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Fibers of continuous real-valued functions on ψ-spaces
Topology and its Applications 195 (2015) 256–264
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Topology and its Applications www.elsevier.com/locate/topol
Fibers of continuous real-valued functions on ψ-spaces Jerry E. Vaughan ∗ , Catherine Payne University of North Carolina at Greensboro, United States
a r t i c l e
i n f o
Article history: Received 10 December 2013 Received in revised form 11 August 2014 Accepted 2 June 2015 MSC: primary 54A25, 03E17 secondary 03E35, 03E55, 03E75, 54C30 Keywords: Maximal almost disjoint families ψ-space Almost disjointness number a Real-valued continuous functions Cardinal cofinality Full fiber Rich MADF Rich cardinals
a b s t r a c t We consider continuous real-valued functions with domain either a ψ-space (studied by S. Mrowka, J. Isbell, and others) or a generalized ψ-space introduced by A. Dow and J. Vaughan. A cardinal κ ≥ ω is called a rich cardinal provided for every infinite, maximal almost disjoint family M of countably infinite subsets of κ (MADF) and for every continuous f : ψ(κ, M) → R defined on the associated space ψ = ψ(κ.M), there exists r ∈ R such that |f −1 (r)| = |ψ|. Dow and Vaughan proved that ω is a rich cardinal if and only if a = c, where a is the smallest cardinality of a MADF on ω. We prove that a = c if and only if, for all ω ≤ κ ≤ c, κ is a rich cardinal, if and only if for every n < ω, ωn is a rich cardinal. We prove every κ > c is rich using a set-theoretic hypothesis weaker than GCH. © 2015 Published by Elsevier B.V.
1. Introduction The classical ψ-space is a well-known topological space, first considered by Alexandroff and Urysohn [1], and later by others (e.g., S. Mrówka [13], and J. Isbell [7, 5I]). The underlying set of points consists of the union of two sets. One set is a countable set X of isolated points (we take X = ω, where ω denotes the set of natural numbers). The other set is a closed discrete set in one-to-one correspondence with an infinite almost disjoint family A on ω. This connection between the topology of ψ-spaces and almost disjoint families (especially maximal almost disjoint families) makes the study of ψ-spaces of interest not only in topology and set-theory but also in Boolean algebra and Banach spaces (cf., [2,11]). A. Dow and J. Vaughan [3,4], extended the study of classical ψ-spaces to the case where the set X of isolated points is of arbitrary cardinality (we take the set X to be an arbitrary cardinal κ ≥ ω), and the * Corresponding author. E-mail addresses: [email protected] (J.E. Vaughan), [email protected] (C. Payne). http://dx.doi.org/10.1016/j.topol.2015.09.032 0166-8641/© 2015 Published by Elsevier B.V.
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closed discrete set is in one-to-one correspondence with an infinite almost disjoint family A of countably infinite subsets of κ. In this paper we consider real-valued continuous functions defined on such ψ-spaces, and ask about the cardinalities of their fibers. We discuss our results following a few definitions. We use standard set-theoretic notation (e.g., see [12]). For an infinite cardinal κ, let [κ]ω denote the set of all countably infinite subsets of κ. A family A ⊂ [κ]ω is called an almost disjoint family (ADF) provided the intersection of any two members of A is finite. A maximal almost disjoint family (MADF) M ⊂ [κ]ω is an infinite ADF that is not properly contained in any other ADF of countably infinite subsets of κ (we often write “M ⊂ [κ]ω MADF”). The topological space ψ(κ, M) is defined as follows. Let M be an ADF (or MADF) on an infinite cardinal κ. The underlying set of the space ψ(κ, M) is κ ∪ M and the open sets are defined by taking the points of κ to be isolated and basic neighborhoods of a point M ∈ M to have the form {M } ∪ (M \ F ), where F is any finite subset of M . When κ and M are understood we write ψ for ψ(κ, M). Given an infinite cardinal κ, κ+ denotes the first cardinal larger than κ, and κω denotes the cardinality of [κ]ω . We follow the usual convention and write ωn instead of ℵn for n ∈ ω and we write ω instead of ω0 . The cardinal ω1 , defined by ω1 = ω + , denotes the first uncountable cardinal. Similarly ωn = (ωn−1 )+ for n ≥ 2, and ℵω = sup{ωn : n ∈ ω}. The cardinal number of the real line (the cardinality of the continuum) is denoted by c, and the cardinal a (called the almost-disjointness number) denotes the smallest cardinality of a MADF on ω. It is well known that ω1 ≤ a ≤ c (e.g., see [5]). By CH we mean the Continuum Hypothesis (2ω = ω1 ) and by GCH we mean the Generalized Continuum Hypothesis (i.e., 2κ = κ+ for all κ ≥ ω). We introduce the following terminology. Definition 1.1. Let f : X → Y . We say that a fiber f −1 (y), where y ∈ Y , is of full size (or is full) provided |f −1 (y)| = |X|. We say that a MADF M ⊂ [κ]ω is a rich MADF provided every real-valued continuous function defined on ψ(κ, M) has at least one full fiber. An infinite cardinal κ is called a rich cardinal provided every M ⊂ [κ]ω MADF is a rich MADF. The notion of rich MADF evolved naturally from special properties of MADF’s constructed in [13] and [3]. In [13], Mrówka constructed a M ⊂ [ω]ω MADF such that |βψ(ω, M) \ ψ(ω, M)| = 1, or equivalently, for every continuous f : ψ → R, there exists r ∈ R such that |ψ \ f −1 (r)| ≤ ω. Thus Mrówka’s special M is a rich MADF. In [3], Dow and Vaughan proved that (assuming GCH) for all κ > ω, there exists Mκ ⊂ [κ]ω MADF such that for every continuous f : ψ(κ, Mκ ) → R, there exists r ∈ R such that |ψ \ f −1 (r)| < |ψ|. These Mκ are rich MADF’s. The results in this paper were motivated by the following theorem (stated here using the above terminology). Theorem 1.2. (Dow–Vaughan [3]) The cardinal ω is a rich cardinal if and only if a = c. Several years ago, we proved that the cardinal ω1 is rich iff a = c (half of this result is in [15]). This is an exact analog of Theorem 1.2. These two results obviously raise the question: When are the cardinals ωn (n ≥ 2) rich? In general, when is a cardinal κ rich? Towards answering these questions, the following result gives two general conditions equivalent to “a = c.” Theorem 1.3 (Corollary 4.4). The following are equivalent: (1) a = c (2) for every ω ≤ κ ≤ c, κ is a rich cardinal. (3) for every n < ω, ωn is a rich cardinal. We prove, under hypothesis weaker than GCH, that every κ ≥ ω is a rich cardinal (see Section 5). Indeed our hypothesis is much weaker than GCH or “CH + SCH” where SCH stands for the singular
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cardinal hypothesis (e.g., see [10, §8]). Moreover we prove in ZFC that if κ = κω , then κ is a rich cardinal (Corollary 3.2). In the absence of GCH, there are cardinals that are not rich. For instance, it is consistent that a < c (hence ω and ω1 are not rich cardinals), and ω2 is a rich cardinal; see Section 4 and Example 4.5. Concerning rich MADFs, we begin with a simple remark. Remark 1.4. If M ⊂ [κ]ω MADF with cf (|M|) > c then M is a rich MADF. Proof. Let M ⊂ [κ]ω MADF with cf (|M|) > c. We need only note that in general if f : X → R is any function (continuous or not) with cf |X| > c, then for some x ∈ R, |f −1 (x)| = |X|. For otherwise, we have |f −1 (r)| < |X| for all r ∈ R. Then since X = {f −1 (r) : r ∈ R}, we have |X| ≤
|f −1 (r)| < |X|
(1)
r∈R
where the strict inequality holds because we are adding fewer than cf |X| sets, each of cardinality less than |X|. Of course, (1) is a contradiction. 2 This remark shows, for example, that c+ is a rich cardinal because (c+ )ω = c+ (see [10, p. 48]). The remark also shows that for cardinals κ > c, the major difficulty regarding whether κ is a rich cardinal is the case cf κω ≤ c (see Section 5, and Question 5.6). To prove our results about the cardinals ωn we need to know the smallest cardinality of a MADF on ωn . We consider this in the next section. 2. The smallest cardinality of a MADF on ωn The main result in this section (Theorem 2.4) was given previously, and in more general settings, in unpublished preprints (see [14], [8]). We include our direct proof here for completeness and for the convenience of the reader. Definition 2.1. ([14],[8]) Let κ be an infinite cardinal. For κ ≥ ω a(κ) = min |M| : M ⊂ [κ]ω is a MADF A general problem is to find a(κ) for all κ (see [14]). For the purposes of this paper, we find a(ωn ) for n ∈ ω. To do this we use the following definitions. Definition 2.2. Let P = {Pn : n ∈ ω} be a family of pairwise disjoint infinite subsets of a set X. We say that H ∈ [X]ω is a diagonal of P provided |{n ∈ ω : H ∩ Pn = ∅}| = ω. We say that a family A ⊂ [X]ω is maximal for diagonal subsets of P provided for every diagonal H of P there exists A ∈ A such that |A ∩ H| = ω. Lemma 2.3. If κ ≥ ω and P = {Pn : n ∈ ω} is a pairwise disjoint family of infinite subsets of κ, then there exists A ⊂ [κ]ω ADF that is maximal for diagonal sets of P and such that for every A ∈ A, A ∩ Pn is finite for all n ∈ ω, and |A| ≤ a(κ). Proof. Let P be as stated in the lemma, and let M ⊂ [κ]ω be a MADF with |M| = a(κ). We partition M into two sets. First, let B = {M ∈ M : |{n ∈ ω : |M ∩ Pn | = ω}| < ω}. For all M ∈ B, let kM ∈ ω such that for all n ≥ kM , (|M ∩ Pn | < ω). Put M = M \ {Pi : i ≤ kM }. Then M ∩ Pn is finite for all n ∈ ω (indeed, M = ∅ is possible). Second, let C = {M ∈ M : |{n ∈ ω : |M ∩ Pn | = ω}| = ω}. For each M ∈ C,
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put EM = {M ∩ Pn : n ∈ ω and |M ∩ Pn | = ω}. Then EM is a countable, infinite pairwise disjoint family of infinite subsets of M . Since M is countable, by [3, Lemma 11.2] there exists AM ⊂ [M ]ω MADF such that EM ⊂ AM , and |AM | = a. Let AM = AM \ EM . Then for all A ∈ AM and all n ∈ ω we have |A ∩ Pn | < ω. Also for M ∈ B and n ∈ ω, M ∩ Pn is finite, as noted above. Define A = M : M ∈ B and M is infinite ∪ AM : M ∈ C . We have already noted that for every A ∈ A and every n ∈ ω, we have A ∩ Pn is finite. We show that A satisfies the other conclusions of the lemma. Since M is ADF and B and C are disjoint, it is easy to see that A ⊂ [κ]ω is an ADF. Moreover, it is clear that |A| ≤ |B| + |C| · a ≤ |M| · a = a(κ). Finally, we show that A is maximal for the diagonal sets of P. Let H ∈ [κ]ω be a diagonal of P. Let J ⊂ H be such that J is infinite and |J ∩ Pn | ≤ 1 for all n ∈ ω. It suffices to find A ∈ A such that |A ∩ J| = ω. Since M is maximal there exists M ∈ M such that |M ∩ J| = ω. If M ∈ B, then |M ∩ J| = ω since |J ∩ ( i≤kM Pi )| < ω, and therefore M ∈ A, and we are done. Otherwise M ∈ C. Since AM ⊂ [M ]ω is maximal, there exists A ∈ AM such that |A ∩ (M ∩ J)| = ω, and since (A ∩ M ∩ J) intersects more than one Pn , we have A = M ∩ Pn for any n ∈ ω. Thus A ∈ / EM ; so A ∈ AM ⊂ A, and this completes the proof. 2 We now determine the minimum cardinality of a MADF on ωn for n ∈ ω. Of course, a(ω) = a by definition. Theorem 2.4. Let 0 ≤ n < ω. There exists M ⊂ [ωn ]ω a MADF such that |M| = ωn · a. Thus a(ωn ) = ωn · a for all n ≥ 0. Proof. The proof is by induction on n. For n = 0, the theorem asserts that there is M ⊂ [ω]ω MADF such that |M| = ω · a = a, and this is true by the definition of a (for n = 1, this is a result in [15]). We assume the theorem is true for all 0 ≤ i ≤ n and prove it for n + 1. To simplify the notation, let κ = ωn+1 . Thus we seek M ⊂ [κ]ω a MADF such that |M| = κ · a = ωn+1 · a. Let {λα : α < κ} be an order preserving list of the limit ordinals in κ (thus λ0 = ω) and put λκ = κ. We construct Aγ ⊂ [κ]ω (for γ ≤ κ) by recursion; so we assume for all α < γ, where γ ≤ κ we have constructed Aα ⊂ [λα ]ω MADF such that (i) |Aα | ≤ ωn · a, and (ii) β < α implies Aβ ⊂ Aα . We construct Aγ using three cases. Case 1: γ = α + 1. Then λγ = λα+1 = λα ∪ {λα + n : n ∈ ω}. Take B ⊂ {λα + n : n ∈ ω}ω MADF with |B| = a, and put Aγ = Aα ∪ B. Clearly Aγ is a MADF on λγ and (i) and (ii) are preserved at step γ. We now assume that γ is a limit ordinal, hence λγ is a limit of limit ordinals. Case 2: cf (γ) = ω. Then there exists a strictly increasing sequence of limit ordinals {λαi : i ∈ ω} with supremum λγ . Since |λγ | < κ then |λγ | ≤ ωn . By the induction hypothesis, there is N ⊂ [λγ ]ω MADF with |N | ≤ ωn · a. Define P0 = λα0 and for n ≥ 1 define Pn = λαn \ λαn−1 . Then P = {Pi : i ∈ ω} is a partition of λγ by disjoint sets.
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By Lemma 2.3, there exists A ⊂ [λγ ]ω an ADF such that A is maximal for diagonals of P, |A| ≤ ωn · a, and for every A ∈ A and every n ∈ ω, |A ∩ Pn | < ω. Put Aγ = {Aαi : i ∈ ω} ∪ A. Then by (i) |Aγ | ≤ Σi<ω |Aαi | + |A| ≤ Σi<ω (ωn · a) + ωn · a ≤ ωn · a Clearly Aγ is a MADF and (i) and (ii) are preserved at step γ. Case 3. cf (γ) > ω. Put Aγ = {Aα : α < γ}. By (ii) Aγ is an ADF on λγ , and is maximal since countable subsets of λγ have an upper bound below λγ . Further by (i), |Aγ | ≤ Σα<γ |Aγ | ≤ Σα<γ (ωn · a)) ≤ ωn · a · |γ| ≤ ωn · a. We have |Aγ | ≤ {
if γ < κ = ωn+1 ω n · a · ωn = ω n · a ωn · a · ωn+1 = ωn+1 · a if γ = κ
This completes the recursion for γ ≤ κ = ωn+1 . To complete the proof, we put M = Aκ . Since cf κ = cf ωn+1 > ω, M ⊂ [κ]ω is a MADF, and moreover |M| = |Aκ | ≤ ωn+1 · a. Since a MADF on ωn+1 = ∪{λα : α < κ} cannot have size less than ωn+1 · a, we have |M| = ωn+1 · a and this completes the proof. 2 3. Some κ ≥ ω2 for which κ is a rich cardinal Theorem 3.1. Let κ ≥ ω. If M ⊂ [κ]ω MADF and |M| = κ, then M is a rich MADF. Proof. For κ = ω, there is no countable MADF; so the result is true by default. Therefore we assume κ > ω. First we consider property (∗). For any A ⊂ [κ]ω we say that A satisfies property (∗) provided (∗) For every E ⊂ κ with |E| < κ, there exists M ∈ A such that M ∩ (κ \ E) is infinite. Clearly if A ⊂ [κ]ω MADF, then A satisfies (∗). We will use the following simple fact: Fact: If A satisfies (∗) and A = A0 ∪ A1 , then there exists i < 2 such that Ai satisfies (∗). If the Fact were not true for a family A = A0 ∪ A1 satisfying (∗), then for each i < 2 there exists a set Ei ⊂ κ such that |Ei | < κ and for all M ∈ Ai , M ∩ (κ \ Ei ) is finite. Put E = E0 ∪ E1 . Then |E| < κ. Since A satisfies (∗), there exist M ∈ A such that M ∩ (κ \ E) is infinite. There exists i < 2 such that M ∈ Ai , but this is a contradiction. Thus the Fact is proved. Let M ⊂ [κ]ω be a MADF with |M| = κ. We need to show that every continuous f : ψ(κ, M) → R has a full fiber. Let f be any such continuous function. Since ψ(κ, M) is pseudocompact, [3, §1], f ((ψ(κ, M)) is a compact subset of R hence contained in a finite closed interval, call it U0 = [a, b]. Since f −1 (U0 ) ∩ M = M, and all MADFs satisfy (∗), we know f −1 (U0 ) ∩ M satisfies (∗). Let c be the midpoint of U0 . Then at least one of f −1 ([a, c]) ∩ M or f −1 ([c, b]) ∩ M satisfies (∗) by the Fact. By recursion we construct a decreasing sequence of closed intervals Un such that Un+1 ⊂ Un , the length of Un+1 equals one-half the length of Un , and f −1 (Un ) ∩ M satisfies (∗). By compactness, there exists x ∈ R such that {x} =
n∈ω
{Un : n ∈ ω}
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To complete the proof of the theorem, we show that |f −1 (x) ∩ M| = κ. Towards this goal, let N ⊂ M be such that |N | < |M| = κ. It suffices to show that (f −1 (x) ∩ M) \ N = ∅. Let E = ∪N . Then |E| ≤ |N | · ω < κ, since the elements of N are countable sets and κ is uncountable. By recursion, for each i ≥ 0 pick Mi ∈ f −1 (Ui ) so that the Mi are distinct and Mi ∩ (κ \ E) is infinite (using property (∗)). Claim. The sequence of numbers {f (Mi ) : i ∈ ω} converges to x in R. Since {Ui : i ∈ ω} is a decreasing family of sets each containing x and the diameters of the Ui converge to zero, any sequence (yi ) such that yi ∈ Ui (i ∈ ω) converges to x. Next pick a sequence (αi ) of distinct ordinals αi ∈ Mi \ E such that |f (αi ) − f (Mi )| < 21i for i ∈ ω. Put D = {αi : i ∈ ω}. By maximality, there exists M ∈ M such that M ∩ D is infinite. Therefore M \ E is infinite; so M ∈ / N . It remains to show that M ∈ f −1 (x). By the Claim, the sequence {f (Mi ) : i ∈ ω} converges to x in R. Since |f (αi ) − f (Mi )| < 21i for i ∈ ω, also the sequence {f (αi ) : α ∈ ω} converges to x. We have a sequence (f (Mi )) that converges to f (M ) and a sequence f (D) that converges to x. Since M ∩ D is infinite, it follows that x = f (M ) or M ∈ f −1 (x). This completes the proof that (f −1 (x) ∩ M) \ N = ∅, and this proves that f −1 (x) is full. Since f was arbitrary, M is a rich MADF. 2 Corollary 3.2. If κ ≥ ω and κω = κ then κ is a rich cardinal. Proof. If κω = κ, then trivially for any M ⊂ [κ]ω , κ ≤ |M| ≤ κω , hence |M| = κ. So M is rich by Theorem 3.1, and as M was arbitrary, κ is a rich cardinal. 2 4. When are κ ≤ c and ωn rich cardinals? Lemma 4.1. Assume a < c and ω < κ < c. If there exists B ⊂ [κ]ω MADF such that |B| < c, then κ is not a rich cardinal. Proof. Let κ and B have the properties stated in the Lemma. To show that κ is not a rich cardinal, we need to construct M ⊂ [κ]ω MADF and a continuous function f : ψ(κ, M) → R such that f has no full fibers. We start with the construction in [3, §11] which gives us N ⊂ [ω]ω MADF with |N | = c, and a continuous function g : ψ(ω, N ) → [0, 1] such that for all r ∈ R, |g −1 (r)| = a. Thus g has no full fibers since a < c. By hypothesis, there exists B ⊂ [κ \ ω]ω MADF on κ \ ω with |B| < c. Then M = N ∪ B is a MADF on κ (of cardinality c since |N | = c). Define a function f : ψ(κ, M) → [0, 1] by f (y) =
g(y) if y ∈ ω ∪ N 0 if y ∈ ((κ \ ω) ∪ B)
The domain of f is ψ(κ, M) since ψ(κ, M) = (ω ∪ N ) ∪ ((κ \ ω) ∪ B). It is clear that for 0 < x ≤ 1, f −1 (x) = g −1 (x) hence |f −1 (x)| = |g −1 (x)| = a < c = |ψ(κ, M)|. For x = 0, f −1 (0) = g −1 (0) ∪ (κ \ ω) ∪ B. We have |f −1 (0)| ≤ |g −1 (0)| + κ + |B| = a + κ + |B| < c. Thus f has no full fibers. To see that f is continuous it suffices to note that (ω ∪ N ) is clopen in ψ(κ, M). This completes the proof. 2
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Lemma 4.2. If a = c then for every ω ≤ κ ≤ c, κ is a rich cardinal. Proof. We prove the contrapositive statement; so we prove that if there exists κ ≤ c which is not a rich cardinal, then a < c. Assuming κ ≤ c and κ is not rich, there exists M ⊂ [κ]ω MADF and a continuous f : ψ(κ, M) → R having no full fiber. By continuity, there exists r ∈ R such that a ≤ |f −1 (r)| (see the proof of [3, Theorem 11.1]). Since f has no full fibers, we have a ≤ |f −1 (r)| < |ψ| = |M| ≤ κω ≤ cω = c. Thus a < c, and that completes the proof. 2 Now we have the tools to determine when the ωn are rich cardinals: Theorem 4.3. (1) ω and ω1 are rich cardinals if and only if a = c. (2) For all 2 ≤ n < ω, if ωn < c, then ωn is rich if and only if a = c. (3) For all 2 ≤ n < ω, if c ≤ ωn , then ωn is a rich cardinal. Proof. By Theorem 1.2, ω is a rich cardinal if and only if a = c. We need to prove (1) for ω1 . First we assume a < c and prove that ω1 is not a rich cardinal. By Theorem 2.4 there exists M ⊂ [ω1 ]ω MADF with |M| = ω1 · a = a < c. Now we have a < c, κ = ω1 ≤ a < c, and there exists B = M ⊂ [ω1 ]ω MADF such that |B| < c. Thus ω1 is not a rich cardinal by Lemma 4.1. To prove the other half of (1) we assume a = c. Since ω1 ≤ c, we have ω1 is a rich cardinal by Lemma 4.2. Proof of (2). We assume that ωn < c. If a = c, then ωn is a rich cardinal by Lemma 4.2. Conversely we need to show that if ωn is a rich cardinal, then a = c. We prove the contrapositive statement: If a < c then ωn is not a rich cardinal. Since we have both a < c, and ωn < c, by Theorem 2.4, there exists B ⊂ [ωn ]ω MADF with |B| = ωn · a < c. By Lemma 4.1 ωn is not a rich cardinal. Proof of (3). Assume c ≤ ωn . For ωn = c, we have cω = c; so by Corollary 3.2, ωn = c is a rich cardinal. For ωn > c, we have cf ωn = ωn > c; so ωn is a rich cardinal by Remark 1.4. This completes the proof of the theorem. 2 In the proof of Theorem 4.3 we implicitly made use of the continuity of f because we called on Theorem 1.2, Corollary 3.2, and Lemma 4.2. Now we prove Theorem 1.3 as a corollary to Theorem 4.3 Corollary 4.4. The following are equivalent: (1) a = c (2) for every ω ≤ κ ≤ c, κ is a rich cardinal. (3) for every n < ω, ωn is a rich cardinal. Proof. We have (1) implies (2) by Lemma 4.2. To see that (2) implies (3), we note that by (2), if ωn ≤ c, then ωn is rich, and if c < ωn , then ωn is rich by Theorem 4.3(3). Thus ωn is rich for all n ∈ ω. Finally, (3) implies (1) because the assumption that ω = ω0 is rich implies a = c by Theorem 1.2. 2 Example 4.5. It is consistent that for every integer 2 ≤ N < ω there exists a model in which ωn is not a rich cardinal for all n < N , and ωn is a rich cardinal for all n ≥ N . Further, there is a model in which ωn is not rich for all n ∈ ω.
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Proof. Fix 2 ≤ N < ω. K. Kunen has a model [12, VIII, 2.3] in which a = ω1 and c = ωN . Since a < c, by Theorem 4.3(2), ωn is not a rich cardinal for any n < N , and ωn is a rich cardinal for n ≥ N , by Theorem 4.3(3). For the second model, take Kunen’s model where a = ω1 and c = ℵω+1 . 2 In conclusion, our results show that the answer to the question “When is ωn a rich cardinal?” depends first of all on whether or not a = c. If a = c, then ωn is rich for all n < ω. If a < c then whether a given ωn is rich depends on whether or not c ≤ ωn . 5. GCH implies all infinite cardinals are rich Our proof that all cardinals are rich works with hypothesis much weaker than the GCH. To state this hypothesis we recall a notation from set theory. The α-th successor, κ+α , of a cardinal κ. is defined as follows: If κ = ℵβ , then κ+α = ℵβ+α . For instance, κ+0 = κ, κ+1 = κ+ , κ+2 = κ++ , etc., κ+ω = sup{κ+n : n < ω} is the first singular cardinal greater than κ, and κ+ω1 = sup{κ+α : α < ω1 } is the first singular cardinal of cofinality ω1 greater that κ. We also use the following unpublished result of Peter Nyikos. Theorem 5.1. (Nyikos [14]) If κ ≥ c then for any M ⊂ [κ]ω MADF, |M| = κω . The above theorem of Nyikos leads to the following improvement in Remark 1.4. Corollary 5.2. If cf κω > c then κ is a rich cardinal. Proof. Let κ be a cardinal such that cf κω > c. Then κ > c. Let M ⊂ [κ]ω MADF. By Theorem 5.1, |M| = κω . Hence cf |M| = cf κω > c, thus by Remark 1.4, M is a rich MADF. Since M was arbitrary, κ is a rich cardinal. We state the main result of this section: Theorem 5.3. If κ ≥ c and κω < κ+ω1 , then κ is a rich cardinal. Proof. Let M ⊂ [κ]ω MADF. By the above result of Nyikos, |M| = κω . By our hypothesis, κω < κ+ω1 . Hence κω = κ+α for some countable ordinal α < ω1 , and since κω does not have countable cofinality, α is a successor ordinal, say α = β + 1, thus κω = κ+(β+1) = (κβ )+ is a successor cardinal, hence a regular cardinal greater than c; so cf κω = κω > c. By Remark 1.4, M is a rich MADF, and since M was arbitrary, κ is a rich cardinal. 2 Corollary 5.4. Assume a = c, and for all κ ≥ c, κω < κ+ω1 , then every κ ≥ ω is a rich cardinal. Proof. Since a = c, cardinals κ ≤ c are rich by Corollary 4.4. For κ ≥ c, κ is rich by Theorem 5.3.
2
Corollary 5.5. Assume GCH (or CH + SCH), then every κ ≥ ω is a rich cardinal. Proof. GCH (or SCH for κ ≥ c) imply κω ∈ {κ, κ+ } = {κ+0 , κ+1 }, hence κ < κ+ω1 .
2
Question 5.6. Does there exist a model having a cardinal κ > c such that κ is not a rich cardinal? Concerning the cardinal κ = ℵω , the first uncountable cardinal with countable cofinality, there are models (for each 0 ≤ α < ω1 ) of M. Gitik and M. Magidor (constructed using large cardinals) where the GCH holds below ℵω and κω = κ+(α+1) in the model [9, §3].
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Recall that a cardinal κ is called a strong limit cardinal provided κ is a limit cardinal and 2λ < 2κ for all λ < κ (e.g., see [6, 7.5]). Corollary 5.7. Every strong limit cardinal κ > ω is a rich cardinal. Proof. Let κ > ω be a strong limit cardinal. If cf κ ≥ ω1 , then [κ]ω = {[λ]ω : λ < κ} so by the strong limit hypothesis, κω = κ. Therefore κ is rich by Corollary 3.2. If cf κ = ω, then by strong limit, κω = 2κ (see [10, Lemma 6.5]), hence cf κω = cf 2κ > κ by Konig’s Theorem [10, page 45]. Since κ > ω is a strong limit cardinal, c = 2ω < κ < cf (κω ). Therefore κ is a rich cardinal by Corollary 5.2. 2 The previous corollary shows that in the models of Gitik and Magidor mentioned above, the cardinal ℵω is a rich cardinal. We thank Peter Nyikos for bringing his preprint [14] to our attention, and the referee for [8]. References [1] P. Alexandroff, P. Urysohn, Mémoire sur les espaces topologiques compacts, Verh. Akad. Wetensch. Amst. 14 (1926) 1–96. [2] J.E. Baumgartner, M. Weese, Partition algebras for almost-disjoint families, Trans. Am. Math. Soc. 274 (2) (1982) 619–630. [3] Alan Dow, Jerry E. Vaughan, Mrówka maximal almost disjoint families for uncountable cardinals, Topol. Appl. 157 (8) (2010) 1379–1394. [4] Alan Dow, Jerry E. Vaughan, Ordinal remainders of ψ-spaces, Topol. Appl. 158 (14) (2011) 1852–1857 (special issue in memory of Melvin Henriksen). [5] Eric van Douwen, The integers and topology, in: K. Kunen, J.E. Vaughan (Eds.), Handbook of Set-Theoretic Topology, North-Holland, Amsterdam, 1984, Chapter 3. [6] Frank R. Drake, Set Theory, North Holland Publishing Co., Amsterdam, 1974. [7] L. Gilman, M. Jerison, Rings of Continuous Functions, D. Van Nostrand Co., Inc., Princeton, 1960. [8] S. Fuchino, S. Geschke, L. Soukup, Almost disjoint families on large underlying sets (preprint), http://www.academia.edu/7361831/Almost_disjoint_families_on_large_underlying_sets. [9] M. Gitik, M. Magidor, The singular cardinal hypothesis revisited, in: H. Judah, et al. (Eds.), Set Theory and the Continuum, Springer-Verlag, Berlin, 1992, pp. 243–279. [10] Thomas Jech, Set Theory, Academic Press, New York, 1978. [11] P. Koszmider, On decompositions of Banach spaces of continuous functions on Mrowka’s spaces, Proc. Am. Math. Soc. 133 (7) (2005) 2137–2146. [12] Kenneth Kunen, Set Theory, North-Holland Pub. Co., Amsterdam, 1980. [13] S. Mrówka, Some set-theoretic constructions in topology, Fundam. Math. 94 (1977) 83–92. [14] Peter Nyikos, Generalized Kurepa and MAD families and topology (preprint), http://www.math.sc.edu~nyikos/ Kurepa.pdf. [15] Catherine Payne, On ψ(κ, M)-spaces with κ = ω1 , Master’s thesis UNC–Greensboro, 2010.
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Topology and its Applications www.elsevier.com/locate/topol
Fibers of continuous real-valued functions on ψ-spaces Jerry E. Vaughan ∗ , Catherine Payne University of North Carolina at Greensboro, United States
a r t i c l e
i n f o
Article history: Received 10 December 2013 Received in revised form 11 August 2014 Accepted 2 June 2015 MSC: primary 54A25, 03E17 secondary 03E35, 03E55, 03E75, 54C30 Keywords: Maximal almost disjoint families ψ-space Almost disjointness number a Real-valued continuous functions Cardinal cofinality Full fiber Rich MADF Rich cardinals
a b s t r a c t We consider continuous real-valued functions with domain either a ψ-space (studied by S. Mrowka, J. Isbell, and others) or a generalized ψ-space introduced by A. Dow and J. Vaughan. A cardinal κ ≥ ω is called a rich cardinal provided for every infinite, maximal almost disjoint family M of countably infinite subsets of κ (MADF) and for every continuous f : ψ(κ, M) → R defined on the associated space ψ = ψ(κ.M), there exists r ∈ R such that |f −1 (r)| = |ψ|. Dow and Vaughan proved that ω is a rich cardinal if and only if a = c, where a is the smallest cardinality of a MADF on ω. We prove that a = c if and only if, for all ω ≤ κ ≤ c, κ is a rich cardinal, if and only if for every n < ω, ωn is a rich cardinal. We prove every κ > c is rich using a set-theoretic hypothesis weaker than GCH. © 2015 Published by Elsevier B.V.
1. Introduction The classical ψ-space is a well-known topological space, first considered by Alexandroff and Urysohn [1], and later by others (e.g., S. Mrówka [13], and J. Isbell [7, 5I]). The underlying set of points consists of the union of two sets. One set is a countable set X of isolated points (we take X = ω, where ω denotes the set of natural numbers). The other set is a closed discrete set in one-to-one correspondence with an infinite almost disjoint family A on ω. This connection between the topology of ψ-spaces and almost disjoint families (especially maximal almost disjoint families) makes the study of ψ-spaces of interest not only in topology and set-theory but also in Boolean algebra and Banach spaces (cf., [2,11]). A. Dow and J. Vaughan [3,4], extended the study of classical ψ-spaces to the case where the set X of isolated points is of arbitrary cardinality (we take the set X to be an arbitrary cardinal κ ≥ ω), and the * Corresponding author. E-mail addresses: [email protected] (J.E. Vaughan), [email protected] (C. Payne). http://dx.doi.org/10.1016/j.topol.2015.09.032 0166-8641/© 2015 Published by Elsevier B.V.
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closed discrete set is in one-to-one correspondence with an infinite almost disjoint family A of countably infinite subsets of κ. In this paper we consider real-valued continuous functions defined on such ψ-spaces, and ask about the cardinalities of their fibers. We discuss our results following a few definitions. We use standard set-theoretic notation (e.g., see [12]). For an infinite cardinal κ, let [κ]ω denote the set of all countably infinite subsets of κ. A family A ⊂ [κ]ω is called an almost disjoint family (ADF) provided the intersection of any two members of A is finite. A maximal almost disjoint family (MADF) M ⊂ [κ]ω is an infinite ADF that is not properly contained in any other ADF of countably infinite subsets of κ (we often write “M ⊂ [κ]ω MADF”). The topological space ψ(κ, M) is defined as follows. Let M be an ADF (or MADF) on an infinite cardinal κ. The underlying set of the space ψ(κ, M) is κ ∪ M and the open sets are defined by taking the points of κ to be isolated and basic neighborhoods of a point M ∈ M to have the form {M } ∪ (M \ F ), where F is any finite subset of M . When κ and M are understood we write ψ for ψ(κ, M). Given an infinite cardinal κ, κ+ denotes the first cardinal larger than κ, and κω denotes the cardinality of [κ]ω . We follow the usual convention and write ωn instead of ℵn for n ∈ ω and we write ω instead of ω0 . The cardinal ω1 , defined by ω1 = ω + , denotes the first uncountable cardinal. Similarly ωn = (ωn−1 )+ for n ≥ 2, and ℵω = sup{ωn : n ∈ ω}. The cardinal number of the real line (the cardinality of the continuum) is denoted by c, and the cardinal a (called the almost-disjointness number) denotes the smallest cardinality of a MADF on ω. It is well known that ω1 ≤ a ≤ c (e.g., see [5]). By CH we mean the Continuum Hypothesis (2ω = ω1 ) and by GCH we mean the Generalized Continuum Hypothesis (i.e., 2κ = κ+ for all κ ≥ ω). We introduce the following terminology. Definition 1.1. Let f : X → Y . We say that a fiber f −1 (y), where y ∈ Y , is of full size (or is full) provided |f −1 (y)| = |X|. We say that a MADF M ⊂ [κ]ω is a rich MADF provided every real-valued continuous function defined on ψ(κ, M) has at least one full fiber. An infinite cardinal κ is called a rich cardinal provided every M ⊂ [κ]ω MADF is a rich MADF. The notion of rich MADF evolved naturally from special properties of MADF’s constructed in [13] and [3]. In [13], Mrówka constructed a M ⊂ [ω]ω MADF such that |βψ(ω, M) \ ψ(ω, M)| = 1, or equivalently, for every continuous f : ψ → R, there exists r ∈ R such that |ψ \ f −1 (r)| ≤ ω. Thus Mrówka’s special M is a rich MADF. In [3], Dow and Vaughan proved that (assuming GCH) for all κ > ω, there exists Mκ ⊂ [κ]ω MADF such that for every continuous f : ψ(κ, Mκ ) → R, there exists r ∈ R such that |ψ \ f −1 (r)| < |ψ|. These Mκ are rich MADF’s. The results in this paper were motivated by the following theorem (stated here using the above terminology). Theorem 1.2. (Dow–Vaughan [3]) The cardinal ω is a rich cardinal if and only if a = c. Several years ago, we proved that the cardinal ω1 is rich iff a = c (half of this result is in [15]). This is an exact analog of Theorem 1.2. These two results obviously raise the question: When are the cardinals ωn (n ≥ 2) rich? In general, when is a cardinal κ rich? Towards answering these questions, the following result gives two general conditions equivalent to “a = c.” Theorem 1.3 (Corollary 4.4). The following are equivalent: (1) a = c (2) for every ω ≤ κ ≤ c, κ is a rich cardinal. (3) for every n < ω, ωn is a rich cardinal. We prove, under hypothesis weaker than GCH, that every κ ≥ ω is a rich cardinal (see Section 5). Indeed our hypothesis is much weaker than GCH or “CH + SCH” where SCH stands for the singular
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cardinal hypothesis (e.g., see [10, §8]). Moreover we prove in ZFC that if κ = κω , then κ is a rich cardinal (Corollary 3.2). In the absence of GCH, there are cardinals that are not rich. For instance, it is consistent that a < c (hence ω and ω1 are not rich cardinals), and ω2 is a rich cardinal; see Section 4 and Example 4.5. Concerning rich MADFs, we begin with a simple remark. Remark 1.4. If M ⊂ [κ]ω MADF with cf (|M|) > c then M is a rich MADF. Proof. Let M ⊂ [κ]ω MADF with cf (|M|) > c. We need only note that in general if f : X → R is any function (continuous or not) with cf |X| > c, then for some x ∈ R, |f −1 (x)| = |X|. For otherwise, we have |f −1 (r)| < |X| for all r ∈ R. Then since X = {f −1 (r) : r ∈ R}, we have |X| ≤
|f −1 (r)| < |X|
(1)
r∈R
where the strict inequality holds because we are adding fewer than cf |X| sets, each of cardinality less than |X|. Of course, (1) is a contradiction. 2 This remark shows, for example, that c+ is a rich cardinal because (c+ )ω = c+ (see [10, p. 48]). The remark also shows that for cardinals κ > c, the major difficulty regarding whether κ is a rich cardinal is the case cf κω ≤ c (see Section 5, and Question 5.6). To prove our results about the cardinals ωn we need to know the smallest cardinality of a MADF on ωn . We consider this in the next section. 2. The smallest cardinality of a MADF on ωn The main result in this section (Theorem 2.4) was given previously, and in more general settings, in unpublished preprints (see [14], [8]). We include our direct proof here for completeness and for the convenience of the reader. Definition 2.1. ([14],[8]) Let κ be an infinite cardinal. For κ ≥ ω a(κ) = min |M| : M ⊂ [κ]ω is a MADF A general problem is to find a(κ) for all κ (see [14]). For the purposes of this paper, we find a(ωn ) for n ∈ ω. To do this we use the following definitions. Definition 2.2. Let P = {Pn : n ∈ ω} be a family of pairwise disjoint infinite subsets of a set X. We say that H ∈ [X]ω is a diagonal of P provided |{n ∈ ω : H ∩ Pn = ∅}| = ω. We say that a family A ⊂ [X]ω is maximal for diagonal subsets of P provided for every diagonal H of P there exists A ∈ A such that |A ∩ H| = ω. Lemma 2.3. If κ ≥ ω and P = {Pn : n ∈ ω} is a pairwise disjoint family of infinite subsets of κ, then there exists A ⊂ [κ]ω ADF that is maximal for diagonal sets of P and such that for every A ∈ A, A ∩ Pn is finite for all n ∈ ω, and |A| ≤ a(κ). Proof. Let P be as stated in the lemma, and let M ⊂ [κ]ω be a MADF with |M| = a(κ). We partition M into two sets. First, let B = {M ∈ M : |{n ∈ ω : |M ∩ Pn | = ω}| < ω}. For all M ∈ B, let kM ∈ ω such that for all n ≥ kM , (|M ∩ Pn | < ω). Put M = M \ {Pi : i ≤ kM }. Then M ∩ Pn is finite for all n ∈ ω (indeed, M = ∅ is possible). Second, let C = {M ∈ M : |{n ∈ ω : |M ∩ Pn | = ω}| = ω}. For each M ∈ C,
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put EM = {M ∩ Pn : n ∈ ω and |M ∩ Pn | = ω}. Then EM is a countable, infinite pairwise disjoint family of infinite subsets of M . Since M is countable, by [3, Lemma 11.2] there exists AM ⊂ [M ]ω MADF such that EM ⊂ AM , and |AM | = a. Let AM = AM \ EM . Then for all A ∈ AM and all n ∈ ω we have |A ∩ Pn | < ω. Also for M ∈ B and n ∈ ω, M ∩ Pn is finite, as noted above. Define A = M : M ∈ B and M is infinite ∪ AM : M ∈ C . We have already noted that for every A ∈ A and every n ∈ ω, we have A ∩ Pn is finite. We show that A satisfies the other conclusions of the lemma. Since M is ADF and B and C are disjoint, it is easy to see that A ⊂ [κ]ω is an ADF. Moreover, it is clear that |A| ≤ |B| + |C| · a ≤ |M| · a = a(κ). Finally, we show that A is maximal for the diagonal sets of P. Let H ∈ [κ]ω be a diagonal of P. Let J ⊂ H be such that J is infinite and |J ∩ Pn | ≤ 1 for all n ∈ ω. It suffices to find A ∈ A such that |A ∩ J| = ω. Since M is maximal there exists M ∈ M such that |M ∩ J| = ω. If M ∈ B, then |M ∩ J| = ω since |J ∩ ( i≤kM Pi )| < ω, and therefore M ∈ A, and we are done. Otherwise M ∈ C. Since AM ⊂ [M ]ω is maximal, there exists A ∈ AM such that |A ∩ (M ∩ J)| = ω, and since (A ∩ M ∩ J) intersects more than one Pn , we have A = M ∩ Pn for any n ∈ ω. Thus A ∈ / EM ; so A ∈ AM ⊂ A, and this completes the proof. 2 We now determine the minimum cardinality of a MADF on ωn for n ∈ ω. Of course, a(ω) = a by definition. Theorem 2.4. Let 0 ≤ n < ω. There exists M ⊂ [ωn ]ω a MADF such that |M| = ωn · a. Thus a(ωn ) = ωn · a for all n ≥ 0. Proof. The proof is by induction on n. For n = 0, the theorem asserts that there is M ⊂ [ω]ω MADF such that |M| = ω · a = a, and this is true by the definition of a (for n = 1, this is a result in [15]). We assume the theorem is true for all 0 ≤ i ≤ n and prove it for n + 1. To simplify the notation, let κ = ωn+1 . Thus we seek M ⊂ [κ]ω a MADF such that |M| = κ · a = ωn+1 · a. Let {λα : α < κ} be an order preserving list of the limit ordinals in κ (thus λ0 = ω) and put λκ = κ. We construct Aγ ⊂ [κ]ω (for γ ≤ κ) by recursion; so we assume for all α < γ, where γ ≤ κ we have constructed Aα ⊂ [λα ]ω MADF such that (i) |Aα | ≤ ωn · a, and (ii) β < α implies Aβ ⊂ Aα . We construct Aγ using three cases. Case 1: γ = α + 1. Then λγ = λα+1 = λα ∪ {λα + n : n ∈ ω}. Take B ⊂ {λα + n : n ∈ ω}ω MADF with |B| = a, and put Aγ = Aα ∪ B. Clearly Aγ is a MADF on λγ and (i) and (ii) are preserved at step γ. We now assume that γ is a limit ordinal, hence λγ is a limit of limit ordinals. Case 2: cf (γ) = ω. Then there exists a strictly increasing sequence of limit ordinals {λαi : i ∈ ω} with supremum λγ . Since |λγ | < κ then |λγ | ≤ ωn . By the induction hypothesis, there is N ⊂ [λγ ]ω MADF with |N | ≤ ωn · a. Define P0 = λα0 and for n ≥ 1 define Pn = λαn \ λαn−1 . Then P = {Pi : i ∈ ω} is a partition of λγ by disjoint sets.
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By Lemma 2.3, there exists A ⊂ [λγ ]ω an ADF such that A is maximal for diagonals of P, |A| ≤ ωn · a, and for every A ∈ A and every n ∈ ω, |A ∩ Pn | < ω. Put Aγ = {Aαi : i ∈ ω} ∪ A. Then by (i) |Aγ | ≤ Σi<ω |Aαi | + |A| ≤ Σi<ω (ωn · a) + ωn · a ≤ ωn · a Clearly Aγ is a MADF and (i) and (ii) are preserved at step γ. Case 3. cf (γ) > ω. Put Aγ = {Aα : α < γ}. By (ii) Aγ is an ADF on λγ , and is maximal since countable subsets of λγ have an upper bound below λγ . Further by (i), |Aγ | ≤ Σα<γ |Aγ | ≤ Σα<γ (ωn · a)) ≤ ωn · a · |γ| ≤ ωn · a. We have |Aγ | ≤ {
if γ < κ = ωn+1 ω n · a · ωn = ω n · a ωn · a · ωn+1 = ωn+1 · a if γ = κ
This completes the recursion for γ ≤ κ = ωn+1 . To complete the proof, we put M = Aκ . Since cf κ = cf ωn+1 > ω, M ⊂ [κ]ω is a MADF, and moreover |M| = |Aκ | ≤ ωn+1 · a. Since a MADF on ωn+1 = ∪{λα : α < κ} cannot have size less than ωn+1 · a, we have |M| = ωn+1 · a and this completes the proof. 2 3. Some κ ≥ ω2 for which κ is a rich cardinal Theorem 3.1. Let κ ≥ ω. If M ⊂ [κ]ω MADF and |M| = κ, then M is a rich MADF. Proof. For κ = ω, there is no countable MADF; so the result is true by default. Therefore we assume κ > ω. First we consider property (∗). For any A ⊂ [κ]ω we say that A satisfies property (∗) provided (∗) For every E ⊂ κ with |E| < κ, there exists M ∈ A such that M ∩ (κ \ E) is infinite. Clearly if A ⊂ [κ]ω MADF, then A satisfies (∗). We will use the following simple fact: Fact: If A satisfies (∗) and A = A0 ∪ A1 , then there exists i < 2 such that Ai satisfies (∗). If the Fact were not true for a family A = A0 ∪ A1 satisfying (∗), then for each i < 2 there exists a set Ei ⊂ κ such that |Ei | < κ and for all M ∈ Ai , M ∩ (κ \ Ei ) is finite. Put E = E0 ∪ E1 . Then |E| < κ. Since A satisfies (∗), there exist M ∈ A such that M ∩ (κ \ E) is infinite. There exists i < 2 such that M ∈ Ai , but this is a contradiction. Thus the Fact is proved. Let M ⊂ [κ]ω be a MADF with |M| = κ. We need to show that every continuous f : ψ(κ, M) → R has a full fiber. Let f be any such continuous function. Since ψ(κ, M) is pseudocompact, [3, §1], f ((ψ(κ, M)) is a compact subset of R hence contained in a finite closed interval, call it U0 = [a, b]. Since f −1 (U0 ) ∩ M = M, and all MADFs satisfy (∗), we know f −1 (U0 ) ∩ M satisfies (∗). Let c be the midpoint of U0 . Then at least one of f −1 ([a, c]) ∩ M or f −1 ([c, b]) ∩ M satisfies (∗) by the Fact. By recursion we construct a decreasing sequence of closed intervals Un such that Un+1 ⊂ Un , the length of Un+1 equals one-half the length of Un , and f −1 (Un ) ∩ M satisfies (∗). By compactness, there exists x ∈ R such that {x} =
n∈ω
{Un : n ∈ ω}
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To complete the proof of the theorem, we show that |f −1 (x) ∩ M| = κ. Towards this goal, let N ⊂ M be such that |N | < |M| = κ. It suffices to show that (f −1 (x) ∩ M) \ N = ∅. Let E = ∪N . Then |E| ≤ |N | · ω < κ, since the elements of N are countable sets and κ is uncountable. By recursion, for each i ≥ 0 pick Mi ∈ f −1 (Ui ) so that the Mi are distinct and Mi ∩ (κ \ E) is infinite (using property (∗)). Claim. The sequence of numbers {f (Mi ) : i ∈ ω} converges to x in R. Since {Ui : i ∈ ω} is a decreasing family of sets each containing x and the diameters of the Ui converge to zero, any sequence (yi ) such that yi ∈ Ui (i ∈ ω) converges to x. Next pick a sequence (αi ) of distinct ordinals αi ∈ Mi \ E such that |f (αi ) − f (Mi )| < 21i for i ∈ ω. Put D = {αi : i ∈ ω}. By maximality, there exists M ∈ M such that M ∩ D is infinite. Therefore M \ E is infinite; so M ∈ / N . It remains to show that M ∈ f −1 (x). By the Claim, the sequence {f (Mi ) : i ∈ ω} converges to x in R. Since |f (αi ) − f (Mi )| < 21i for i ∈ ω, also the sequence {f (αi ) : α ∈ ω} converges to x. We have a sequence (f (Mi )) that converges to f (M ) and a sequence f (D) that converges to x. Since M ∩ D is infinite, it follows that x = f (M ) or M ∈ f −1 (x). This completes the proof that (f −1 (x) ∩ M) \ N = ∅, and this proves that f −1 (x) is full. Since f was arbitrary, M is a rich MADF. 2 Corollary 3.2. If κ ≥ ω and κω = κ then κ is a rich cardinal. Proof. If κω = κ, then trivially for any M ⊂ [κ]ω , κ ≤ |M| ≤ κω , hence |M| = κ. So M is rich by Theorem 3.1, and as M was arbitrary, κ is a rich cardinal. 2 4. When are κ ≤ c and ωn rich cardinals? Lemma 4.1. Assume a < c and ω < κ < c. If there exists B ⊂ [κ]ω MADF such that |B| < c, then κ is not a rich cardinal. Proof. Let κ and B have the properties stated in the Lemma. To show that κ is not a rich cardinal, we need to construct M ⊂ [κ]ω MADF and a continuous function f : ψ(κ, M) → R such that f has no full fibers. We start with the construction in [3, §11] which gives us N ⊂ [ω]ω MADF with |N | = c, and a continuous function g : ψ(ω, N ) → [0, 1] such that for all r ∈ R, |g −1 (r)| = a. Thus g has no full fibers since a < c. By hypothesis, there exists B ⊂ [κ \ ω]ω MADF on κ \ ω with |B| < c. Then M = N ∪ B is a MADF on κ (of cardinality c since |N | = c). Define a function f : ψ(κ, M) → [0, 1] by f (y) =
g(y) if y ∈ ω ∪ N 0 if y ∈ ((κ \ ω) ∪ B)
The domain of f is ψ(κ, M) since ψ(κ, M) = (ω ∪ N ) ∪ ((κ \ ω) ∪ B). It is clear that for 0 < x ≤ 1, f −1 (x) = g −1 (x) hence |f −1 (x)| = |g −1 (x)| = a < c = |ψ(κ, M)|. For x = 0, f −1 (0) = g −1 (0) ∪ (κ \ ω) ∪ B. We have |f −1 (0)| ≤ |g −1 (0)| + κ + |B| = a + κ + |B| < c. Thus f has no full fibers. To see that f is continuous it suffices to note that (ω ∪ N ) is clopen in ψ(κ, M). This completes the proof. 2
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Lemma 4.2. If a = c then for every ω ≤ κ ≤ c, κ is a rich cardinal. Proof. We prove the contrapositive statement; so we prove that if there exists κ ≤ c which is not a rich cardinal, then a < c. Assuming κ ≤ c and κ is not rich, there exists M ⊂ [κ]ω MADF and a continuous f : ψ(κ, M) → R having no full fiber. By continuity, there exists r ∈ R such that a ≤ |f −1 (r)| (see the proof of [3, Theorem 11.1]). Since f has no full fibers, we have a ≤ |f −1 (r)| < |ψ| = |M| ≤ κω ≤ cω = c. Thus a < c, and that completes the proof. 2 Now we have the tools to determine when the ωn are rich cardinals: Theorem 4.3. (1) ω and ω1 are rich cardinals if and only if a = c. (2) For all 2 ≤ n < ω, if ωn < c, then ωn is rich if and only if a = c. (3) For all 2 ≤ n < ω, if c ≤ ωn , then ωn is a rich cardinal. Proof. By Theorem 1.2, ω is a rich cardinal if and only if a = c. We need to prove (1) for ω1 . First we assume a < c and prove that ω1 is not a rich cardinal. By Theorem 2.4 there exists M ⊂ [ω1 ]ω MADF with |M| = ω1 · a = a < c. Now we have a < c, κ = ω1 ≤ a < c, and there exists B = M ⊂ [ω1 ]ω MADF such that |B| < c. Thus ω1 is not a rich cardinal by Lemma 4.1. To prove the other half of (1) we assume a = c. Since ω1 ≤ c, we have ω1 is a rich cardinal by Lemma 4.2. Proof of (2). We assume that ωn < c. If a = c, then ωn is a rich cardinal by Lemma 4.2. Conversely we need to show that if ωn is a rich cardinal, then a = c. We prove the contrapositive statement: If a < c then ωn is not a rich cardinal. Since we have both a < c, and ωn < c, by Theorem 2.4, there exists B ⊂ [ωn ]ω MADF with |B| = ωn · a < c. By Lemma 4.1 ωn is not a rich cardinal. Proof of (3). Assume c ≤ ωn . For ωn = c, we have cω = c; so by Corollary 3.2, ωn = c is a rich cardinal. For ωn > c, we have cf ωn = ωn > c; so ωn is a rich cardinal by Remark 1.4. This completes the proof of the theorem. 2 In the proof of Theorem 4.3 we implicitly made use of the continuity of f because we called on Theorem 1.2, Corollary 3.2, and Lemma 4.2. Now we prove Theorem 1.3 as a corollary to Theorem 4.3 Corollary 4.4. The following are equivalent: (1) a = c (2) for every ω ≤ κ ≤ c, κ is a rich cardinal. (3) for every n < ω, ωn is a rich cardinal. Proof. We have (1) implies (2) by Lemma 4.2. To see that (2) implies (3), we note that by (2), if ωn ≤ c, then ωn is rich, and if c < ωn , then ωn is rich by Theorem 4.3(3). Thus ωn is rich for all n ∈ ω. Finally, (3) implies (1) because the assumption that ω = ω0 is rich implies a = c by Theorem 1.2. 2 Example 4.5. It is consistent that for every integer 2 ≤ N < ω there exists a model in which ωn is not a rich cardinal for all n < N , and ωn is a rich cardinal for all n ≥ N . Further, there is a model in which ωn is not rich for all n ∈ ω.
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Proof. Fix 2 ≤ N < ω. K. Kunen has a model [12, VIII, 2.3] in which a = ω1 and c = ωN . Since a < c, by Theorem 4.3(2), ωn is not a rich cardinal for any n < N , and ωn is a rich cardinal for n ≥ N , by Theorem 4.3(3). For the second model, take Kunen’s model where a = ω1 and c = ℵω+1 . 2 In conclusion, our results show that the answer to the question “When is ωn a rich cardinal?” depends first of all on whether or not a = c. If a = c, then ωn is rich for all n < ω. If a < c then whether a given ωn is rich depends on whether or not c ≤ ωn . 5. GCH implies all infinite cardinals are rich Our proof that all cardinals are rich works with hypothesis much weaker than the GCH. To state this hypothesis we recall a notation from set theory. The α-th successor, κ+α , of a cardinal κ. is defined as follows: If κ = ℵβ , then κ+α = ℵβ+α . For instance, κ+0 = κ, κ+1 = κ+ , κ+2 = κ++ , etc., κ+ω = sup{κ+n : n < ω} is the first singular cardinal greater than κ, and κ+ω1 = sup{κ+α : α < ω1 } is the first singular cardinal of cofinality ω1 greater that κ. We also use the following unpublished result of Peter Nyikos. Theorem 5.1. (Nyikos [14]) If κ ≥ c then for any M ⊂ [κ]ω MADF, |M| = κω . The above theorem of Nyikos leads to the following improvement in Remark 1.4. Corollary 5.2. If cf κω > c then κ is a rich cardinal. Proof. Let κ be a cardinal such that cf κω > c. Then κ > c. Let M ⊂ [κ]ω MADF. By Theorem 5.1, |M| = κω . Hence cf |M| = cf κω > c, thus by Remark 1.4, M is a rich MADF. Since M was arbitrary, κ is a rich cardinal. We state the main result of this section: Theorem 5.3. If κ ≥ c and κω < κ+ω1 , then κ is a rich cardinal. Proof. Let M ⊂ [κ]ω MADF. By the above result of Nyikos, |M| = κω . By our hypothesis, κω < κ+ω1 . Hence κω = κ+α for some countable ordinal α < ω1 , and since κω does not have countable cofinality, α is a successor ordinal, say α = β + 1, thus κω = κ+(β+1) = (κβ )+ is a successor cardinal, hence a regular cardinal greater than c; so cf κω = κω > c. By Remark 1.4, M is a rich MADF, and since M was arbitrary, κ is a rich cardinal. 2 Corollary 5.4. Assume a = c, and for all κ ≥ c, κω < κ+ω1 , then every κ ≥ ω is a rich cardinal. Proof. Since a = c, cardinals κ ≤ c are rich by Corollary 4.4. For κ ≥ c, κ is rich by Theorem 5.3.
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Corollary 5.5. Assume GCH (or CH + SCH), then every κ ≥ ω is a rich cardinal. Proof. GCH (or SCH for κ ≥ c) imply κω ∈ {κ, κ+ } = {κ+0 , κ+1 }, hence κ < κ+ω1 .
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Question 5.6. Does there exist a model having a cardinal κ > c such that κ is not a rich cardinal? Concerning the cardinal κ = ℵω , the first uncountable cardinal with countable cofinality, there are models (for each 0 ≤ α < ω1 ) of M. Gitik and M. Magidor (constructed using large cardinals) where the GCH holds below ℵω and κω = κ+(α+1) in the model [9, §3].
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Recall that a cardinal κ is called a strong limit cardinal provided κ is a limit cardinal and 2λ < 2κ for all λ < κ (e.g., see [6, 7.5]). Corollary 5.7. Every strong limit cardinal κ > ω is a rich cardinal. Proof. Let κ > ω be a strong limit cardinal. If cf κ ≥ ω1 , then [κ]ω = {[λ]ω : λ < κ} so by the strong limit hypothesis, κω = κ. Therefore κ is rich by Corollary 3.2. If cf κ = ω, then by strong limit, κω = 2κ (see [10, Lemma 6.5]), hence cf κω = cf 2κ > κ by Konig’s Theorem [10, page 45]. Since κ > ω is a strong limit cardinal, c = 2ω < κ < cf (κω ). Therefore κ is a rich cardinal by Corollary 5.2. 2 The previous corollary shows that in the models of Gitik and Magidor mentioned above, the cardinal ℵω is a rich cardinal. We thank Peter Nyikos for bringing his preprint [14] to our attention, and the referee for [8]. References [1] P. Alexandroff, P. Urysohn, Mémoire sur les espaces topologiques compacts, Verh. Akad. Wetensch. Amst. 14 (1926) 1–96. [2] J.E. Baumgartner, M. Weese, Partition algebras for almost-disjoint families, Trans. Am. Math. Soc. 274 (2) (1982) 619–630. [3] Alan Dow, Jerry E. Vaughan, Mrówka maximal almost disjoint families for uncountable cardinals, Topol. Appl. 157 (8) (2010) 1379–1394. [4] Alan Dow, Jerry E. Vaughan, Ordinal remainders of ψ-spaces, Topol. Appl. 158 (14) (2011) 1852–1857 (special issue in memory of Melvin Henriksen). [5] Eric van Douwen, The integers and topology, in: K. Kunen, J.E. Vaughan (Eds.), Handbook of Set-Theoretic Topology, North-Holland, Amsterdam, 1984, Chapter 3. [6] Frank R. Drake, Set Theory, North Holland Publishing Co., Amsterdam, 1974. [7] L. Gilman, M. Jerison, Rings of Continuous Functions, D. Van Nostrand Co., Inc., Princeton, 1960. [8] S. Fuchino, S. Geschke, L. Soukup, Almost disjoint families on large underlying sets (preprint), http://www.academia.edu/7361831/Almost_disjoint_families_on_large_underlying_sets. [9] M. Gitik, M. Magidor, The singular cardinal hypothesis revisited, in: H. Judah, et al. (Eds.), Set Theory and the Continuum, Springer-Verlag, Berlin, 1992, pp. 243–279. [10] Thomas Jech, Set Theory, Academic Press, New York, 1978. [11] P. Koszmider, On decompositions of Banach spaces of continuous functions on Mrowka’s spaces, Proc. Am. Math. Soc. 133 (7) (2005) 2137–2146. [12] Kenneth Kunen, Set Theory, North-Holland Pub. Co., Amsterdam, 1980. [13] S. Mrówka, Some set-theoretic constructions in topology, Fundam. Math. 94 (1977) 83–92. [14] Peter Nyikos, Generalized Kurepa and MAD families and topology (preprint), http://www.math.sc.edu~nyikos/ Kurepa.pdf. [15] Catherine Payne, On ψ(κ, M)-spaces with κ = ω1 , Master’s thesis UNC–Greensboro, 2010.