂ X \ { a, b } and | D | = d ( X ). Put A = ∪ α ∈ d ( X ) D α . Fix any bijection f : A → d ( X ) such that f ( x α ) = α for each α ∈ d ( X ) and x α ∈ D α . On the set Z consider the smallest equivalence relation ∼ which satisfies the followingconditions: • x α ∼ a f ( x α ) for any α ∈ d ( X ) and x ∈ D ; • b α ∼ b ξ for any α, ξ ∈ d ( X ).Let D a,b ( X ) be the quotient space Z/ ∼ . Analogously, one can generalize the extension E a,b ( X )for non-separable spaces.Let Z ∈ { D a,b ( X ) , E a,b ( X ) } . Then every continuous map g : Z → Y ∈ Const( X ) a,b is constant.The space D a,b ( X ) is regular and d ( D a,b ( X )) = d ( X ). If the space X is countably compact, then D a,b ( X ) has property D . Also, analogues of Lemma 2.2 and Corollary 2.3 hold for the generalizedextension D a,b ( X ).The space E a,b ( X ) is also regular and d ( E a,b ( X )) = d ( X ), w ( E a,b ( X )) = w ( X ), χ ( E a,b ( X )) = χ ( X ), ψ ( E a,b ( X )) = ψ ( X ), s ( E a,b ( X )) = max { s ( X ) , d ( X ) } , e ( E a,b ( X )) = max { e ( X ) , d ( X ) } , and c ( E a,b ( X )) = max { c ( X ) , d ( X ) } . Acknowledgements
The authors acknowledge Piotr Borodulin-Nadzieja for his comments on Question 4. In parti-cular, he convinced us that this question cannot be solved using tight gaps.
References [1] S. Armentrout,
A Moore space on which every real-valued continuous function is constant , Proc. Amer. Math.Soc. (1961), 106–109.[2] T. Banakh, S. Bardyla, A. Ravsky, Embedding topological spaces into Hausdorff κ -bounded spaces , Topologyand its Applications, (2020), 107277.[3] T. Banakh, S. Bardyla, A. Ravsky, Embeddings into countably compact Hausdorff spaces , preprint,arXiv:1906.04541.[4] S. Bardyla, A. Osipov,
On regular κ -bounded spaces admitting only constant continuous mappings into T spaces of pseudo-character ≤ κ , Acta Math. Hung. (accepted), arXiv:1912.10423v2[5] A. Blaszczyk, A. Szyma´nski, Some non-normal subspaces of the ˇCechStone compactification of a discrete space ,in: Z. Frolk (Ed.), Abstracta. 8th Winter School on Abstract Analysis, Czechoslovak Academy of Sciences, Praha,1980, pp. 35-38. [6] H. Brandemburg and A. Mysior,
For every Hausdorff space Y there exists a non-trivial Moore space on whichall continuous functions into Y are constant , Pacific J. Math., :1 (1984), 1–8.[7] K.C. Ciesielski and J. Wojciechowski, Cardinality of regular spaces admitting only constant continuous func-tions , Topology Proceedings, (2016), 313-329.[8] E. van Douwen, A regular space on which every continuous real-valued function is constant , Nieuw Archief voorWiskunde, (1972), 143–145.[9] E. van Douwen, The integers and topology , in: K. Kunen, J.E. Vaughan (Eds.),
Handbook of set-theoretictopology , North-Holland, Amsterdam, (1984) 111–167.[10] R. Engelking,
General topology , Heldermann Verlag, Berlin, 1989.[11] S. Franklin, M. Rajagopalan,
Some examples in topology , Trans. Amer. Math. Soc. :2 (1971), 305-314.[12] T.E. Gantner,
A regular space on which every continuous real-valued function is constant , Amer. Math.Monthly (1971), 52-53.[13] M. Goldstern, Finite support iterations of σ -centered forcing notions, a discussion on MathOverflow availableat https://mathoverflow.net/questions/84124/ .[14] E. Hewitt, On two problems of Urysohn , Annals of Mathematics, :3 (1946), 503–509.[15] S. Iliadis, and V. Tzannes, Spaces on which every continuous map into a given space is constant , Can. J.Math., :6 (1986), 1281–1298.[16] F. B. Jones, Hereditary separable, non-completely regular spaces , Topology Conf., Virginia Polytech. Inst. andState Univ., Lecture Notes in Math., no. 375, Springer-Verlag, Berlin and New York, 1974, pp. 149–152.[17] I. Juhsz,
Cardinal functions in topology , Mathematical Centre Tracts, No. 34. Mathematisch Centrum, Ams-terdam, 1971. xiii + 149 pp.[18] Kenneth Kunen,
Set Theory, Studies in Logic: Mathematical Logic and Foundations , Vol. 34, College Publi-cations, London, 2011, viii + 401 pp.[19] P. Nyikos,
On first countable, countably compact spaces III. The problem of obtaining separable noncompactexamples , Open problems in topology, North-Holland, Amsterdam, 1990, pp. 127-161.[20] P. Nyikos,
First countable, countably compact,noncompact spaces , Open Problems in Topology II, Elsevier,2007, pp. 217-224.[21] P. Nyikos, J. Vaughan
On first countable, countably compact spaces I: ( ω , ω ∗ ) -gaps , Trans. Amer. Math. Soc., :2 (1983), 463–469.[22] A. Ostaszewski, On countably compact, perfectly normal spaces , J. Lond. Math. Soc. s2-14 no. 3, (1976)505-516.[23] E. Pearl,
Problems from Topology Proceedings , Topology Atlas, Toronto, 2003[24] V. Tzannes,
A Moore strongly rigid space , Can. Math. Bull., :4 (1991), 547–552.[25] V. Tzannes, Two Moore spaces on which every continuous real-valued function is constant , Tsukuba J. Math., :1 (1992), 203–210.[26] V. Tzannes, A Hausdorff countably compact space on which every continuous real-valued function is constant ,Topology Proceedings, (1996), 239-244.[27] J.N. Younglove, A locally connected, complete Moore space on which every real-valued continuous function isconstant , Proc. Amer. Math. Soc., (1969), 527–530. Universit¨at Wien, Institut f¨ur Mathematik, Kurt G¨odel Research Center, Augasse 2–6, UZA 1 -Building 2, 1090 Wien, Austria.
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