On complemented copies of the space c 0 in spaces C p (X×Y)
Jerzy Kąkol, Witold Marciszewski, Damian Sobota, Lyubomyr Zdomskyy
aa r X i v : . [ m a t h . GN ] J u l ON COMPLEMENTED COPIES OF THE SPACE c IN SPACES C p ( X × Y ) J. KĄKOL, W. MARCISZEWSKI, D. SOBOTA, AND L. ZDOMSKYY
Abstract.
Cembranos and Freniche proved that for every two infinite compact Hausdorffspaces X and Y the Banach space C ( X × Y ) of continuous real-valued functions on X × Y endowed with the supremum norm contains a complemented copy of the Banach space c .We extend this theorem to the class of C p -spaces, that is, we prove that for all infiniteTychonoff spaces X and Y the space C p ( X × Y ) of continuous functions on X × Y endowedwith the pointwise topology contains either a complemented copy of R ω or a complementedcopy of the space ( c ) p = { ( x n ) n ∈ ω ∈ R ω : x n → } , both endowed with the producttopology. We show that the latter case holds always when X × Y is pseudocompact.On the other hand, assuming the Continuum Hypothesis (or even a weaker set-theoreticassumption), we provide an example of a pseudocompact space X such that C p ( X × X ) does not contain a complemented copy of ( c ) p .As a corollary to the first result, we show that for all infinite Tychonoff spaces X and Y the space C p ( X × Y ) is linearly homeomorphic to the space C p ( X × Y ) × R , although, asproved earlier by Marciszewski, there exists an infinite compact space X such that C p ( X ) cannot be mapped onto C p ( X ) × R by a continuous linear surjection. This provides apositive answer to a problem of Arkhangel’ski for spaces of the form C p ( X × Y ) .Our another corollary—analogous to the classical Rosenthal–Lacey theorem for Banachspaces C ( X ) with X compact and Hausdorff—asserts that for every infinite Tychonoffspaces X and Y the space C k ( X × Y ) of continuous functions on X × Y endowed withthe compact-open topology admits a quotient map onto a space isomorphic to one of thefollowing three spaces: R ω , ( c ) p or c . Introduction
Recall that a Banach space E is a Grothendieck space or has the
Grothendieck property ifevery weakly ∗ convergent sequence in the dual E ∗ of E converges weakly, i.e. every sequence ( ϕ n ) n ∈ ω of continuous functionals on E satisfying the condition that lim n →∞ ϕ n ( x ) = 0 forevery x ∈ E satisfies also the condition that lim n →∞ ψ ( ϕ n ) = 0 for every ψ in E ∗∗ , thebidual space of E . Grothendieck [21] proved that spaces of the form ℓ ∞ (Γ) are Grothendieckspaces. Later, many other Banach spaces were recognized to be Grothendieck, e.g. vonNeumann algebras (Pfitzner [38]), the space H ∞ of bounded analytic functions on the unitdisc (Bourgain [11]), spaces of the form C ( K ) for K an F-space (Seever [41]), etc. Onthe other hand, the space c of all sequences convergent to is not Grothendieck, since aseparable Banach space is Grothendieck if and only if it is reflexive. It follows that closed The research for the first named author is supported by the GAČR project 20-22230L and RVO:67985840. The third and fourth authors thank the Austrian Science Fund FWF (Grants I 2374-N35,I 3709-N35, M 2500-N35) for generous support for this research. linear subspaces of Grothendieck spaces need not be Grothendieck, although this propertyis preserved by complemented subspaces. Cembranos [12] proved that a space C ( K ) isGrothendieck if and only if it does not contain any complemented copy of the space c .The following results due to Cembranos [12] and Freniche [19] are the main motivationfor our paper. Theorem 1.1 ([12], [19]) . Let K and L be infinite compact spaces and let E be an infinitedimensional Banach space.(1) C ( K, E ) contains a complemented copy of (the Banach space) c and hence it isnot a Grothendieck space.(2) The Banach space C ( K × L ) contains a complemented copy of c . The second statement of Theorem 1.1 follows from the first one combined with the factthat C ( K × L ) is isomorphic to C ( K, C ( L )) .A strengthening of Theorem 1.1.(2) was obtained by Kąkol, Sobota and Zdomskyy [27]for spaces of the form C p ( K × L ) (note that it follows from the Closed Graph Theoremthat a complemented copy of ( c ) p in C p ( K × L ) is actually a complemented copy of c in C ( K × L ) ). Here, by C p ( X ) and C k ( X ) we mean the space C ( X ) of continuous real-valued functions over a Tychonoff space X endowed with the pointwise and compact-opentopology, respectively. C ∗ p ( X ) denotes the vector subspace of C p ( X ) consisting of boundedfunctions. It is well known that for every infinite compact space K the space C ( K ) containsa copy of c , while it is also easy to see that C p ( K ) over an infinite compact K containsa (closed) subspace isomorphic to ( c ) p = { ( x n ) n ∈ ω ∈ R ω : x n → } endowed with theproduct topology of R ω . Theorem 1.2 ([27]) . For every infinite compact spaces K and L the space C p ( K × L ) contains a complemented copy of the space ( c ) p . In [27] many other spaces C p ( X ) were recognized to contain a complemented copy of ( c ) p , i.a. those C p ( X ) where X is a compact space such that the Banach space C ( X ) isnot Grothendieck. In fact, for compact spaces K , the existence of a complemented copy of ( c ) p in the space C p ( K ) appeared to be equivalent to the property that the Banach space C ( K ) does not have the so-called ℓ -Grothendieck property, a variant of the Grothendieckproperty defined as follows: for a given compact space K we say that a Banach space C ( K ) has the ℓ -Grothendieck property if every weakly ∗ convergent sequence of Radon measureson K with countable supports (equivalently, with finite supports) is weakly convergent;see [27] for details. Trivially, if C ( K ) is Grothendieck, then it has the ℓ -Grothendieckproperty; a counterexample for the reverse implication was first constructed by Plebanek([39], see also [6]), in [27, Section 7] another example and a more detailed discussion onthis topic were provided.The research in [27] was motivated, i.a., by the following theorem of Banakh, Kąkol andŚliwa [5], especially by the equivalence (1) ⇔ (2). We refer the reader to the paper [27] fora detailed discussion concerning properties of sequences of measures from (2). Theorem 1.3 ([5]) . For a Tychonoff space X the following conditions are equivalent: OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) (1) C p ( X ) satisfies the Josefson–Nissenzweig property ( JNP in short), i.e. there is asequence ( µ n ) n of finitely supported signed measures on X such that k µ n k = 1 forall n ∈ ω , and µ n ( f ) → n for each f ∈ C p ( X ) .(2) C p ( X ) contains a complemented subspace isomorphic to ( c ) p ;(3) C p ( X ) has a quotient isomorphic to ( c ) p ;(4) C p ( X ) admits a linear continuous map onto ( c ) p .Moreover, for pseudocompact X the above conditions are equivalent to:(5) C p ( X ) contains an infinite dimensional complemented metrizable subspace. Consequently, if the space C p ( X ) contains a complemented copy of ( c ) p , the same holdsfor C p ( X × Y ) for all non-empty Tychonoff spaces Y . The converse implication does nothold (e.g. for X = βω ).The condition (1) in Theorem 1.3 is a variant of the celebrated Josefson–Nissenzweigtheorem for C p ( X ) spaces. Recall here that the Josefson–Nissenzweig theorem asserts thatfor each infinite-dimensional Banach space E there exists a sequence ( x ∗ n ) n in the dualspace E ∗ convergent to in the weak ∗ topology of E ∗ and such that k x ∗ n k = 1 for every n ∈ ω , see e.g. [14] (a proof for the case E = C ( K ) was provided in [27], too).The main results of the present paper generalize Theorems 1.1.(2) and 1.2 to pseudo-compact spaces in the following way. Theorem 1.4.
Let X and Y be infinite Tychonoff spaces. Then:(1) If the space X × Y is pseudocompact, then C p ( X × Y ) contains a complementedcopy of ( c ) p .(2) If the space X × Y is not pseudocompact, then C p ( X × Y ) contains a complementedcopy of R ω . Consequently, if for infinite Tychonoff spaces X and Y the product X × Y is pseudo-compact, then C p ( X × Y ) contains a complemented copy of ( c ) p , so by the Closed GraphTheorem the Banach space ( C ( X × Y ) , k . k ∞ ) contains a complemented copy of the Banachspace c . This provides a stronger version of Cembranos–Freniche theorem. However, ifthe product X × Y is not pseudocompact, then consistently C p ( X × Y ) may fail to havethe complemented copy of ( c ) p . As our next theorem shows, this may happen even for thesquares. We refer the reader to the paragraph before Theorem 5.8 for the exact formula-tion of the set-theoretic assumption we use in the proof of the next theorem (and which iscalled ( † ) by us). Let us only mention here that this assumption ( † ) is satisfied if eitherthe Continuum Hypothesis or Martin’s axiom holds true. Theorem 1.5.
It is consistent that there exists an infinite pseudocompact space X suchthat the spaces C p ( X × X ) and C ∗ p ( X × X ) do not contain a complemented copy of ( c ) p . The proofs of Theorems 1.4 and 1.5 are provided in Sections 2 and 5, respectively.Unfortunately, we do not know whether the negation of the conclusion of Theorem 1.5may consistently hold. We are also not aware of any model of ZFC where ( † ) , our set-theoretic assumption used to prove Theorem 1.5, fails. J. KĄKOL, W. MARCISZEWSKI, D. SOBOTA, AND L. ZDOMSKYY
Problem 1.6.
Is it consistent that for any infinite pseudocompact space X the space C p ( X × X ) (respectively, C ∗ p ( X × X ) ) contains a complemented copy of ( c ) p ? Recall that every countably compact space is pseudocompact, thus the following questionseems natural in the context of Theorems 1.2 and 1.5.
Problem 1.7.
Is it consistent that there exists an infinite countably compact space X suchthat the space C p ( X × X ) (respectively, C ∗ p ( X × X ) ) does not contain a complemented copyof ( c ) p ? The methods applied to prove Theorem 1.5 cannot be used to answer Problem 1.7 inaffirmative since they produce spaces which are very far from being countably compact.However, recall that there exist countably compact spaces whose squares are not pseudo-compact, see e.g. [17, Example 3.10.9].Theorem 1.4 has two important applications concerning the following remarkable prob-lem in C p -theory posed by Arkhangel’ski, see [1, 2], and the famous Rosenthal–Laceytheorem. Problem 1.8 (Arkhangel’ski) . Is it true that for every infinite (compact) space K thespace C p ( K ) is linearly homeomorphic to C p ( K ) × R ? For a wide class of spaces the answer to Problem 1.8 is affirmative, e.g. if the space X contains a non-trivial convergent sequence, or X is not pseudocompact (see [2, Section4]), yet, in general, the answer is negative even for compact spaces X , see Marciszewski[33]. It appears however that for finite products of Tychonoff spaces the answer is stillpositive—the following corollary follows immediately from Theorem 1.4. Corollary 1.9.
Let X and Y be infinite Tychonoff spaces. Then C p ( X × Y ) is linearlyhomeomorphic to the product C p ( X × Y ) × R . The famous Rosenthal–Lacey theorem [40], [32], see also [24, Corollary 1], asserts thatfor each infinite compact space K the Banach space C ( K ) admits a quotient map onto c or ℓ ; we refer the reader to a survey paper [18, Theorem 18] for a detailed discussionon the theorem. The case of C p -spaces remains however open, namely, it is still unknownwhether for every infinite compact space K the space C p ( K ) admits a quotient map ontoan infinite-dimensional metrizable space, see [29]. Nevertheless, Theorem 1.4 yields thefollowing corollary. Corollary 1.10.
Let X and Y be infinite Tychonoff spaces. Then(1) C p ( X × Y ) admits a quotient map onto R ω or ( c ) p .(2) C k ( X × Y ) admits a quotient map onto a space isomorphic to one of the followingspaces: R ω , ( c ) p or c . The proofs of Theorem 1.4 and its consequences, e.g. Corollary 1.10, are provided inSection 2.
Notation and terminology.
Our notation and terminology are standard, i.e., we followmonographs of Tkachuk [42] (function spaces), Engelking [17] (general topology), Halmos
OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) [25] (measure theory) and Jech [23] (set theory). In particular, we assume that all topo-logical spaces we consider are Tychonoff , that is, completely regular and Hausdorff.The cardinality of a set X is denoted by | X | . By ω we denote the first infinite cardinalnumber, i.e., the cardinality of the space of natural numbers N . Usually we identify ω with N , so ω is an infinite countable discrete topological space, thus, e.g., such notions like theČech–Stone compactifiactions βω of ω have sense. We denote the remainder βω \ ω of thiscompactification by ω ∗ . The continuum, i.e. the cardinality of the real line R , is denotedboth by c and ω . If X is a set and κ is a (finite or infinite) cardinal number, then by [ X ] κ we denote the family of all subsets of X of cardinality κ ; in particular, [ X ] ω denotes thefamily of all infinite countable subsets of X . We put [ X ] <ω = S n ∈ ω [ X ] n , so [ X ] <ω is thefamily of all finite subsets of X . Finally, ω ω denotes the family of all functions from ω into ω .All other necessary and possibly non-standard notions will be defined in relevant placesof the text. 2. Proof of Theorem 1.4 and its consequences
For a space X and a point x ∈ X let δ x : C p ( X ) → R , δ x : f f ( x ) , be the Diracmeasure concentrated at x . The linear hull L p ( X ) of the set { δ x : x ∈ X } in R C p ( X ) canbe identified with the dual space of C p ( X ) . Elements of the space L p ( X ) will be called finitely supported signed measures (or simply signed measures ) on X .Each µ ∈ L p ( X ) can be uniquely written as a linear combination of Dirac measures µ = X x ∈ F α x δ x for some finite set F ⊂ X and some non-zero real numbers α x . The set F is called the support of the signed measure µ and is denoted by supp( µ ) . The measure P x ∈ F | α x | δ x willbe denoted by | µ | and the real number k µ k = X x ∈ F | α x | coincides with the norm of µ (in the dual Banach space C ( βX ) ∗ ).A sequence ( µ n ) n of finitely supported signed measures on X such that k µ n k = 1 for all n ∈ ω , and lim n µ n ( f ) = 0 for each f ∈ C p ( X ) is called a Josefson–Nissenzweig sequence or,in short, a JN -sequence on X . A JN -sequence ( µ n ) n is supported on a subset A of a space X if the supports of all measures µ n are contained in A .We say that C p ( X ) has the Josefson–Nissenzweig property or, in short, JNP if X admitsa JN -sequence.The following proposition was proved in [27]. We provide a brief sketch of the proof forthe sake of completeness. Proposition 2.1.
The product βω × βω has a JN -sequence ( µ n ) n supported on ω × ω .Moreover, the supports of µ n have pairwise disjoint projections onto each axis. J. KĄKOL, W. MARCISZEWSKI, D. SOBOTA, AND L. ZDOMSKYY
Proof.
For every n ∈ N put Ω n = {− , } n and Σ n = n × { n } . To simplify the notation,we will usually write i ∈ Σ n instead of ( i, n ) ∈ Σ n .Define Ω = S n ∈ ω Ω n and Σ = S n ∈ ω Σ n , and endow these two sets with the discretetopology. This enables us to look at the the product space Ω × Σ as a countable union ofpairwise disjoint discrete rectangles Ω k × Σ m of size m k —the rectangles Ω n × Σ n , lyingalong the diagonal, will bear a special meaning, namely, they will be the supports ofmeasures from a JN -sequence ( µ n ) n on the space β Ω × β Σ defined as follows: µ n = X s ∈ Ω n ,i ∈ Σ n s ( i ) n n δ ( s,i ) , n ∈ ω. It turns out that the sequence ( µ n ) n defined above is a JN -sequence, see [27, Section7] for details. Note that ω is homeomorphic to both Ω and Σ , so βω , β Ω and β Σ aremutually homeomorphic. Consequently, βω × βω has the JN -sequence with the requiredproperties. (cid:3) In [27, Section 7] this result was used to prove that, for any infinite compact spaces
K, L ,the space C p ( K × L ) has the JNP . By a modification of this argument we can obtain thefollowing strengthening:
Theorem 2.2.
For any infinite spaces
X, Y , if the product X × Y is pseudocompact, thenit admits a JN -sequence ( µ n ) n . Moreover, we can require that ( µ n ) n is supported on theproduct D × E , where D ⊂ X and E ⊂ Y are countable discrete, and the supports of µ n have pairwise disjoint projections onto each axis.Proof. Let D and E be countable discrete subsets of X and Y , respectively. Let ϕ : ω → D and ψ : ω → E be bijections. By the Stone Extension Property of βω , there are continuousmaps Φ : βω → βX and Ψ : βω → βY such that Φ ↾ ω = ϕ and Ψ ↾ ω = ψ .We denote by Θ the product map Φ × Ψ : βω × βω → βX × βY. Clearly Θ maps ω × ω injectively into X × Y .Let ( µ n ) n ∈ ω be a JN -sequence of measures on ( βω ) supported on ω , given by Proposition2.1. For each n ∈ ω , we consider the image of µ n under Θ , i.e., the measure ν n on X × Y defined as follows: ν n = X z ∈ supp( µ n ) µ n (cid:0) { z } (cid:1) · δ Θ( z ) , it follows that k ν n k = 1 and supp( ν n ) is finite. We will show that the sequence ( ν n ) converges to on every f ∈ C ( X × Y ) which will demonstrate that X × Y admits a JN -sequence.Recall that by Glicksberg’s theorem [17, 3.12.20(c)], pseudocompactness of X × Y impliesthat βX × βY is the Čech–Stone compactification of X × Y , i.e., every continuous functionon X × Y is continuously extendable over βX × βY . OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) Fix f ∈ C ( X × Y ) and let F be its continuous extension over βX × βY . Then thecomposition F ◦ Θ is a continuous function on βω × βω and ν n ( f ) = µ n ( F ◦ Θ) for each n ∈ ω . Therefore lim n ν n ( f ) = 0 .The additional properties of the supports of the measures ν n follow easily from thedefinition of ν n and the corresponding properties of supports of the measures µ n . (cid:3) We are in the position to prove the main result of this paper, Theorem 1.4.
Proof of Theorem 1.4 . If the space X × Y is pseudocompact, then by the above theoremand Theorem 1.3 the space C p ( X × Y ) has a complemented copy of ( c ) p .On the other hand, it is well known that if a space X is not pseudocompact, then C p ( X ) has a complemented copy of R ω , cf. [2, Section 4] or [28, Theorem 14]. This completes theproof of part (2) of Theorem 1.4. (cid:3) Observe that if a subspace Y of a space X has a JN -sequence, then this sequence is also a JN -sequence on X . By Theorem 1.3, it follows that if C p ( Y ) contains a complemented copyof ( c ) p , then C p ( X ) also contains such a copy. Therefore Theorem 1.4.(1) immediatelygives the following Corollary 2.3.
If a space Z contains a topological copy of a pseudocompact product S × T ofinfinite spaces S and T , then C p ( Z ) contains a complemented copy of ( c ) p . In particular, ifspaces X and Y contain infinite compact subsets, then C p ( X × Y ) contains a complementedcopy of ( c ) p . Note that there exist infinite spaces X without infinite compact subsets such that thesquare X × X is pseudocompact, see Example 3.4.Recall that a subset B of a locally convex space E is bounded if for every neighbourhoodof zero U in E there exists a scalar λ > such that λB ⊂ U. Note the following fact connected with the next Corollary 2.6.
Lemma 2.4.
For a space X the following assertions are equivalent:(1) The space C k ( X ) is covered by a sequence of bounded sets.(2) The space C p ( X ) is covered by a sequence of bounded sets.(3) X is pseudocompact.Proof. (1) ⇒ (2) is clear, since the compact-open topology is stronger than the pointwiseone of C ( X ) .(2) ⇒ (3): Assume C p ( X ) is covered by a sequence of bounded sets but X is notpsudocompact. Then C p ( X ) contains a complemented copy of R ω . On the other hand, R ω cannot be covered by a sequence ( S n ) n of bounded sets. Indeed, we may assume that each S n is absolutely convex and closed and S n ⊂ S n +1 for all n ∈ ω . By the Baire theoremsome S n is a neighbourhood of zero in R ω . Consequently R ω must be a normed space bya theorem of Day, see [22, Proposition 6.9.4], a contradiction (since R ω is not normable).(3) ⇒ (1): If X is pseudocompact and S = { f ∈ C ( X ) : | f ( x ) | ≤ , x ∈ X } , J. KĄKOL, W. MARCISZEWSKI, D. SOBOTA, AND L. ZDOMSKYY then the sequence of bounded sets ( nS ) n in the compact-open topology covers C k ( X ) . (cid:3) Remark . Since the image of a bounded subset of a topological vector space under acontinuous linear operator is bounded, from the above lemma we can easily deduce that if T : C k ( X ) → C k ( Y ) ( T : C p ( X ) → C p ( Y ) ) is a linear continuous surjection and the space X is pseudocompact, then Y is also pseudocompact. This fact for the pointwise topologyis well known, cf. [36, Proposition 6.8.6].Since the pseudocompactness is not transferred even to finite products (see [17]), thereexist spaces X and Y such that both C k ( X ) and C k ( Y ) are covered by a sequence ofbounded sets but C k ( X × Y ) lacks this property, by Lemma 2.4.A sequence ( S n ) n of bounded sets in a locally convex space E is fundamental if everybounded set in E is contained in some set S n . Every ( DF ) (in particular, every normed)space C k ( X ) admits a fundamental sequence of bounded sets. By Warner [44] the space C k ( X ) admits a fundamental sequence of bounded sets if and only if the following conditionholds:( ∗ ) Given any sequence ( G n ) n of pairwise disjoint non-empty open subsets of X thereis a compact set K ⊂ X such that { n ∈ ω : K ∩ G n = ∅} is infinite. This characterization easily implies that if the space C k ( X ) has a fundamental sequenceof bounded sets, and the space X is infinite, then X contains an infinite compact subspace.Therefore, by Corollary 2.3 we obtain Corollary 2.6.
Let X and Y be two infinite spaces. If C k ( X ) and C k ( Y ) admit fun-damental sequences of bounded sets, then C p ( X × Y ) contains a complemented copy of ( c ) p .Remark . Note that if X is a pseudocompact space without infinite compact subsets(cf. Section 3), then by Warner’s characterization, C k ( X ) does not have a fundamentalsequence of bounded sets, although it is covered by a sequence of bounded sets, by Lemma2.4.Theorem 1.4 applies also to get the following Corollary 2.8.
Let X be an infinite pseudocompact space such that C ( βX ) is a Grothendieckspace. Then for no infinite spaces Y and Z does exist a continuous linear surjection from C p ( X ) onto the space C p ( Y × Z ) .Proof. Assume that for some infinite spaces Y and Z there exists a continuous linearsurjection T : C p ( X ) → C p ( Y × Z ) . Then the space Y × Z is pseudocompact, see Remark2.5. By Theorem 1.4 C p ( Y × Z ) contains a complemented copy of ( c ) p . Hence C p ( X ) maps onto ( c ) p by a continuous linear map, and then by Theorem 1.3 the space C p ( X ) hasa complemented copy of ( c ) p . This implies that C ( βX ) contains a complemented copy of c , a contradiction with the Grothendieck property of C ( βX ) . (cid:3) For the proof of Corollary 1.10 will need some auxiliary facts.The next proposition is known, the case of the pointwise topology can be found in[42], the general result covering the cases of both topologies is stated in [34, Exercise 2,
OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) p. 36], but without a proof, hence, for the sake of completeness, we include the prooffor the compact-open topology. Recall that a subset Y of a topological space X is C-embedded ( C ∗ -embedded ) in X if every (bounded) continuous real-valued function on Y can be continuously extended over X . Given a compact subspace K of a space X and ε > , we denote the set { f ∈ C k ( X ) : f ( K ) ⊂ ( − ε, ε ) } , a basic neighborhood of zero in C k ( X ) , by U X ( K, ε ) . Proposition 2.9.
Let Y be a subspace of a topological space X , and R : C k ( X ) → C k ( Y ) ( R : C p ( X ) → C p ( Y ) ) be the restriction operator defined by R ( f ) = f ↾ Y . R is open ifand only if Y is closed and C-embedded in X .Proof. As we mentioned, the proof for the case of the pointwise topology can be found in[42, S.152], therefore we will only give the proof for the case of the compact-open topology.Assume first that R is open. Then obviously it is a surjection, which means that Y isC-embedded in X . Suppose, towards a contradiction, that Y is not closed in X and picka point x ∈ Y \ Y . We will show that the image R ( U X ( { x } , is not open in C k ( Y ) .Consider arbitrary basic neighborhood of zero U Y ( K, ε ) in C k ( Y ) , given by a compact set K ⊂ Y and ε > . The set L = { x } ∪ K is compact in X , hence it is C-embedded in X ,see [20, 3.11].Let f : L → R be the (continuous) function which takes value at x and value on K , and let F be its continuous extension over X . Put g = F ↾ Y . Obviously, for any h ∈ C k ( X ) such that h ↾ Y = g ↾ Y , the functions h and F must agree on the closure of Y , so h ( x ) = F ( x ) = 1 . This shows that g ∈ U Y ( K, ε ) \ R ( U X ( { x } , . Now, assume that Y is closed and C-embedded in X . To prove that R is open its enoughto verify that, for any compact K ⊂ X and ε > , we have R ( U X ( K, ε )) = U Y ( K ∩ Y, ε ) . In the last equality, one inclusion is obvious. Take any g ∈ U Y ( K ∩ Y, ε ) . We will show that g is an image under R of some f ∈ U X ( K, ε ) . Let G be a continuousextension of g over X . Put L = { x ∈ K : | G ( x ) | ≥ ε } . Clearly, L is a compact subset of X disjoint with Y . Hence we can use the well knownfact (see [20, 3.11]), that a compact set can be separated from a closed set by a continuousfunction, i.e., we can find a continuous h : X → [0 , which sends L to and Y to . Onecan easily verify that f = hG has the required properties. (cid:3) Recall that if X is a pseudocompact space and K is a compact space, then X × K ispseudocompact (see [17, Corolllary 3.10.27]), so C p ( X × K ) contains a complemented copyof ( c ) p , provided that both X and K are infinite. For the compact-open topology in thisfunction space we have the following special case of Corollary 1.10. Proposition 2.10.
Assume that X is an infinite pseudocompact space and K is an infinitecompact space.(1) If X contains an infinite compact subset, then C k ( X × K ) contains a complementedcopy of the Banach space c .(2) If X has no infinite compact subsets, then C k ( X × K ) contains a complementedcopy of ( c ) p .Proof. Part (1): By Domański’s and Drewnowski’s theorem [16], the space C k ( X, C k ( K )) contains a complemented copy of c . On the other hand, by [34, Corollary 2.5.7] the spaces C k ( X × K ) and C k ( X, C k ( K )) are linearly homeomorphic.Part (2): Since the product X × K is pseudocompact, we can apply Theorem 2.2 toobtain a JN -sequence ( µ n ) n on X × K such that ( µ n ) n is supported on the product D × E ,where D ⊂ X and E ⊂ Y are countable discrete, and the supports of µ n have pairwisedisjoint projections onto each axis. Since the sets D and E are discrete and countable,we can find families { U d : d ∈ D } and { V e : e ∈ E } of pairwise disjoint sets, such that,for each d ∈ D , U d is an open neighborhood of d in X and for each e ∈ E , V e is an openneighborhood of e in Y . Let A n = supp( µ n ) and A = S n ∈ ω A n . Given a = ( d a , e a ) ∈ A ,put W a = U d a × V e a . Clearly, the family { W a : a ∈ A } of neighborhoods of points of A consists of pairwise disjoint sets. Moreover, if we define, for each n ∈ ω , W n = [ { W a : a ∈ A n } , then the properties of supports A n imply that the sets W n have pairwise disjoint projectionsonto each axis. For each a ∈ A , take a continuous function g a : X × K → [0 , such that g a ( a ) = 1 and g takes value on ( X × K ) \ W a . For every n ∈ ω define f n : X × K → [ − , by f n = X a ∈ A n µ n ( a ) | µ n ( a ) | g a . The functions f n and the sets W n have the following properties for all n ∈ ω :(a) µ n ( f n ) = 1 ;(b) the support of f n is contained in W n ;(c) the support of µ n is contained in W n ;(d) the projections of the sets W k , k ∈ ω , onto X are pairwise disjoint.Consider the linear operator S : ( c ) p → C k ( X × K ) defined by S (( t n )) = X n ∈ ω t n f n OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) for ( t n ) ∈ ( c ) p . By properties (b) and (d), the supports of f n are pairwise disjoint,therefore S is well defined. Given a compact subset L of X × K , the projection of L onto X is finite, hence, by properties (b) and (d), L intersects only finitely many supports of f n , say only for n in some finite set F . Therefore, for any ε > , if | t n | < ε , for n ∈ F , then S (( t n )) ∈ U X × K ( L, ε ) , which means that the operator S is continuous. Put Z = S (( c ) p ) .We will show that Z is an isomorphic copy of ( c ) p which is complemented in C k ( X × K ) .Let T : C k ( X × K ) → ( c ) p be defined by T ( f )( n ) = µ n ( f ) for f ∈ C k ( X × K ) and n ∈ ω . Obviously, the operator T is continuous. Using properties (a)–(d) one can easilyverify that ( T ↾ Z ) ◦ S = Id ( c ) p and S ◦ ( T ↾ Z ) = Id Z , hence the spaces Z and ( c ) p are isomorphic. Let P = S ◦ T : C k ( X × K ) → C k ( X × K ) . The identities Z = S (( c ) p ) and S ◦ ( T ↾ Z ) = Id Z imply that P is a continuous projectionof C k ( X × K ) onto Z . (cid:3) We are ready to provide a proof of Corollary 1.10.
Proof of Corollary 1.10 . Part (1) is a direct consequence of Theorem 1.4.For the proof of part (2) we shall consider several cases:
Case 1.
If both spaces X and Y contain infinite compact subsets, say K and L , re-spectively, then K × L is C-embedded in X × Y (see [20, 3.11]), and the restriction map C k ( X × Y ) → C k ( K × L ) is a continuous and open surjection by Proposition 2.9. Now wecan use Theorem 1.1 (2). Case 2.
If neither X nor Y admits an infinite compact subset, then for C p ( X × Y ) = C k ( X × Y ) applies part (1). Case 3.
If one of the spaces X or Y is not pseudocompact, the product X × Y is also notpseudocompact, and the space C k ( X × Y ) contains a complemented copy of R ω , see [28,Theorem 14]. Case 4.
The only remaining case to be checked is that when both X and Y are pseudo-compact, and one of these spaces contains an infinite compact subset but the other onelacks infinite compact subsets. Without loss of generality we can assume that Y containsan infinite compact subset K . We will verify that X × K is C-embedded in X × Y . Let f bea continuous function on X × K . Since X × K is pseudocompact, by Glicksberg’s theorem[17, 3.12.20(c)], f can be extended to a continuous function g on βX × K . By compactnessof βX × K , we can extend g continuously over βX × Y obtaining a function h . Now, therestriction of h to X × K is the desired extension of f . Proposition 2.9 implies that therestriction operator R : C k ( X × Y ) → C k ( X × K ) is open, hence also surjective. FromProposition 2.10 we obtain a projection P of C k ( X × K ) onto a subspace Z isomorphic to ( c ) p . Since P is open (as a map onto Z ), the composition P ◦ R is an open continuoussurjection of C k ( X × Y ) onto a copy of ( c ) p . (cid:3) Haydon’s construction of a pseudocompact space with no infinitecompact subsets
Theorem 1.4 may suggest a question whether C p ( X × X ) contains a complemented copyof ( c ) p for any infinite pseudocompact space X . In Section 5, assuming the ContinuumHypothesis (or even a weaker set-theoretic assumption), we will answer this question neg-atively. Our examples use the following general scheme of constructing pseudocompactspaces with no infinite compact subsets given by Haydon in [26].For each A ∈ [ ω ] ω , choose an ultrafilter u A ∈ ω ∗ in the closure of A in βω . Let X = ω ∪ { u A : A ∈ [ ω ] ω } be topologized as a subspace of βω . To simplify the notation we will denote the family ofall such spaces X by HS .We start with the following fact collecting some properties of every space X ∈ HS . Proposition 3.1.
Every space X ∈ HS has the following properties:(1) X is pseudocompact of cardinality continuum;(2) all compact subspaces of X are finite;(3) C p ( X ) does not have the JNP ;(4) C p ( X ) admits an infinite dimensional metrizable quotient isomorphic to the sub-space ( ℓ ∞ ) p = { ( x n ) ∈ R ω : sup n | x n | < ∞} of R ω endowed with the product topologyand the Banach space ( C ( X ) , k . k ∞ ) (i.e. endowed with the sup-norm topology) isisometric to the Banach space ℓ ∞ ;(5) for every infinite compact space K both spaces C p ( K × X ) and C k ( K × X ) containa complemented copy of ( c ) p .Proof. Obviously | X | ≤ ω . To see that X must contain continuum many ultrafilters,it is enough to take an almost family A ⊂ [ ω ] ω (i.e., distinct members of A have finiteintersection) of cardinality continuum. Then the closures of elements of A in βω havepairwise disjoint intersections with ω ∗ . This proves the second item of (1).The first item of (1) has been shown by Haydon in the third example of [26]. Indeed,if C p ( X ) is not pseudocompact, then there exists f ∈ C ( X ) unbounded. As ω is dense in X one gets a sequence ( n k ) in ω such that | f ( n k ) | > k . For the set A = { n k : k ∈ ω } thecorresponding u A in βω belongs to the closure in βω of { n k : k > m } for each m ∈ ω .Hence | f ( u A ) | > m for each m ∈ ω , a contradiction.Proof of (2): Since infinite compact subsets of βω have cardinality c , every compactsubset of X is finite by (1).Proof of (3): Note that βX = βω and the Banach space ( C ( X ) , k . k ∞ ) is isometric to ℓ ∞ = C ( βω ) . On the other hand, as C ( βω ) is a Grothendieck space, C ( X ) does not havea complemented copy of the Banach space c . Consequently, by applying the closed graphtheorem [37, Theorem 4.1.10], the space C p ( X ) does not contain a complemented copy of ( c ) p . Now by Theorem 1.3 we know that C p ( X ) does not have JNP .Proof of (4): Since the space X is pseudocompact and contains discrete ω which isC ∗ -embedded into X , we apply Theorem 1 from [4] to deduce that C p ( X ) has a quotient OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) C p ( X ) /W isomorphic to the subspace ( ℓ ∞ ) p of R ω endowed with the product topology,where W = \ n { f ∈ C p ( X ) : X x ∈ F n f ( x ) = 0 } and ( F n ) n is any sequence of non-empty, finite and pairwise disjoint subsets of ω with lim n | F n | = ∞ . The second part of the item (4) is clear since C ( βω ) and ( C ( X ) , k . k ∞ ) areisometric.Proof of (5): Since K × X is pseudocompact, the first claim of (5) follows from Theorem1.4. The other claim follows from Proposition 2.10. (cid:3) The following proposition characterizes those subspaces X of βω containing ω which arein the class HS . Proposition 3.2.
For a subspace X of βω containing ω , the following conditions areequivalent:(1) X ∈ HS ;(2) X is pseudocompact and of cardinality continuum;(3) X is of cardinality continuum, and every infinite subset of ω has an accumulationpoint in X .Proof. The implication (1) ⇒ (2) was explained in the previous proposition.The implication (2) ⇒ (3) is obvious, a pseudocompact X cannot contain an infiniteclopen discrete subset.To verify (3) ⇒ (1), first observe that each ultrafilter in X ∩ ω ∗ is in the closure ofcontinuum many sets A ∈ [ ω ] ω . Also, given A ∈ [ ω ] ω , the closure of A in X has cardinalitycontinuum, which is witnessed by an almost disjoint family A of subsets of A of cardinality | A | = 2 ω . Therefore, by a standard back-and-forth inductive argument, we can constructa bijection f between [ ω ] ω and X ∩ ω ∗ such that u A = f ( A ) is in the closure of A . To thisend, enumerate [ ω ] ω as { A α : α < ω } , and X ∩ ω ∗ as { u α : α < ω } . Inductively, for each α < ω , define f on { A β : β < α } and f − on { u β : β < α } . (cid:3) In [33] Marciszewski constructed an example of a space X ∈ HS such that there is nocontinuous linear surjection from the space C p ( X ) onto C p ( X ) × R . In particular, C p ( X ) isnot linearly homeomorphic to C p ( X ) × R , hence C p ( X ) does not contain any complementedcopy of c or R ω . To achieve this it is enough to require that all ultrafilters u A in X areweak P-points and they are pairwise non-isomorphic.Recall that a point p ∈ ω ∗ is a weak P -point if p is not in the closure of any countableset D ⊂ ω ∗ \ { p } , see [35]. Two ultrafilters u, v ∈ βω are called isomorphic if h ( u ) = v for some homeomorphism h of βω ; equivalently, there is a bijection f : ω → ω such that u = { f − ( A ) : A ∈ v } . The Rudin–Keisler preorder on βω is the binary relation ≤ RK on βω given by v ≤ RK u if there is f : ω → ω such that u ⊃ { f − ( A ) : A ∈ v } , i.e., v = f ( u ) , where f ( u ) = { A ⊂ ω : f − ( A ) ∈ u } . This preorder becomes an order if we identify isomorphic ultrafilters,see [13, Corollary 9.3]. We say that ultrafilters u, v ∈ ω ∗ are incompatible with respect tothe Rudin–Keisler preorder if there is no w ∈ ω ∗ such that w ≤ RK u and w ≤ RK v . An RK -antichain in ω ∗ is a subset of ω ∗ consisting of pairwise incompatible ultrafilters withrespect to the Rudin–Keisler preorder. The Continuum Hypothesis or even a weaker set-thoretic assumptions, like Martin’s Axiom, imply that there exists an RK -antichain of sizecontinuum, consisting of weak P-points in ω ∗ , cf. [13, Theorem 9.13, Lemma 9.14] or [7,Theorem 2]. Blass and Shelah [9] proved the consistency, relative to ZFC , of the statementthat any two ultrafilters in ω ∗ are compatible with respect to the Rudin–Keisler preorder. Proposition 3.3.
Let X = ω ∪ { u A : A ∈ [ ω ] ω } be a space in HS such that for distinct A, B ∈ [ ω ] ω the ultrafilters u A , u B are not isomorphic. Then the square X × X is notpseudocompact.Proof. Take two disjoint infinite sets
A, B ⊂ ω and any bijection f : A → B . Observe thatthe graph Γ = { ( x, f ( x )) : x ∈ A } of f is an open discrete subspace of X × X . We willshow that Γ is closed in X × X .Suppose, towards a contradiction that, there exists ( u , u ) ∈ Cl X × X (Γ) \ Γ . For any n ∈ ω , the intersections Γ ∩ ( { n } × X ) and Γ ∩ ( X × { n } ) contain at most one point.Therefore, they are closed in { n } × X and X × { n } , respectively. Since the latter sets areopen in X × X , it follows that u , u ∈ ω ∗ . Since Γ ⊂ A × B , we have ( u , u ) ∈ Cl X × X ( A × B ) = Cl X ( A ) × Cl X ( B ) . The disjoint sets A and B have disjoint closures in X , so u = u .Let g : ω → ω be any bijection extending f . We claim that the ultrafilter { g − ( C ) : C ∈ u } is equal to u . Suppose that these two ultrafilters are not equal. Then we can find D ⊂ ω such that D ∈ u and ( ω \ D ) ∈ { g − ( C ) : C ∈ u } , i.e., E = g ( ω \ D ) ∈ u . Observe that the product D × E is disjoint with the graph of g ,hence is also disjoint with Γ . Therefore the product Cl X ( D ) × Cl X ( E ) is a neighborhoodof ( u , u ) disjoint with Γ (because Γ ⊂ ω × ω and (Cl X ( D ) × Cl X ( E )) ∩ ω × ω = D × E, and the latter is disjoint with Γ ), a contradiction. We obtained that the ultrafilters u and u are isomorphic, contradicting our assumption on X .Finally, we verified that the infinite discrete subspace Γ is clopen in X × X , hence thespace X × X is not pseudocompact. (cid:3) On the other hand, we have the following example.
Example 3.4.
There exists a space Y ∈ HS such that the square Y × Y is pseudocompact.Consequently, C p ( Y × Y ) contains a complemented copy of ( c ) p . Proof.
For any space X ∈ HS , the square ω × ω is dense in X × X , therefore it is enoughto construct a space Y ∈ HS such that any infinite subset of ω × ω has an accumulationpoint in Y × Y , see the proof of Proposition 3.1.(1). OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) Let { A α : α < ω } be an enumeration of [ ω ] ω , and { C α : α < ω } be an enumeration of [ ω × ω ] ω . For each α < ω , choose an ultrafilter u α from ω ∗ in the closure of A α , and anaccumulation point ( p α , q α ) of C α in βω × βω . Let Y = ω ∪ { u α , p α , q α : α < ω } . One can easily verify that Y satisfies condition (c) from Proposition 3.2, so it is in HS , andany infinite subset of ω × ω has an accumulation point in Y × Y . The last claim followsfrom Theorem 1.4. (cid:3) The Bounded Josefson–Nissenzweig Property
In this section we define a “bounded” version of the Josefson–Nissenzweig property. Asequence ( µ n ) of finitely supported signed measures on X such that k µ n k = 1 for all n ∈ N ,and lim n µ n ( f ) = 0 for each bounded f ∈ C p ( X ) is called a bounded Josefson–Nissenzweigsequences or, in short, a BJN -sequence . We say that C p ( X ) has the Bounded Josefson–Nissenzweig Property or, in short, the BJNP , if X admits a BJN -sequence.Obviously, for pseudocompact spaces X , these bounded versions coincide with the stan-dard ones. It is also trivial that the JNP implies the
BJNP and each JN -sequence is a BJN -sequence.One can easily construct examples of
BJN -sequences, which are not JN -sequences. Ac-tually, if X is not pseudocompact and C p ( X ) has the JNP , then X has a BJN -sequencewhich is not a JN -sequence. Indeed, if X is such a space, take a JN -sequence ( µ n ) n on X and a continuous unbounded function f on X. For each n , pick x n ∈ X such that | f ( x n ) | > n . Define ν n = µ n + (1 /n ) δ x n . One can easily verify that the sequence ( ν n ) n (after normalizing) is a BJN -sequence but not a JN -sequence.In Example 4.2 we construct a space with the BJNP but without the JNP. For a verifica-tion of the properties of the example we will need an auxiliary fact concerning JN -sequences.Recall that a subset A of a topological space X is bounded if for every f ∈ C p ( X ) the set f ( A ) is bounded in R . Proposition 4.1.
Let ( µ n ) n ∈ ω be a JN -sequence of measures on a space X . Then the union S n ∈ ω supp( µ n ) of supports of µ n is bounded in X .Proof. We sketch the proof of the proposition using operators between linear spaces. In[27, Section 4] we provide a more direct and elementary proof.Let S = { } ∪ { /n : n = 1 , , . . . } be equipped with the Euclidean topology. Wecan define a continuous linear operator T : C p ( X ) → C p ( S ) in the following way: for f ∈ C p ( X ) , put T ( f )(0) = 0 and T ( f )(1 / ( n + 1)) = µ n ( f ) for n ∈ ω . From the definition of a JN -sequence immediately follows that the operator T iswell-defined. Observe that, for any n ≥ , the support supp T (1 /n ) of /n in X with respectto T (see [36, Chapter 6.8]) is equal to supp( µ n ) . Obviously the set A = { /n : n ≥ } is bounded in S , S being compact. By Theorem 6.8.3 in [36], the support supp T ( A ) = [ n ≥ supp T (1 /n ) = [ n ∈ ω supp( µ n ) is bounded in X . (cid:3) Example 4.2.
There exists a countable space X such that C p ( X ) has the BJNP , but doesnot have the
JNP . Proof.
Recall the well known method of obtaining a countable space N F associated with afilter F on a countable set T (we consider only free filters on T , i.e., filters containing allcofinite subsets of T ). N F is the space T ∪ {∞} , where ∞ 6∈ T , equipped with the following topology: Allpoints of T are isolated and the family { A ∪ {∞} : A ∈ F } is a neighborhood base at ∞ .Let F be the filter on ω consisting of sets of density , i.e., F = n A ⊂ ω : lim n | A ∩ { , , . . . , n }| n = 1 o . We will show that the space X = N F has the required properties (this space in the contextof function spaces was investigated in [15]).Let S n = { n + 1 , n + 2 , . . . , n +1 } , n ∈ ω , and λ n = X (cid:26) n +1 δ k : k ∈ S n (cid:27) − δ ∞ . Let us check that ( λ n ) n ∈ ω is a BJN -sequence. Take a bounded continuous function f on N F ; for simplicity, we can assume that | f | ≤ . Fix an ε > , and find a set A ∈ F suchthat the image under f of the neighborhood A ∪ {∞} of ∞ has the diameter less than ε .Choose k ∈ ω such that, | A ∩ { , , . . . , m }| m > − ε, for m > k . Observe that, for n > k , we have n +1 > k , | S n | = 2 n and S n ⊂ { , , . . . , n +1 } , hence theabove inequality applied for m = 2 n +1 , gives us the estimate | S n \ A | < n +1 ε . Hence, for OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) n > k , we have | λ n ( f ) | = (cid:12)(cid:12) X (cid:26) n +1 f ( k ) : k ∈ S n (cid:27) − f ( ∞ ) (cid:12)(cid:12) ≤≤ X (cid:26) n +1 | f ( k ) − f ( ∞ ) | : k ∈ S n (cid:27) == X (cid:26) n +1 | f ( k ) − f ( ∞ ) | : k ∈ S n ∩ A (cid:27) ++ X (cid:26) n +1 | f ( k ) − f ( ∞ ) | : k ∈ S n \ A (cid:27) << | S n ∩ A | · ε n +1 + | S n \ A | · n +1 < n · ε n +1 + 2 n +1 ε · n +1 = 52 ε , which shows that the sequence ( λ n ( f )) converges to .It is clear that for any JN -sequence ( µ n ) in a space X , the union S of supports of µ n isinfinite, and by Proposition 4.1 S is also bounded. Since all bounded subsets of N F arefinite (cf. [15, Example 7.1]), C p ( N F ) does not have the JNP . (cid:3) Recall that C ∗ p ( X ) is the subspace of C p ( X ) consisting of bounded functions. Theorem 4.3.
For a space X the following conditions are equivalent:(1) C p ( X ) has the BJNP ;(2) C ∗ p ( X ) has a complemented copy of ( c ) p ;(3) C ∗ p ( X ) has a quotient isomorphic to ( c ) p ;(4) C ∗ p ( X ) admits a continuous linear surjection onto ( c ) p .Proof. The implication (1) ⇒ (2) follows from Lemma 1 in [5] and its proof. It is enoughto apply this lemma for K = βX , and observe that the subspace L of C p ( X ) given bythis lemma contains C ∗ p ( X ) and the range of the projection P from L onto the subspaceisomorphic to ( c ) p is contained in C ∗ p ( X ) .Implications (2) ⇒ (3) ⇒ (4) are obvious.(4) ⇒ (1). Let T : C ∗ p ( X ) → ( c ) p be a continuous linear surjection. For each n ∈ ω , let p n : ( c ) p → R be the projection onto n th axis, and λ n be the finitely supported signedmeasure on X corresponding to the functional p n ◦ T . Let C ∗ ( X ) be the Banach space ofall bounded continuous functions on X , equipped with the standard supremum norm. Bythe Closed Graph Theorem, T can be treated as continuous linear surjection between theBanach spaces C ∗ ( X ) and c , which gives the estimate k λ n k ≤ k T k for all n . By the OpenMapping Theorem we can also get a constant c > such that k λ n k ≥ c for all n . Now itis clear that ( λ n / k λ n k ) n ∈ ω is a BJN -sequence on X . (cid:3) Corollary 4.4. If C p ( X ) has the BJNP , then C ( βX ) contains a complemented copy of theBanach space c . Corollary 4.5. If C ∗ p ( X ) does not have any complemented copy of ( c ) p , then C p ( X ) doesnot have it either. Proof. If C p ( X ) does have a complemented copy of ( c ) p , then C p ( X ) has the JNP and so C ∗ p ( X ) has the BJNP , which implies that C ∗ p ( X ) has a complemented copy of ( c ) p . (cid:3) Example 4.2 combined with Theorems 1.3 and 4.3 gives us the following
Corollary 4.6.
There exists a countable space X such that C ∗ p ( X ) has a complementedcopy of ( c ) p , but C p ( X ) does not have such a copy. Consequently, the Banach space C ( βX ) contains a complemented copy of the Banach space c . A pseudocompact space X such that its square X × X does not have aBJN-sequence First we will prove several auxiliary results. The proof of the following standard fact issimilar to that of [33, Lemma 3.1].
Lemma 5.1.
Let { S n : n ∈ ω } be a family of pairwise disjoint subsets of ω . Then, forevery subset C of βω of the cardinality less than ω , there exists an infinite subset A ⊂ ω such that C ∩ (cid:0) Cl βω [ n ∈ A S n (cid:1) ⊂ [ n ∈ A Cl βω S n . Proof.
Let A ⊂ [ ω ] ω be an almost disjoint family of size | A | = c . It is clear that (cid:0) Cl βω [ n ∈ A S n (cid:1) ∩ (cid:0) Cl βω [ n ∈ A ′ S n (cid:1) = [ n ∈ A ∩ A ′ Cl βω S n for any distinct A, A ′ ∈ A . Thus the family (cid:8)(cid:0) Cl βω [ n ∈ A S n (cid:1) \ [ n ∈ A Cl βω S n : A ∈ A (cid:9) is disjoint, and hence one of its elements is disjoint from C as the latter has size < c = | A | .This completes our proof. (cid:3) Lemma 5.2.
Let
X, Y be spaces in HS , and ( A n ) n ∈ ω , ( A n ) n ∈ ω be two sequences of non-empty subsets of ω such that, for each k, n ∈ ω, k = n , i=1,2, A ik ∩ A in = ∅ . Put U = [ n ∈ ω Cl X × Y ( A n × A n ) and let f , f : ω → ω be functions such that, for i = 1 , , f i ( A in ) = { n } , f i takes oddvalues on ω \ S n ∈ ω A in , and is injective on this complement.If ( u , u ) ∈ Cl X × Y ( U ) \ U , then u , u ∈ ω ∗ and f ( u ) = f ( u ) ∈ ω ∗ , hence theultrafilters u and u are compatible with respect to the Rudin–Keisler preorder. Moreover, u ∈ Cl X (cid:0) [ n ∈ ω A n (cid:1) \ [ n ∈ ω Cl X ( A n ) and u ∈ Cl Y (cid:0) [ n ∈ ω A n (cid:1) \ [ n ∈ ω Cl Y ( A n ) . OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) Proof.
Let ( u , u ) ∈ Cl X × Y ( U ) \ U . For any n ∈ ω , the intersections U ∩ ( { n } × Y ) and U ∩ ( X × { n } ) are either empty or, for some k ∈ ω , are equal to { n } × Cl Y ( A k ) or Cl X ( A k ) ×{ n } , respectively. Therefore they are closed in { n }× Y and X ×{ n } , respectively.Since the sets { n } × Y and X × { n } are open in X × Y , it follows that u , u ∈ ω ∗ .Let V = S n ∈ ω A n × A n . Obviously, V is dense in U , hence ( u , u ) belongs to Cl X × Y ( V ) \ U . Set p i = f i ( u i ) , i = 1 , . If p = p , then we can find C ⊂ ω such that C ∈ p and ( ω \ C ) ∈ p . We claim that the neighborhood O = Cl X ( f − ( C )) × Cl Y ( f − ( ω \ C )) of ( u , u ) is disjoint with V . Indeed, suppose that there are n ∈ ω and ( m , m ) ∈ ( A n × A n ) ∩ (cid:0) Cl X ( f − ( C )) × Cl Y ( f − ( ω \ C )) (cid:1) , i.e., m ∈ Cl X ( f − ( C )) and m ∈ Cl Y ( f − ( ω \ C )) . Since ω is a discrete subspace of X and Y , we conclude that m ∈ f − ( C ) and m ∈ f − ( ω \ C ) , which means n = f ( m ) ∈ C and n = f ( m ) ∈ ω \ C , a contradiction. Therefore O ∩ V = ∅ , which together with ( u , u ) ∈ O and ( u , u ) ∈ Cl X × Y ( V ) again leads to a contradiction.Hence p = p = p , and it remains to observe that p ∈ ω ∗ . Indeed, if { n } ∈ p , then wehave two cases:If n is odd, then | f − i ( n ) | ≤ , so f − i ( n ) cannot belong to u i ∈ ω ∗ , i = 1 , .If n = 2 k , then A ik ∈ u i , i = 1 , , hence ( u , u ) ∈ Cl X × Y ( A k × A k ) ⊂ U, which again gives a contradiction.Finally, it is clear that u ∈ Cl X ( S n ∈ ω A n ) . If u ∈ Cl X ( A n ) for some n , we would getthat { n } = f ( A n ) ∈ f ( u ) = p , thus contradicting p ∈ ω ∗ . Analogously for u . (cid:3) Lemma 5.3.
Let ( µ n ) n ∈ ω be a BJN -sequence on a space X . If Y is a clopen subset of X such that the sequence ( | µ n | ( Y )) n ∈ ω does not converge to , then there exist an increasingsequence ( n k ) and a BJN -sequence ( ν k ) k ∈ ω in X such that supp ν k ⊂ Y ∩ supp µ n k for every k ∈ ω , in particular ( ν k ) k ∈ ω is a BJN -sequence on Y .Proof. Pick an a > and an increasing sequence ( n k ) such that | µ n k | ( Y )) > a for all k ∈ ω .Let ν k = ( µ n k ↾ Y ) / | µ n k | ( Y ) . Since any continuous bounded function on Y can be extended to a continuous boundedfunction on X by declaring value outside Y , it easily follows that ( ν k ) k ∈ ω is a BJN -sequence on X and in Y . (cid:3) Lemma 5.4.
Let X and Y be spaces in HS such that all ultrafilters in X \ ω and Y \ ω are weak P-points. If C p ( X × Y ) has the BJNP ( JNP ), then there exist a
BJN -sequence (a JN -sequence) ( µ n ) n ∈ ω in X × Y , and sequences ( A n ) n ∈ ω , ( A n ) n ∈ ω of finite subsets of X, Y ,respectively, such that(1) A ik ∩ A in = ∅ for k, n ∈ ω, k = n , i = 1 , ; (2) A k ∩ A n = ∅ for k, n ∈ ω, k = n ;(3) supp µ n ⊂ A n × A n for each n ∈ ω ;(4) S n ∈ ω A n ∪ S n ∈ ω A n is a discrete subspace of βω .Proof. First we will consider the
BJNP case. Let ( ν k ) k ∈ ω be a BJN -sequence on X × Y .Given n ∈ ω , the subspaces { n } × Y and X × { n } are homeomorphic to X and Y ,respectively, C p ( { n } × Y ) and C p ( X × { n } ) do not have the BJNP by Proposition 3.1.Since these subspaces are clopen, Lemma 5.3 implies that lim k | ν k | ( { n } × Y ) = 0 , lim k | ν k | ( X × { n } ) = 0 . Let π X : X × Y → X and π Y : X × Y → Y be projections. For a subset B of ω , we denote π − X ( B ) ∪ π − Y ( B ) by C ( B ) . Clearly, for any finite B ⊂ ω , we have lim k | ν k | ( C ( B )) = 0 .By induction we choose an increasing sequence ( k i ) i ∈ ω and a sequence ( B i ) i ∈ ω of pairwisedisjoint finite subsets of ω such that ( i ) ( ω × ω ) ∩ supp ν k i ⊂ C ( S i − j =0 B j ) ∪ ( B i × B i ) ; ( ii ) | ν k i | ( C ( S i − j =0 B j )) < / ( i + 1) .For i ∈ ω , let Z i = ( X × Y ) \ C (cid:0) i − [ j =0 B j (cid:1) . Define λ i = ( ν k i ↾ Z i ) / k ν k i ↾ Z i k . Using condition ( ii ) above, one can easily verify that lim i k λ i − ν k i k = 0 , hence ( λ i ) i ∈ ω is a BJN -sequence on X × Y . The first condition implies that ( ω × ω ) ∩ supp λ i ⊂ B i × B i . Let C = (cid:0) π X (cid:0) [ i ∈ ω supp λ i (cid:1) ∪ π Y (cid:0) [ i ∈ ω supp λ i (cid:1)(cid:1) \ ω ⊂ ω ∗ . By Lemma 5.1 we can find an infinite subset T ⊂ ω such that C ∩ (cid:0) Cl βω [ i ∈ T B i (cid:1) = ∅ . Then the subsequence ( λ i ) i ∈ T has the property that the set D := π X (cid:0) [ i ∈ T supp λ i (cid:1) ∪ π Y (cid:0) [ i ∈ T supp λ i (cid:1) is a discrete subspace of βω . Indeed, the set D ′ = D ∩ ω ∗ is a discrete subspace of βω because it is countable and consists of weak P -points. In addition, Cl βω ( D ∩ ω ) is disjointwith D ′ by our choice of T because D ∩ ω ⊂ S i ∈ T B i . Thus D is a union of two discretesubsets of βω so that the closure of any of them does not intersect another one, and henceit is also discrete. OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) Set D X = π X ( S i ∈ T supp λ i ) and D Y = π Y ( S i ∈ T supp λ i ) , so that D = D X ∪ D Y . Fix apoint p ∈ D X , and find a clopen set V in X such that V ∩ D X = { p } . Let U = V × Y . U is clopen in X × Y and U ∩ [ i ∈ T supp λ i ⊂ { p } × Y. Observe that each continuous function on { p }× Y can be extended to a bounded continuousfunction on U , therefore, by Lemma 5.3, we conclude that lim i ∈ T | λ i | ( { p } × Y ) = 0 . In the same way we can show that, for any q ∈ D Y , we have lim i ∈ T | λ i | ( X × { q } ) = 0 .Now, we can repeat our inductive construction from the first part of the proof. Wechoose an increasing sequence ( i n ) n ∈ ω , i n ∈ T , and sequences ( A n ) n ∈ ω , ( A n ) n ∈ ω of finitesubsets of D X , D Y , respectively, such that the following conditions are satisfied (where E n = S l =1 , S n − j =0 A lj ): ( a ) ( A n ∪ A n ) ∩ E n = ∅ ; ( b ) supp λ i n ⊂ (cid:0) π − X ( X ∩ E n ) ∪ π − Y ( Y ∩ E n ) (cid:1) ∪ ( A n × A n ) ; ( c ) | λ i n | (cid:0) π − X ( X ∩ E n ) ∪ π − Y ( Y ∩ E n ) (cid:1) < / ( n + 1) .For n ∈ ω , let S n = ( X × Y ) \ (cid:0) π − X ( X ∩ E n ) ∪ π − Y ( Y ∩ E n ) (cid:1) . Define µ n = ( λ i n ↾ S n ) / k λ i n ↾ S n k . Conditions (1) and (2) follow from ( a ) . Condition ( c ) implies that lim n k λ i n − µ n k = 0 , hence ( µ n ) n ∈ ω is a BJN -sequence on X × Y . From condition ( b ) we deduce that supp µ n ⊂ A n × A n . Condition (4) follows from inclusions A n ⊂ D X , A n ⊂ D Y .Observe, that the union of supports of measures µ n that we constructed above is con-tained in the union of supports of measures ν k from our initial BJN -sequence. If we startwith a JN -sequence ( ν k ) k ∈ ω , then, by Proposition 4.1, S k ∈ ω supp( ν k ) is bounded in X × Y ,therefore, if we repeat the above argument, the resulting BJN -sequence ( µ n ) n ∈ ω will be alsoa JN -sequence. (cid:3) Before we prove Theorem 5.8, which is the main result of this section, we will show itsweaker version, since it has an essentially simpler proof and a less technical set-theoreticassumption.
Theorem 5.5.
Assume that there exist two incompatible with respect to the Rudin–Keislerpreorder weak P-points u, v in ω ∗ . Then there exist infinite pseudocompact spaces X, Y in HS such that C p ( X × Y ) does not have the BJNP , and hence it does not have the
JNP . Proof.
It is an easy and well known observation that, given an ultrafilter w ∈ ω ∗ and a set A ∈ [ ω ] ω , there is an ultrafilter w ′ in the closure of A in βω , which is isomorphic to w .Therefore, we may take spaces X = ω ∪ { u A : A ∈ [ ω ] ω } , Y = ω ∪ { v A : A ∈ [ ω ] ω } in HS such that all ultrafilters u A are isomorphic to u , and all ultrafilters v A are isomorphicto v . We will show that the product X × Y does not have a BJN -sequence.Suppose the contrary, and let ( µ n ) n ∈ ω , ( A n ) n ∈ ω , ( A n ) n ∈ ω be sequences of measures and sets given by Lemma 5.4. Let A i = [ n ∈ ω A in , i = 1 , . Recall that basic clopen sets in X and Y are closures of subsets of ω . For i = 1 , , since A i is discrete, using a simple induction, we can find a family { U ip : p ∈ A i } of pairwise disjointsubsets of ω , such that the closure of U ip is a neighborhood of p . For i = 1 , and n ∈ ω , let V in = [ { U ip : p ∈ A in } . For fixed i , the family { V in : n ∈ ω } is disjoint, and we have supp µ n ⊂ Cl X × Y ( V n × V n ) . From our construction of spaces X and Y and Lemma 5.2 we infer that the set U = [ n ∈ ω Cl X × Y ( V n × V n ) is clopen in X × Y . Let f n : supp µ n → [ − , be defined by f n ( z ) = µ n ( { z } ) / | µ n ( { z } ) | for z ∈ supp µ n . Since supp µ n is finite, we can extend each f n to a continuous function F n : Cl X × Y ( V n × V n ) → [ − , . Since sets Cl X × Y ( V n × V n ) are clopen, the union of all F n is continuous on U , and wecan extend this union to a bounded continuous F on X × Y , declaring value outside U .Clearly, for all n ∈ ω , we have µ n ( F ) = 1 , a contradiction. (cid:3) For a set A by △ A we denote the diagonal { ( a, a ) : a ∈ A } in A × A . Lemma 5.6.
Let
E, F be non-empty finite subsets of a set X , and µ be a signed measureon E × F . Then there exist disjoint G ⊂ E and H ⊂ F such that | µ | ( G × H ) ≥ | µ | (( E × F ) \ △ X ) / . OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) Proof.
Let A = E ∩ F . First, we will show that there exist disjoint subsets B, C of A suchthat | µ | ( B × C ) ≥ | µ | (( A × A ) \ △ A ) / . The case when | A | ≤ is easy, so we also assume that | A | > . Suppose the contrary,then, for any partition ( B, C ) of A , we have | µ | ( B × C ) < | µ | (( A × A ) \ △ A ) / . Case 1. | A | = 2 n + 1 . Let P be the family of all partitions ( B, C ) of A such that | B | = n, | C | = n + 1 . We have | P | = (cid:0) n +1 n (cid:1) and each point from ( A × A ) \ △ A appears in (cid:0) n − n − (cid:1) sets B × C , where ( B, C ) ∈ P . Hence we have (cid:18) n − n − (cid:19) | µ | (( A × A ) \ △ A ) = X {| µ | ( B × C ) : ( B, C ) ∈ P } < (cid:18) n + 1 n (cid:19) ( | µ | (( A × A ) \ △ A ) / . Since (cid:18) n + 1 n (cid:19) / (cid:18) n − n − (cid:19) = 2(2 n + 1) / ( n + 1) = 4 − / ( n + 1) ≤ , we arrive at a contradiction. Case 2. | A | = 2 n . Repeat the argument from the first case using partitions into sets ofsize n and the estimate (cid:0) nn (cid:1) / (cid:0) n − n − (cid:1) = 2(2 n − /n ≤ .Next, observe that ( E × F ) \ △ X = (cid:0) ( E \ F ) × F (cid:1) ∪ (cid:0) A × ( F \ E ) (cid:1) ∪ (cid:0) ( A × A ) \ △ A (cid:1) , and the sets A, E \ F, F \ E are all disjoint. If the variation | µ | of one of the rectangles ( E \ F ) × F and A × ( F \ E ) is at least equal to | µ | (( E × F ) \ △ X ) / , then we can takethis rectangle as G × H . If this is not the case, then we have | µ | (( A × A ) \ △ A ) > (4 / · | µ | (( E × F ) \ △ X ) , which implies that | µ | ( B × C ) > (1 / · (4 / · | µ | (( E × F ) \ △ X ) , and we can take G = B and H = C . (cid:3) Lemma 5.7.
Let ( µ n ) n ∈ ω be a BJN -sequence on a space X . If Z is a subset of X such that lim n k µ n ↾ Z k = 0 , then there exist an increasing sequence ( n k ) and a BJN -sequence ( ν k ) k ∈ ω in X such that supp ν k ⊂ supp µ n k \ Z for every k ∈ ω . Proof.
Take an increasing sequence ( n k ) that k µ n k ↾ Z k < for all k . Let Y = X \ Z .Define ν k = ( µ n k ↾ Y ) / k µ n k ↾ Y k . We have lim k k ν k − µ n k k = 0 , and hence one has that ( ν k ) k ∈ ω is a BJN -sequence on X . (cid:3) Finally, we present the main result of this section. It is proved under the followingset-theoretic assumption( † ) There exists a function A u A assigning to each A ∈ [ ω ] ω a weak P -point u A ∋ A on ω such that for every pair h f , f i of functions from ω to ω , there exists A ⊂ [ ω ] ω of size | A | < c such that for all A A and A ∈ [ ω ] ω \ { A } we have f ( u A ) = f ( u A ) , provided that f ( u A ) , f ( u A ) ∈ ω ∗ . We do not know whether ( † ) can be proved outright in ZFC. However, at the end ofthis section we present a constellation of cardinal characteristics of the continuum whichimplies ( † ) and holds in many standard models of set theory as well as is implied by theContinuum Hypothesis or Martin’s axiom. Theorem 5.8.
Let [ ω ] ω ∋ A u A ∈ Cl βω ( A ) ∩ ω ∗ be a witness for ( † ) and X = ω ∪ { u A : A ∈ [ ω ] ω } . Then C p ( X × X ) does not have the BJNP , and hence it does not have the
JNP .Proof.
Suppose the contrary, and let ( µ n ) n ∈ ω , ( A n ) n ∈ ω , and ( A n ) n ∈ ω be sequences of mea-sures and sets given by Lemma 5.4. Let A = [ i =1 , [ n ∈ ω A in . By condition (4) of Lemma 5.4, the set A is discrete.First we will show that( ∗ ) lim n | µ n | (( A n × A n ) \ △ X ) = 0 . Assume towards the contradiction that there exist an a > and an increasing sequence ( n k ) such that | µ n k | (( A n k × A n k ) \ △ X )) > a for all k ∈ ω . Using Lemma 5.6, for each k ∈ ω , we can find disjoint sets B ik ⊂ A in k , i = 1 , ,such that | µ n k | ( B k × B k ) > a/ . Let B i = S k ∈ ω B ik for i = 1 , . From conditions (1), and (2) of Lemma 5.4, it follows thatthe sets B , B ⊂ A are disjoint.Since the set A is discrete, we can find a disjoint family { U p : p ∈ A } of subsets of ω ,such that the closure of U p is a neighborhood of p .For i = 1 , and k ∈ ω , let V ik = [ { U p : p ∈ B ik } and W ik = [ { U p : p ∈ A in k } . These sets have the following properties:
OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) (1) For fixed i , the family { W ik : k ∈ ω } is disjoint;(2) B k × B k ⊂ Cl X × X ( V k × V k ) ⊂ Cl X × X ( W k × W k ) ;(3) supp µ n k ⊂ Cl X × X ( W k × W k ) .Moreover, the sets V i = S k ∈ ω V ik , i = 1 , , are disjoint, hence have disjoint closures in X .Consider maps f i : ω → ω such that f − i ( { k } ) = V ik and f i ↾ ( ω \ V i ) is injective andtakes odd values, where i = 1 , . Set U ′ = [ k ∈ ω Cl X × X ( V k × V k ) and note that if ( u , u ) ∈ Cl X × X ( U ′ ) \ U ′ , then u = u and, by Lemma 5.2, f ( u ) = f ( u ) ∈ ω ∗ . Applying ( † ) to ( f , f ) we get A ⊂ [ ω ] ω of size | A | < c such that for all A A and A ∈ [ ω ] ω \ { A } we have f ( u A ) = f ( u A ) , provided that f ( u A ) , f ( u A ) ∈ ω ∗ . Itfollows from the above that Cl X × X ( U ′ ) \ U ′ ⊂ { u A : A ∈ A } . Applying Lemma 5.1 we can find I ∈ [ ω ] ω such that the set Cl X (cid:0) [ k ∈ I V ik (cid:1) \ [ k ∈ I Cl X ( V ik ) is disjoint from { u A : A ∈ A } , where i = 1 , . Thus, the set U = [ k ∈ I Cl X × X ( V k × V k ) is clopen in X × X . Indeed, since U is clearly open, it remains to prove that it is alsoclosed. Otherwise there exists ( u , u ) ∈ Cl X × X ( U ) \ U . Since for every n I we havethat Cl X × X ( V n × V n ) is a clopen subset of X × X disjoint from U , it is also disjoint from Cl X × X ( U ) , and hence Cl X × X ( U ) \ U ⊂ Cl X × X ( U ′ ) \ U ′ , which yields ( u , u ) ∈ { u A : A ∈ A } . In particular u ∈ { u A : A ∈ A } . However, u ∈ Cl X (cid:0) [ k ∈ I V k (cid:1) \ [ k ∈ I Cl X ( V k ) by the last clause of Lemma 5.2, which is impossible by our choice of I .For every k ∈ I set C k = supp µ n k ∩ U . Properties (1)–(3) imply the inclusions supp µ n k ∩ ( B k × B k ) ⊂ C k ⊂ Cl X × X ( V k × V k ) . Let f k : C k → [ − , be defined by f k ( z ) = µ n k ( { z } ) / | µ n K ( { z } ) | for z ∈ C k . Since C k is finite, we can extend each f k to a continuous function F k : Cl X × X ( V k × V k ) → [ − , . Since sets Cl X × Y ( V k × V k ) are clopen, the union of all F k , k ∈ I , is continuous on U , andwe can extend this union to a bounded continuous F on X × X , declaring value outside U . Clearly, for all k ∈ I , we have µ n k ( F ) > a/ , a contradiction.Now, condition ( ∗ ) and Lemma 5.7 imply that X × X admits a BJN -sequence ( ν k ) k ∈ ω supported on △ X . Clearly, △ X is homeomorphic to X , and each continuous bounded func-tion on △ X extends to a continuous bounded function on X × X . We obtain a contradictionwith the lack of BJNP for C p ( X ) , see Proposition 3.1. (cid:3) Now, we present a sufficient condition for ( † ) . Recall that u denotes the minimal cardi-nality of a base of neighborhoods for an ultrafilter in ω ∗ , and d is the minimal cardinalityof a cover of ω ω by compact sets. These cardinal characteristics can consistently attainarbitrary uncountable regular values ≤ c , independently one from another, see [10]. If u < d , then d is regular, see [43] and [8] for more information on these as well as othercardinal characteristics of the continuum.Recall that u ∈ ω ∗ is a P-point , if for any countable family B of open neighbourhoods of u in ω ∗ the intersection ∩ B contains u in its interior. Clearly, every P -point is also a weak P -point. P -points exist under various set-theoretic assumptions, e.g., under d = c everyfilter on ω generated by < c many sets can be enlarged to a P -point, see [30]. On the otherhand, there are models of ZFC without P -points (see [45]). In contrast with P -points theweak P -points exist in ZFC by [31].A family A ⊂ [ ω ] ω is called strongly centered , if ∩ A ′ ∈ [ ω ] ω for any finite A ′ ⊂ A .A sequence of elements of [ ω ] ω is strongly centered if the set of its elements is stronglycentered. Lemma 5.9. If d = c ≤ u + , then ( † ) holds. Moreover, there is a map A u A witnessingfor ( † ) such that u A is a P -point for every A ∈ [ ω ] ω .Proof. We need the following auxiliary
Sublemma 5.10.
Suppose that κ is a cardinal such that κ ≤ u and κ < c = d . Supposethat { u ξ : ξ < κ } ⊂ ω ∗ and {h f ξ , f ξ i : ξ < κ } is a family of pairs of maps from ω to ω .Then for every A ∈ [ ω ] ω there exists a P -point u ∋ A such that f ξ ( u ) = f ξ ( u ξ ) for all ξ < κ , provided that f ξ ( u ) , f ξ ( u ξ ) , ∈ ω ∗ .Proof. Without loss of generality we may assume that f ξ ( u ξ ) ∈ ω ∗ for all ξ < κ . Fix A ∈ [ ω ] ω . Set A − = A and suppose that for some η < κ we have already constructeda strongly centered sequence A η = h A ξ : − ≤ ξ < η i of infinite subsets of ω with thefollowing property (we associate ξ with the set { α : α < ξ } ): ( ∗ ) η For every ξ < η , if f ξ (cid:0) T γ ∈ a A γ (cid:1) ∈ [ ω ] ω for all a ∈ [ ξ ∪ {− } ] <ω , then A ξ =( f ξ ) − ( B ξ ) for some B ξ ∈ [ ω ] ω \ f ξ ( u ξ ) (hence f ξ ( A ξ ) f ξ ( u ξ ) ).If f η (cid:0) T γ ∈ a A γ (cid:1) is finite for some a ∈ [ η ∪ {− } ] <ω , then we simply set A η = ω . If f η (cid:0) T γ ∈ a A γ ) is infinite for all a ∈ [ η ∪ {− } ] <ω and f η (cid:0) T γ ∈ a η A γ ) f η ( u η ) for some finite OMPLEMENTED COPIES OF c IN SPACES C p ( X × Y ) a η ∈ [ η ∪ {− } ] <ω , then we set A η = ( f η ) − ( f η (cid:0) \ γ ∈ a η A γ (cid:1) ) . Otherwise, notice that for the family Y = n f η (cid:0) \ γ ∈ a A γ (cid:1) : a ∈ [ η ∪ {− } ] <ω o ⊂ f η ( u η ) the family { Cl βω ( Y ) \ Y : Y ∈ Y } cannot be a base of neighborhoods of f η ( u η ) in ω ∗ , because it has size < u , and thus thereexists X ∈ f η ( u η ) with | Y \ X | = ω for all Y ∈ Y . Set B η = ω \ X and A η = ( f η ) − ( B η ) .It follows from the above that A η +1 = h A ξ : − ≤ ξ ≤ η i is strongly centered and satisfies ( ∗ ) η +1 . This completes our recursive construction of a sequence A κ = h A ξ : − ≤ ξ < κ i .Since κ < c = d , there exists a P -point u ⊃ { A ξ : − ≤ ξ < κ } , see [30, 1.2 Theorem]. Forevery ξ ∈ κ such that f ξ ( u ) ∈ ω ∗ we have that f ξ ( A ξ ) ∈ f ξ ( u ) \ f ξ ( u ξ ) , which completesour proof. (cid:3) Now we proceed with the proof of Lemma 5.9. Let { A α : α < c } and {h f α , f α i : α < c } be enumerations of [ ω ] ω and the family of all pairs of functions from ω to ω , respectively.Recursively over α ∈ c we shall construct a P -point u α ∋ A α as follows: Assuming that u β is already constructed for all β ∈ α , let us fix an enumeration (cid:8) h u ξ , ( f ξ , f ξ ) i : ξ < κ (cid:9) of (cid:8) h u β , ( f γ , f γ i : β, γ ∈ α (cid:9) , where κ = | α × α | . By Sublemma 5.10 there exists a P -point u α ∋ A α such that f ξ ( u α ) = f ξ ( u ξ ) for all ξ < κ , provided that f ξ ( u α ) , f ξ ( u ξ ) , ∈ ω ∗ . In other words, ∀ β, γ ∈ α (cid:0) f γ ( u α ) , f γ ( u β ) ∈ ω ∗ ⇒ f γ ( u α ) = f γ ( u β ) (cid:1) . The map A α u α constructed this way clearly witnesses for ( † ) . (cid:3) Remark . If there exists an RK -antichain of size c consisting of weak P -points, then ( † ) holds. Indeed, let A be such an antichain. Using a standard transfinite inductiveconstruction, we can easily obtain a family { u A : A ∈ [ ω ] ω } ⊂ ω ∗ such that, for any A ∈ [ ω ] ω , A ∈ u A , and the ultrafilters u A and u B are isomorphic to distinct elements of A for A = B . Clearly, the assignment A u A is as required in ( † ) . However, by Lemma 5.9, ( † ) holds in all models of c = d ≤ ω , in particular it holds in the Blass–Shelah modelconstructed in [9] in which the RK -preorder is downwards directed. Thus the existence ofsuch RK -antichans as mentioned above is sufficient but not necessary for ( † ) and thus alsofor Theorem 5.8.Theorem 1.5 follows now easily. Proof of Theorem 1.5 . The theorem follows from Theorems 5.8 and 4.3, Corollary 4.5,and the consistency of ( † ) . (cid:3) References [1] A. V. Arkhangel’ski,
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Israel Journal of Mathematics (1982),28–48. Faculty of Mathematics and Computer Science, A. Mickiewicz Univesity, Poznań, Poland,and Institute of Mathematics, Czech Academy of Sciences, Prague, Czech Republic.
E-mail address : [email protected] Institute of Mathematics and Computer Science, University of Warsaw, Poland Warszawa,Poland.
E-mail address : [email protected] Universität Wien, Institut für Mathematik, Kurt Gödel Research Center, Augasse 2-6,UZA 1 — Building 2, 1090 Wien, Austria.
E-mail address : [email protected] Universität Wien, Institut für Mathematik, Kurt Gödel Research Center, Augasse 2-6,UZA 1 — Building 2, 1090 Wien, Austria.
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