NON-NORMALITY POINTS OF βx \ X

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1 NON-NORMALITY POINTS OF βx \ X LYNNE YENGULALP 1. INTRODUCTION A point p in a space X is called a non-normality point if X \ {p} is not normal and is called a butterfly point if there are closed subsets H, K of X such that H K = {p} and p cl(h \ {p}) cl(k \ {p}). For an infinite cardinal κ, we denote the set of uniform ultrafilters on the discrete space of size κ by U(κ) and the set of non-uniform ultrafilters by NU(κ). Under CH, any point p βω \ ω is a non-normality point of βω \ ω. The case where p is a p-point was done by Gillman [2] and when p is not a p-point independently by Rajagopalan [4] and Warren [8]. For a cardinal κ ω, Beslagic and van Douwen [1] showed that it is consistent that each p U(κ) be a non-normality point of U(κ). They achieved this by embedding the non-normal space NU(cf(2 κ )) into U(κ) \ {p} as a closed subspace. For non-discrete spaces, X, one may ask: when are all points y βx \ X non-normality points of βx or βx \ X? If X is metrizable, non-compact and has no isolated points, then every point y in βx \ X is a non-normality point of βx (Terasawa [7], Logunov [5]). They each showed that any y βx \X is a butterfly point of βx. Since βx \{y} is C -embedded in βx, it is easily seen that if y is a butterfly point of βx, it is also a non-normalilty point. In this paper we show that it is consistent that if X is locally compact, metrizable and has no isolated points then every point y in βx \X is a non-normality point of βx \X. Since βx \ (X {y}) is not necessarily C -embedded in βx, it will not suffice that p be a butterfly point. We instead embed NU(θ) as a closed subset of βx \ (X {y}) for some regular cardinal θ. Key words and phrases. non-normality point, butterfly point. 1

2 2. NON-NORMALITY POINTS OF βx \ X Lemma 2.1. [6] If θ is regular and not a strong limit cardinal, then N U(θ) is not normal. Notation: For a collection U of subsets of X we write U = U. We say a collection of subsets, V, densely refines a collection U if for all V V there is U U such that V U and cl X (V ) = cl X (U ). The following lemma defines a π -base for X similar to the one used in [7]. In fact, we modify much of the structure from [7] that was used to create butterfly sets and use it instead to embed NU(θ) into βx \ (X {y}). Lemma 2.2. Let X be a locally compact metric space without isolated points. There exists a collection B = n ω B n of open subsets of X with the following properties. (1) B n is pairwise disjoint, locally finite and cl(b n) = X. (2) B n+1 refines B n and {B B n+1 : B B} = 4 for all B B n. (3) For B B n there are B 0, B 1 B n+1 such that clb 0 clb 1 = and clb 0, clb 1 B. (4) If U = {U, V } is an open cover of X, there is a pairwise disjoint locally finite collection V B densely refining U. (5) For each n ω, B n = w(x). Proof. Let O be an open cover of X consisting of sets U such that clu is compact. Let B 0 be a locally finite open refinement of size w(x). In fact, it must be that B 0 = w(x). Otherwise, since clb is compact metric for each B B 0, there is a countable collection of open subsets of X that is a base for points in clb. Since B 0 covers, if B 0 < w(x) the union of each of these bases would be a basis for X of size < w(x), a contradiction. Let κ = w(x). Well order B 0 as {B α : α κ}. Define B α = B α \ γ<α clb γ and set B 0 = {B α : α κ}. Notice that since B 0 is locally finite, B 0 is locally finite as well and each B α is open. Furthermore, since clb α clb α, clb α is compact. Fix α κ. Since clb α is compact and metric, there is a countable base for clb α. Let A α = {A i clb α : i ω} be such a base such that A 0 = clb α and A i is open with respect to clb α. Notice that int(a i ) for all i ω. Let Wα 0 = {B α }. Assume we have defined for each i n a collection Wα i of open (w.r.t. X ) subsets of B α such that:

3 i) Wα i is a pairwise disjoint finite collection such that cl( Wα) n = clb α. ii) Wα i+1 refines Wα i and {B W i+1 : B B} = 4 for all B Wα i. iii) For B Wα i there are B 0, B 1 Wα n+1 such that clb 0 clb 1 = and clb 0, clb 1 B. iv) For each B Wα i, either B A i or B B α \ cla i. Fix W Wα n. Case 1. W A n+1 = or W \ A n+1 =. Let B 0 and B 1 be any two non-empty open subsets of W such that clb 0 clb 1 = and clb 0 clb 1 W. Since X has no isolated points, this can be done. Then let B 2 and B 3 be non-empty open subsets of W such that B 2 B 3 is dense in W \ (clb 0 clb 1 ). Case 2. W A n+1 and W \ A n+1. Let B 0 be a non-empty open subset of W such that clb 0 W A n+1 and let B 2 = (W int(a n+1 ))\clb 0. Then let B 1 be a non-empty open subset of W such that clb 1 W \ A n+1 and let B 3 = W \ (cla n+1 clb 1 ). Again, since X has no isolated points, this can be done. Set Wα n+1 = {B i : i = 0, 1, 2, 3}. By construction, Wα n+1 has properties (i) - (iv). Let B n = α κ Wn α. Properties (i)-(iii) for Wα n imply properties (1)-(3) for B n. It is left to be shown that (4) holds. Let {U, V } be an open cover of X. Fix α κ. If B α U or B α V then let V α = Wα 0 = {B α }. Otherwise, since clb α \ U is closed as a subset of clb α, it is compact in clb α. So, consider U = {A i A α : A i V }. Since clb α \ U V, U is an open (w.r.t. B α ) cover of clb α \ U and hence has a finite subcover {A ik : k = 1,..., m}. Let n = max{i k : k = 1,..., m}. Then, Wα n has the property that W A i or W B α \cla i for all i n. So, either there exists i k for some k = 1,..., m such that W A ik V, or W k=1,...,m (B α \ cla ik ) U. Let V α = Wα n. Now, let V = α κ V α. Since V α = Wα n, it is finite. Moreover, since V α B α and B 0 is locally finite, V is locally finite. Since cl( Wα) n = B α and cl( B 0 ) = X, cl( V) = X. Finally, V refines {U, V } by construction. From now on, X is a locally compact metric space without isolated points and y βx \ X. We may view y as an ultrafilter of zero sets on X.

4 Definition 2.3. A point y βx \ X is uniform if w(z) = w(x) for all Z y. Definition 2.4. A uniform ultrafilter, p, on a κ is (ℵ 0, κ)-regular (or just regular) if there exists {S α : α κ} p such that for all A [κ] ℵ 0, {Sα : α A} =. Let κ ω be the collection of functions from κ to ω. Given a filter p on κ, we define an equivalence relation p as follows. For f, g κ ω, f p g if {α κ : f(α) = g(α)} p. We define a partial order, < p, on κ ω as follows. For f, g κ ω, f < p g if {α κ : f(α) < g(α)} p. We write κ ω/p for κ ω/ p and [f] for the equivalence class of f in κ ω/p. For [f], [g] κ ω/p such that f < p g, if f [f] and g [g] then it is easy to see that f < p g. So, < p induces a partial order, < on κ ω/p. If p is an utlrafilter, for any f, g κ ω one of {α κ : f(α) < g(α)}, {α κ : f(α) = g(α)} or {α κ : f(α) > g(α)} is in p. Hence < is a linear order on κ ω/p. Lemma 2.5. If p is a regular ultrafilter on κ then cf( κ ω/p) > κ Proof. Let p be a regular ultrafilter on κ. Let {S α : α κ} p be such that for all A [κ] ℵ 0, {Sα : α A} =. Therefore, for each γ κ, I γ = {α : γ S α } is finite. Let {f α : α κ} be representatives from an increasing sequence in κ ω/p. For γ κ define f(γ) = max{f α (γ) : α I γ } + 1. We wish to show that f α < p f for all α κ. Let α κ and δ S α. Since α I δ, f(δ) > f α (δ) and therefore S α {δ : f(δ) > f α (δ)}. Since S α p, f α < p f. Hence, {[f α ] : α κ} cannot be cofinal in κ ω/p. Lemma 2.6. Let X = X 1 X 2 where X 1 X 2 =, and let f : X Y be a closed map. If f [f[x 1 ]] = X 1, (we say X 1 is a full preimage) then f X1 is closed map. Proof. Let H X 1 be closed in X 1. There exists H X closed in X such that H = H X 1. Since f is closed, f[h ] is a closed subset of Y. So, to show that f[h] is closed in f[x 1 ] we argue that f[h] = f[h ] f[x 1 ]. First, f[h] = f[h X 1 ] f[h ] f[x 1 ]. For the other direction, let y f[h ] f[x 1 ]. Since y f[h ], f 1 (y) H. Since y f[x 1 ] and f [f[x 1 ]] = X 1, f 1 (y) X 1. Hence f 1 (y) H X 1 = H and therefore x f[h].

5 Theorem 2.7. (GCH + all ultrafilters are regular) Let X be a locally compact metric space with no isolated points. Then each uniform y βx \ X is a non-normality point of βx \ X. Proof. Let κ = w(x). List B 0 = {B α : α κ} and B i = {B ασ : α κ, σ i 4} such that B ασ B αν if σ extends ν. From 2.2 (3) we may assume that for α κ and σ i 4, cl X B ασ 0 cl X B ασ 1 = and cl X B ασ 0, cl X B ασ 1 B ασ. For Z y let U 0 (Z) = {B B 0 : B Z } and let N 0 = {U 0 (Z) : Z y}. In a metric space, closed sets are z-sets. For U B 0, either cl X U y or X \ (U ) = cl X (B i \ U) y, since y is an z-ultrafilter. Hence N 0 is an ultrafilter on B 0 and cl X U y for all U N 0. For any V B, let S(V) = {α κ :there is σ such thatb ασ V}. Since N 0 is an ultrafilter, p = {S(V) : V N 0 } is an ultrafilter on κ. Moreover, since y is uniform, for any Z y, w(z) = κ. Hence U 0 (Z) = κ and therefore p is uniform. Since we assume all uniform ultrafilters are regular, by lemma 2.5 cf( κ ω/p) > κ. Consequently, since we assume GCH, cf( κ ω/p) = κ +. Let {[f α ] : α κ + } be an unbounded sequence in κ ω/p. Denote f α (γ) by n(α, γ). For i ω, let ξ i = B i and for ω α < θ let ξ α = {B γσ : γ κ, σ n(α,γ) 4}. For α κ and Z y let U α (Z) = {B ξ α : B Z } and let N α = {U α (Z) : Z y}. As with N 0, for each α κ \ {0}, N α is an ultrafilter on ξ α and cl X U y for each U N α. For S p and α κ +, let ξ α (S) = γ S {B γσ : σ n(α,γ) 4}. Note that for any α, δ κ +, cl X (ξ α (S)) = cl X (ξ δ (S)). From the definition of p, for any S p, ξ 0 (S) N 0 and hence S0 Z for some Z y. Now, as we noted before, cl X(ξ 0 (S)) = cl X (ξ α (S)). Therefore since cl X (ξ 0 (S)) Z, we have that cl X (ξ α (S)) Z. Hence ξ α (S) N α for each S p. Claim. For α < β < κ + and U N α, there is U N β such that cl X U cl X U. Proof. For α < β θ, f α < p f β and hence there is S p such that f α (γ) < f β (γ) for all γ S. If U N α, ξ β refines V = U ξ α (S). Let Z = cl X V. Since V N α, Z y. Moreover, U β (Z) N β. Now, since ξ β refines V, cl X (U β (Z)) cl X V. Let U = U β (Z) Defining H α.

6 Let H α = {cl βx (U ) : U N α }. Suppose α < β. From the claim above, if U N α, there is U N β such that cl X (U ) cl X (U ) and U N β. Therefore, cl βx (U ) cl βx (U ). Since H β is the intersection over all such U, we have that H β cl βx (U ). But, U N α was arbitrary, so H β H α. Since {H α : α κ + } is a nested sequence of closed subsets of the compact space βx, α κ + H α. Claim. α κ + H α = {y}. Proof. Let W τ y. We will find α κ + such that H α W. Let V, U τ y be such that cl βx V U cl βx U W. Let U = X U and V = X \ cl βx V. Then, {U, V } is an open cover of X. Let ξ B be a refinement as guaranteed in Lemma 0.3 (4). For each α κ, since B is locally finite, {B B : B B α } is finite. Let g(α) = max({n ω : there is σ n 4 such that B ασ ξ} {0}). Let ξ = γ κ {B γσ : σ g(γ) 4}. Note, ξ refines ξ and in turn refines {U, V }. Also, g κ ω and hence there is α κ + such that f α > p g. So, there is S p such that f α (γ) > g(γ) for all γ S. In other words, ξ α (S) refines γ S {B γσ : σ g(γ) 4} ξ. Now, Z = cl βx V X = X \ V y, hence U α (Z) N α. As noted before, ξ α (S) N α, so, V = U α (Z) ξ α (S) N α. Let B V. Then, since B U α (Z), B Z = B (X \ V ). But since B ξ α (S), there is B ξ such that B B. Therefore B (X \ V ) and since ξ refines {U, V }, it must be the case that B U. Hence B U U and therefore V U cl βx U. Finally, since V N α, H α cl βx V cl βx U W. The following collections are defined in [7]. We noticed that these collections play the role of the reaping sets in [1]. Defining the L i α s For α Θ and i = 0, 1 define L i α = {B γσ i : γ κ, σ n(α,γ) 4}. Claim. For all α θ, cl βx ( L 0 α) cl βx ( L 1 α) =. Proof. For each γ κ and σ i 4, cl X B γσ 0 cl X B γσ 1 =. Also, B γσ B γβ = for σ β n(α,γ) 4, and for i = 0, 1 we have cl X B γσ i B γσ and cl X B γβ i B γβ. Therefore cl X B γσ i cl X B γβ j = for i, j = 0, 1. So, {clx B γσ 0 : σ n(α,γ) 4} {cl X B γσ 0 : σ n(α,γ) 4} =. Now,

7 since {B γ : γ κ} is a locally finite family and since cl X B γσ i B γ for each σ n ω n 4 and i = 0, 1, we have that cl X ( L 0 α) cl X ( L 1 α) = {clx B γσ 0 : σ n(α,γ) 4, γ κ} {cl X B γσ 0 : σ n(α,γ) 4, γ κ} =. Finally, since cl X ( L 0 α) cl X ( L 1 α) =, cl βx ( L 0 α) cl βx ( L 1 α) =. Since cl βx ( L 0 α) cl βx ( L 1 α) =, y can be in at most one of cl βx ( L 0 α) or cl βx ( L 1 α). Without loss of generality, assume y / cl βx ( L 0 α) for each α θ. A version of the following claim, in particular when Φ is constant, is proven in [7]. Claim. For any α < Θ and Φ : D [α, Θ) 2, the collection {H α } {cl βx ( L Φ(γ) γ ) : γ D} has nonempty intersection. Proof. Let α < Θ and Φ : D 2 for some D [α, Θ). To prove the claim we show that {cl βx U : U N α } {cl βx ( L Φ(γ) γ ) : γ α} has the finite intersection property. Let U 1,..., U n N α and let γ 1,..., γ m D be such that γ m γ 1 α. Since N α is an ultrafilter, there is U N α such that U 1 i n U i. Since f α < p f γ1 < p < p f γm, there is S p such that f α (µ) < f γ1 (µ) < < f γm (µ) for all µ S, in other words n(α, µ) < n(γ 1, µ) < < n(γ m, µ). Since ξ α (S) N α, ξ α (S) U. Hence there is µ S and σ n(α,µ) 4 such that B µσ ξ α (S) U. Define σ n(γm,µ)+1 4 as follows: σ n(α,µ) = σ, σ (n(γ i, µ) + 1) = Φ(γ i ) for each 1 i m and σ (k) = 0 otherwise. Then, B µσ B µσ, since σ extends σ hence B µσ U. Furthermore, B µσ L Φ(γ i) γ i since σ extends σ n(γi,µ)+1 = σ n(γi,µ) Φ(γ i ) and B µ,σ n(γi,µ) Φ(γ i ) L Φ(γ i) γ i. Let θ = κ +. We follow the argument found in [1] to embed NU(θ) into βx \ (X {y}). The induction Denote by θ the discrete space of size θ. We define an embedding, g, of θ into βx \ X such that (1) y cl βx g[a] if and only if A = θ. (2) If A, B [θ] <θ and A B = then cl βx g[a] cl βx g[b] =.

8 Then, we extend g to βg : βθ βx \ X and prove that U(θ) = g [{y}]. Therefore βx \ (X {y}) contains a closed copy of NU(θ) and is therefore not normal. Since we assume GCH we have that θ <θ = θ. List θ {(A, B) : A, B [θ] <θ and A B = } as {T η : η θ} in such a way that if T η = (A, B), then η sup(a B) and if T η θ, then η T η. For α θ let D α = {η : T η = (A, B) and α A B} {η : α T η }. Note that D α [α, θ). For each α θ we define Φ α : D α 2 and choose g(α) to be any element of ({H α } {cl βx ( L Φα(γ) γ ) : γ D}). We define Φ α by induction. Let η θ and assume we have defined Φ α η Dα. If T η θ, let Φ β (η) = 0 for all β < T η. If T η = (A, B), let Φ β (η) = 0 for all β A and let Φ β (η) = 1 for all β B. Let K α = ({H α } {cl βx ( L Φα(γ) γ ) : γ D α }) =. By the claim, K α for eachα θ, so we may choose g(α) K α. To show (1), let A θ be such that A < θ. There is γ θ such that A [0, γ). Let η be such that T η = γ. Note, η γ. For any α < γ = T η, Φ α (η) = 0. So, for α A, K α L 0 η. But, x / cl βx ( L 0 η). Hence, x / cl βx g[a]. For the other direction, let A θ be such that A = θ. Since θ is regular, A is unbounded in θ. Let U N. There is γ θ such that H γ U. For α A such that α γ, g(α) H α H γ U. Hence x cl βx g[a]. To show (2), let A, B [θ] <θ be such that A B =. Let η be such that T η = (A, B). Then, for each α A, Φ α (η) = 0 and for each α B, Φ α (η) = 1. Hence g(α) K α cl βx ( L 0 η) for α A and g(α) K α cl βx ( L 1 η) for α B. But, cl βx ( L 0 η) cl βx ( L 1 η) =. Hence cl βx g[a] cl βx g[b] =. Note, (2) implies g is one-to-one. Since θ is discrete, g is continuous. Extend g to βg : βθ βx \ X. Since βθ is compact, βg is a closed map. In order to show that βg maps N U(θ) homeomorphically to a closed subset of βx \ (X {y}), we must verify the following: (1) βg[βθ] \ {y} = βg[nu(θ)] (2) βg[u(θ)] {y} (3) βg NU(θ) is one-to-one

9 If (1) holds, NU(θ) is mapped onto a closed subset of βx \ (X {y}). If (1) and (2) hold, then NU(θ) is a full preimage. Since βg is a closed continuous map by 2.6, βg NU(θ) is a closed continuous map. Therefore if (3) holds, βg NU(θ) is a homeomorphism. (1) Let q NU(θ). There is A θ such that A < θ and A q. Since βg is continuous, g(q) cl βx g[a]. Hence, g(q) y. Let z βg[βθ]\{y}. Let U be an open neighborhood of z such that y / cl βx U. Since βg is continuous, A = U βg[θ]. Let A = g [A ]. Since y / cl βx U, A < Θ, otherwise y cl βx A cl βx U. If q g (z) then q cl βθ A. Hence q NU(θ) and therefore z βg[nu(θ)]. (2) The preceding argument also shows that for any q βθ, if βg(q) y, then q NU(θ). Hence βg[u(θ)] {y}. (3) Let q q NU(θ). There are A, B [θ] <θ such that A B = and q cl βθ A and q cl βθ B. By continuity, g(q) cl βx g[a] and g(q ) cl βx g[b]. But, by (2) cl βx g[a] cl βx g[b] =. Hence g(q) g(q ). Now, since θ = κ + = 2 κ is regular and not a strong limit, by 2.1, N U(θ) is not normal. Hence y is a non-normality point of βx \ X. Corollary 2.8. (GCH + all ultrafilters are regular) Let X be a locally compact metric space with no isolated points. Then each y βx \ X is a non-normality of βx \ X. Proof. We have seen that if y βx \ X is uniform then it is a nonnormality point of βx \ X. Suppose that y βx \ X is not uniform. That is, there exists Z y such that w(z) < w(x). Let Z y be such that w(z) is minimum. Then, there is a cover of Z consisting of sets clb from a subcollection, Z, of B 0 of size w(z). Let Y = {clb : B Z}. Since B 0 is locally finite, Y is closed. Also, y cl βx H. Since X is normal and H is closed, H is C -embedded in X. Therefore, βh = cl βx H and y H is uniform on H. So, by the theorem, y is a non-normality point of the set cl βx H \ H, and hence is a non-normality point of βx \ X.

10 REFERENCES [1] Beslagic, A.; van Douwen, E. K. Spaces of nonuniform ultrafilters in spaces of uniform ultrafilters. Topology and its Applications 35 (1990) [2] Gillman, L. The space βn and the continuum hypothesis. General Topology and its relations to Modern Analysis and Algebra 2 (Proc. Second Prague Topological Sympos., 1966) (1967), [3] Kunen, K.; Parsons, L. Projective covers of ordinal subspaces. Proceedings of the 1978 Topology Conference (Univ. Oklahoma, Norman, Okla., 1978), II. Topology Proc. 3 (1978), no. 2, (1979). [4] Rajagopalan, M. βn \ N \ {p} is not normal, J. Indian Math. Soc. 36 (1972), [5] Logunov, Sergei On non-normality points and metrizable crowded spaces. Comment. Math. Univ. Carolin. 48 (2007), no. 3, [6] Malyhin, V.I. Nonnormality of certain subspaces of βx, where X is a discrete space. (Russian) Dokl. Akad. Nauk SSSR 211 (1973), [7] Terasawa, Jun. βx \{p} are non-normal for non-discrete spaces X. Topology Proc. 31 (2007) No. 1, 1 9. [8] Warren, N. M. Properties of the Stone-Cech compactifications of discrete spaces, Proc. Amer. Math. Soc. 33 (1972), DEPARTMENT OF MATHEMATICS, UNIVERSITY OF KANSAS, LAWRENCE, KANSAS address: spencel1@math.ku.edu