Houston Journal of Mathematics. c 2013 University of Houston Volume 39, No. 3, Dedicated to the memory of Mikhail Misha Matveev

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1 Houston Journal of Mathematics c 2013 University of Houston Volume 39, No. 3, 2013 ON THE HAUSDORFF NUMBER OF A TOPOLOGICAL SPACE MADDALENA BONANZINGA Communicated by Yasunao Hattori Dedicated to the memory of Mikhail Misha Matveev Abstract. The Hausdorff number of a space X is H(X) = min{τ : for every subset A X such that A τ there exist neighbourhoods U a for all a A so that a A U a = }. Some known cardinal inequalities for Hausdorff spaces are generalized in terms of the Hausdorff number. 1. Introduction Let X be a topological space. Put H(X) = min{τ : for every subset A X such that A τ there exist neighbourhoods U a for all a A so that a A U a = }. We will call this number H(X) (finite or infinite) the Hausdorff number of X. Of course, with X 2, X is Hausdorff iff H(X) 2. A space X is said to be n-hausdorff if H(X) n (where n 2). We give, for every finite n, examples of (n + 1)-Hausdorff spaces which are not n-hausdorff. But the principal aim of this paper is to extend some known cardinal inequalities for Hausdorff spaces to spaces with finite Hausdorff number. Also in this paper we want to compare the n-hausdorff condition with other weak separation axioms, in particular T 1. The approach we follow in this paper is is similar to that in [2] for n-urysohn spaces. For a subset A of a topological space X we will denote by [A] <λ ([A] λ ) the family of all subsets of A of cardinality < λ (= λ). We consider cardinal invariants of topological spaces (see [4] and [8]) and all cardinal functions are multiplied by ω. In particular, d(x) is the density of X, 2000 Mathematics Subject Classification. 54A25, 54A35. Key words and phrases. Hausdorff space, the Hausdorff number of a space, the weak Hausdorff number of a space, n-hausdorff pseudocharacter. 1013

2 1014 MADDALENA BONANZINGA l(x) is the Lindelöf number of X, ψ(x) s the pseudocharacter of X, χ(x) is the character of X, nw(x) is the netweight of X, pw(x) is the pseudoweight of X, c(x) is the cellularity of X, s(x) is the spread of X, hc(x) and hl(x) denote respectively hereditary cellularity and hereditary Lindelof degree. Note that hc(x) = s(x) (of course: for every X, s(x) hl(x)). The diagonal degree of a space X is (X) = min{ U : U is a family of open sets in X 2 such that U = }, where denotes the diagonal of X 2. The Hausdorff pseudocharacter of X, denoted Hψ(X), is the smallest infinite cardinal κ such that for each point x there is a collection {V (α, x) : α < κ} of open neighborhoods of x such that if x y, then there exists α, β < κ such that V (α, x) V (β, y) = [10]. This cardinal function is defined only for Hausdorff spaces. The following holds: ψ(x) Hψ(X) χ(x). Note that every countable Hausdorff space has countable Hausdorff pseudocharacter. So, for example ω {p} where p is from ω has countable Hausdorff pseudocharacter but is not first-countable. Recall that a family U of open sets of a space X is point-finite if for every x X, the set {U U : x U} is finite [4]. Tkachuck [16] defined p(x) = sup{τ : there is a point-finite family γ in X such that γ = τ}. We may introduce the following definition: Definition 1. A family U of open sets of a space X is point-( n) finite where n N if for every x X, the set {U U : x U} has cardinality n. For each n N, p n (X) = sup{ U : U is a point-( n) family in X}. Of course c(x) = p 1 (X) p 2 (X)..., but in fact the following is true: Proposition 2. p n (X) = c(x), n N. Proof. Let τ be an infinite cardinal. We will prove by induction on n N that (*) If there exists a point-( n) family of cardinality τ, then there exists a cellular family of cardinality τ. This will show that for every n N, p n (X) c(x). Of course (*) is true for n = 1. Assume that (*) holds for n-1 and prove it for n. Let U be a point-( n) family of cardinality τ. Let U = { F : F [U] n, F }. Then U is a cellular family such that U U. If U = U, then we are done. If U < U, put U 0 = {U U : F [U] n such that F U and F }. Then U \ U 0 is a point-( n-1) family of cardinality τ (because U 0 < U ) and then, by hypothesis, there exists a cellular family of cardinality τ.

3 ON THE HAUSDORFF NUMBER OF A TOPOLOGICAL SPACE 1015 The referee pointed out that in the previous proof one takes into account that p n (X) and c(x) are defined the sup and then the sup = max question aries. 2. General properties and some examples Proposition 3. For every space X, H(X) X +. In particular for a T 1 space X, H(X) X. Proof. Let X be a T 1 space, A X, A = X and f : A X : x f(x) x a onto mapping. Then for every a A, X \ f(a) is a open neighborhood of a and a A X \ f(a) =. Then H(X) A. A space with the antidiscrete topology shows that the Hausdorff number of a space X can be equal to X +. Example 4. An (n + 1)-Hausdorff space which is not n-hausdorff, where n N. Let n N and X = (n + 1) {ω (n + 1)} be the space topologized as follows: all points from ω {i}, with i (n + 1) are isolated; a basic neighbourhood of i n + 1 takes the form U(i, N) = {i} {(m, j) : j i, m N}, where N N. X is an (n + 1)-Hausdorff not n-hausdorff space. More generally, we have the following fact: Example 5. For every cardinal λ, there is a space X such that X λ, and X is λ-hausdorff, but X is not κ-hausdorff for any κ < λ. Let λ 3 be a cardinal, and let τ be a cardinal such that cf(τ) λ. Let A be a set of cardinality λ. Put Y = [A] <λ τ and X = A Y. Topologize X as follows: the points of Y are isolated in X; a basic neighborhood of a A takes the form U α (a) = {a} { M, β : a M [A] <λ and α < β < τ}. X is λ-hausdorff, but X is not κ-hausdorff for any κ < λ. Notice that every T 2 space is a T 1 space having finite Hausdorff number but a T 1 space having finite Hausdorff number need not be T 2 as the space X with n = 2 in Example 4 shows. Also notice that the notion of n-hausdorff, with n 2 and T 1 are independent. Indeed, fix n N, n > 1; the space X in Example 4 is a (n + 1)-Hausdorff non T 1 space. For every n N, the space in Example 13 (see below) is a T 1 non n-hausdorff space. Recall that a space X is Hausdorff iff the diagonal of X 2 is closed. Now we give a characterization of 3-Hausdorfness in terms of the diagonal 3 in the cube of the space.

4 1016 MADDALENA BONANZINGA Proposition 6. X is 3-Hausdorff iff 3 is closed in Y 3, where Y = {(x, y, z) X 3 : x, y, z are distinct}. Proof. Let X be a 3-Hausdorff space and let x, y, z X be distinct. Then there exists open in X disjoint U x x, U y y, U z z. Then U = U x U y U z is open in X 3 and U 3 =. The other implication is similar. Theorem 7. Let (X, T ) be a 3-Hausdorff space. For every x X, denote T x = {U T : x U}. Then one of the following holds: (1) T x = {x}; (2) T x = 2, T x = {x, y} = T y ; (3) T x 2 and for every y T x \ {x}, T y = {y}. Proof. Claim 1: Let x, y X distinct. If y T x, then T y T x. (Follows immediately from the definition of T x.) Claim 2: If y and z are distinct points of T x \ {x} then y and z have disjoint neighborhoods in X, and therefore T y T z =. Proof of claim 2. Since X is 3-Hausdorff, there are neighborhoods V x x, V y y, V z z such that V x V y V z =. Since y, z T x, y, z V x. Then V x V y and V x V z are disjoint neighborhoods of y and z in X. Claim 3: If y T x then T y {x, y}. (Follows from Claim 1 and Claim 2). Now, it remains to show that if T x \ {x} =, then we have either case 2 or case 3. By Claim 3, for each y T x \ {x} we have only two possibilities: (a) T y = {x, y} and (b) T y = {y}. If for some y T x \ {x} we have possibility (a), then by Claim 1 T x T y = {x, y} and thus we have case 2. If for all y T x \ {x} we have possibility (b) then we have case 3. Note that for T 1 spaces X, case (1) in the previous theorem holds for every x X. Now we are going to introduce a variation of the Hausdorff number; the reader will see that in some applications, this variation is more convenient than the original Hausdorff number. Definition 8. Let X be a space. Put H (X) = min{τ : for every A X such that A τ there exist B A such that B < τ and neighborhoods U b of b, b B, so that b B U b = }. We will call this number H (X) (finite or infinite) the weak Hausdorff number. Remark 9. The property to have weak Hausdorff number less or equal to ω and property T 1 are independent. Let X = D {p} be a space with arbitrarily big

5 ON THE HAUSDORFF NUMBER OF A TOPOLOGICAL SPACE 1017 cardinality topologized as follows: D is an open discrete subspace and the only neighborhood of p is the whole X. X is a space such that H (X) = 3 which is not T 1. Further the space X from Example 13 is a T 1 space such that H (X) > ω. Of course H (X) H(X), for every topological space X. We have that, for every n N, if H(X) = n H (X) = n or H (X) = n + 1. This fact motivates the previous terminology. Notice that the space (X, T ) where X = {1, 2, 3} and T = {, X, {1}, {2, 3}}, is such that H(X) = H (X) = 3. More generally, we have the following easy fact: Proposition 10. For every space X, H (X) = H(X) or H (X) = H(X) +. The space in Example 4 proves that a space X such that H (X) = ω may fail to be n-hausdorff, for some n N. The next example improves this result: Example 11. A space X such that H (X) = ω and H(X) n, for each n N. Let X = n N X n, where denotes the discrete sum, and H(X n ) = n for each n N. Then H (X) = H(X) = ω. Example 12. For every limit cardinal κ, there is a space X such that H(X) = H (X) = κ. Let κ be a cardinal. Consider the space X = ℵα κx ℵα, where H(X ℵα ) = ℵ α for each ℵ α < κ. Then H (X) = H(X) = κ. Example 13. For every regular cardinal κ there is a space X such that H(X) = κ and H (X) = κ +. Consider the space X = κ A, where κ is a regular cardinal and A 2 k topologized as follows: the points of κ are isolated and basic neighborhoods of a A take the form a (κ \ H), where H is a subset of κ such that H < κ. X is T 1, d(x) = κ, H (X) = κ + and H(X) = cf(κ) (because H(X) = ψ(x, a), where a is any point in A and ψ is the pseudocharacter). Since κ is regular, we have that H(X) = κ. Question 14. Is there a singular cardinal κ and a space X such that H(X) = κ and H (X) = κ +?

6 1018 MADDALENA BONANZINGA 3. Special subsets of n-hausdorff spaces Now we are going to show that every n-hausdorff space contains a large (in some sense) T 1 subspace. Lemma 15. For every X and every x X, {x} < H(X). Proof. Let H(X) = k. By contradiction assume {x} = τ k. Then A = {x} is a subset of X such that for every neighbourhood U a of a, where a A, we have x a A U a. Since A = τ k, we have H(X) > k; a contradiction. Recall the following definition: Definition 16. [11] A map F : X P(X) is a set mapping if x X, x / F (x). A subset M X is free (under the set mapping F ) or F-free if x M, F (x) M =. Also Theorem 17. ([13], see also [11]) If X = κ, λ < cf(κ) and F : X P(X) is a set mapping such that x X, F (x) < λ, then there exists M X such that M = κ and M is F-free. Remark 18. In fact, in the proof of the previous theorem it is shown that there exists a F-free subset Y of X having the same cardinality as X. By Lemma 15, we have Corollary 19. Let X be an infinite space such that H(X) < X and F : X P(X) be such that F (x) = {x} \ {x}. Then there exists a F-free subset Y of X having the same cardinality as X. Note that if F : X P(X) is such that x X {x} \ {x} P(X), a subset Y X is F-free iff Y {F (x) : x Y } = iff Y {{x} \ {x} : x Y } = iff x Y, {x} Y = {x} iff Y is T 1. Then we have the following two propositions: Proposition 20. Let X be an infinite space such that H(X) < X. Then there exists a maximal T 1 subset Y of X having the same cardinality as X. It is well known that every infinite Hausdorff space contains an infinite discrete subspace. Proposition 21. Every infinite n-hausdorff space X, where n 2, contains a discrete infinite subspace.

7 ON THE HAUSDORFF NUMBER OF A TOPOLOGICAL SPACE 1019 Proof. Let X be an infinite n-hausdorff space, n N. Of course, H(X) < X and then by Proposition 20, there exists a maximal T 1 subset Y X having the same cardinality as X. We claim that there are disjoint open sets in Y one of which is non empty and the other infinite. Fix a subset A Y of cardinality n. By n-hausdorfness, there exist open neighborhoods V x of x A such that x A V x =. If one of V x is finite, then x is isolated, so since Y is T 1 we may assume that x / V y, whenever x, y A and x y. For every B [A] n 1, put W B = x B V x. Then W B W B =, wherever B, B [A] n 1, B B. Case 1: If all W B, B [A] n 1 are finite, put X 1 = x A V x \ ( B [A] W B). n 1 Then there exist x 1 A such that U x1 = V x1 \ B [A] W n 1 B is an open neighborhood of x 1 in X 1 and Ũx 1 = V x1 \ U x1 is an infinite open subset of X 1 disjoint from U x1. Case 2: If there is B [A] n 1 such that W B is infinite, put X 1 = x A V x \ ( B B W B). Put {x 1 } = A \ B. Then U x1 = V x1 \ B [A] W n 1 B is an open neighborhood of x 1 in X 1 and Ũx 1 = V x1 \ U x1 is an infinite open subset of X 1 disjoint from U x1. Then we may consider Ũx 1, fix n distinct points in it and proceed like before. We may do infinitely many steps and then we obtain an infinite discrete subspace {x 1,..., x n,...}. Question 22. Let X be a space such that H (X) < X. Does there necessarily exist a maximal T 1 subset Y of X having the same cardinality as X? 4. Some generalizations of cardinal inequalities 4.1. Generalizations of some elementary inequalities. In 1937 Pospisil proved that Proposition 23. (see [8]) For every Hausdorff space X, X 2 2d(X). The next proposition extends the previous inequality from Hausdorff spaces to spaces having weak Hausdorff number less or equal to ω (hence in particular to spaces with finite Hausdorff number). Proposition 24. Let X be a space such that H (X) ω. Then X 2 2d(X). Proof. Let D be dense in (X, T ) and D = d(x). For x X, put A x = {U D : x U and U T }. Consider the mapping φ from X to P(P(D)) defined by x A x. We want to show that the cardinality of each fiber is finite. This will imply that X P(P(D)) = 2 2d(X).

8 1020 MADDALENA BONANZINGA By contradiction, assume there exists an infinite fiber φ 1 (A x ). Since H (X) ω, there exist n N, F = {x 1,..., x n } φ 1 (A x ) and neighborhoods U xi T, i {1,..., n}, of x i such that {U xi : 1 i n} =. Then A x1 =... = A xn = A x. For 1 i n, U xi D A x and therefore U xi D = V i D where V i is an open neighborhood of x. Let V = V 1... V n. Then V D. This contradicts {Uxi : 1 i n} =. Corollary 25. Let X be a space such that H(X) is finite. Then X 2 2d(X). The space from Example 13 shows that H (X) ω is the weakest possible restriction on H under which the inequality X 2 2d(X) remains true. Indeed in Example 13 we may assume A > 2 2κ and then the space X is such that X > 2 2κ and d(x) = κ. The space X of Example 13 in case κ = ω, however, does not disprove the following hypothesis which is a hybrid of Proposition 23 and inequality for Hausdorff spaces X 2 2s(X) (see [8], Theorem 4.12) because s(x) in Example 13 is as big as A : Question 26. Is X 2 2H(X)d(X)s(X) true for every space X? In 1937 Pospisil proved that Proposition 27. (see [8]) For every Hausdorff space X, X d(x) χ(x). The following proposition extends the previous inequality from Hausdorff spaces to spaces having weak Hausdorff number less or equal to ω (hence in particular to spaces with finite Hausdorff number). Proposition 28. Let X be a space such that H (X) ω. Then X d(x) χ(x). Proof. Let D be a dense subset of X such that D = d(x) and {B(x)} x X a neighborhood system for X. Suppose for every x X, B(x) χ(x) ℵ 0. Denote by A 0 the family of all subsets of D whose cardinality is χ(x); clearly A 0 d(x) χ(x). For every x X and U B(x) pick d(x, U) U D and put D(x) = {d(x, U) : U B(x)} D; clearly D(x) χ(x). For x X, put A 0 (x) = {U D(x) : U B(x)}; clearly A 0 (x) is a subset of A 0 of cardinality χ(x). Consider the mapping φ from X to [A 0 ] χ(x) defined by x A 0 (x). We want to show that the cardinality of each fiber of φ is finite. This will imply that X (d(x) χ(x) ) χ(x) = d(x) χ(x). By contradiction, assume there exists an infinite fiber φ 1 (A 0 (x)). Since H (X) ω, there exist F = {x 1,..., x n } φ 1 (A 0 (x)) and neighborhoods U xi T, i {1,..., n}, of x i such that {U xi : 1 i n} =. Then

9 ON THE HAUSDORFF NUMBER OF A TOPOLOGICAL SPACE 1021 A 0 (x 1 ) =... = A 0 (x n ) = A 0 (x). For 1 i n, U xi D(x i ) A 0 (x) and therefore U xi D(x i ) = V i D(x) where V i is an open neighborhood of x. Let V = V 1... V n. Then V D(x). This contradicts {U xi : 1 i n} =. The space from Example 13 shows that H (X) ω is the weakest possible restriction on H under which the inequality X d(x) χ(x) remains true. Indeed in Example 13 we may assume A > κ κ ; then the space X is such that X > κ κ and d(x) = χ(x) = κ. In particular if X is a space such that H(X) is finite, then X d(x) χ(x). However we may prove even more. First we consider the following generalization of Hausdorff pseudocharacter: Definition 29. The n-hausdorff pseudocharacter of X (where n 2), denoted n-hψ(x), is the smallest infinite cardinal κ such that for each point x there is a collection {V (α, x) : α < κ} of open neighborhoods of x such that if x 1,..., x n are distinct points from X, then there exists α 1,..., α n < κ such that n i=1 V (α i, x i ) =. This cardinal function is defined for n-hausdorff spaces. Of course Hψ(X) k n-hψ(x) k, for every n N. Then, for every n N, ω {p} where p ω, is a countable n-hausdorff space such that n-hψ(x) = ω which is not first countable. Then we may prove the following generalization of Proposition 28. Proposition 30. Let n N, n > 1. For every space X with H(X) n, X d(x) n-hψ(x). Proof. Let D be a dense subset of X such that D = d(x), {B(x)} x X be a neighborhood system for X and let n-hψ(x) κ. For each x X, let {V x,α : α < κ} be a collection of open neighborhoods of x such that for every n distinct points x 1,..., x n X, there exists α 1,..., α n < κ such that n i=1 V x i,α i =. Denote by A 0 the family of all subsets of D whose cardinality is κ; clearly A 0 d(x) κ. For every x X and α < κ pick d(x, α) V x,α D and put D(x) = {d(x, α) : α < κ} D; clearly D(x) κ. For x X, put A 0 (x) = {U D(x) : U B(x)}; clearly A 0 (x) is a subset of A 0 of cardinality κ. Consider the mapping from X to [A 0 ] κ defined by x A 0 (x). We want to show that the cardinality of each fiber is less than n. This will imply that X (d(x) κ ) κ = d(x) κ. By contradiction, assume there exists φ 1 (A 0 (x)) having cardinality n. Let F = {x 1,..., x n } an n-element subset of φ 1 (A 0 (x)). Pick V xi,α i for all i

10 1022 MADDALENA BONANZINGA {1,..., n} such that n i=1 V x i,α i =. Then A 0 (x 1 ) =... = A 0 (x n ) = A 0 (x). For 1 i n, V xi,α i D(x i ) A 0 (x) and therefore V xi,α i D(x i ) = V i D(x) where V i is an open neighborhood of x. Let V = V 1... V n. Then V D(x). This contradicts {V xi,α i : 1 i n} = Generalization of Arhargel skii inequality. Arhangel skii proved that Theorem 31. (see [8]) For every Hausdorff space X, X 2 l(x)χ(x). The following proposition extends this inequality from Hausdorff spaces to T 1 spaces having weak Hausdorff number ω (hence in particular to T 1 spaces with finite Hausdorff number). Theorem 32. Let X be a T 1 space such that H (X) ω. Then X 2 l(x)χ(x). Proof. Let κ = l(x)χ(x) and let H (X) ω. For each x X fix a local base B x at x such that B x κ. By transfinite induction, construct a family {F α : 0 α < κ + } of subsets of X such that 1. F α 2 κ ; 2. F α is closed in X; 3. if β < α, then F β F α ; 4. if there exists a family U { B x : x β<α F β} such that U κ and X \ U, then F α \ U. To construct such a family, fix a point x 0 X and put F 0 = {x 0 }. Let 0 α < κ + and assume that F β has been constructed for each β < α. Put B = {B x : x β<α F β}. Since for each β < α, F β 2 κ we have that β<α F β α 2 κ = 2 κ. Then B = {B x : x β<α F β} 2 κ κ = 2 κ. Let B + = {X \ U : U B and U κ}. Since {U B : U κ} = [B] κ B κ (2 κ ) κ = 2 κ, we have that B + 2 κ. Choose one point in each non empty element of B + and denote the set of such points by E. Of course, E 2 κ. Put F α = E ( β<α F β). Since H (F α ) ω, by Proposition 28 we have that F α d(f α ) χ(fα) ; also, d(f α ) E ( β<α F β) 2 κ and χ(f α ) κ. Then F α d(f α ) χ(fα) (2 κ ) κ = 2 κ, that is F α 2 κ. This completes the construction of the family {F α : 0 α < κ + }. Put F = α<κ + F α. We claim that F is closed in X. Let x F. Since t(x) χ(x) l(x) χ(x) = κ, we have that t(x) κ. Then there exists B F such that B κ and x B. So, for every y B, there exists α y < κ + such that y F αy. Of course α y : y B} κ. Since cf(κ + ) = κ +, every family of κ ordinals in κ + is bounded and then there exists α < κ + such that α y α,

11 ON THE HAUSDORFF NUMBER OF A TOPOLOGICAL SPACE 1023 for every y B. Since {F α : 0 α < κ + } is an increasing family of sets, we have that F αy F α, for every y B. Then B y B F α y F α. Since F α is closed in X, we have that x B F α F, that is x F. This proves that F is closed in X. Now we prove that X = F. Suppose, by contradiction, there exists y X \ F. Since X is T 1, for every x F fix U x B x such that y / U x. The family {U x : x F } covers F. Since F is closed in X and l(x) κ, we have that l(f ) κ. Then there exists A F, A κ such that U = {U x : x A} covers F. Since A F = α<κ F α, A κ, cf(κ + ) = κ + and {F + α : α < κ + } is an increasing family of sets, by the same argument we used before, there exists δ < κ + such that A F δ. Then U {B x : x F δ = β<δ+1 F β}. Since y / U, we have that X \ U. Also F δ+1 F U ; this contradicts 4. Then X = F. So, we have X = F = α<κ F α κ 2 κ = 2 κ. + Note that in the previous theorem the condition on H was used only when referring to Proposition 28. Also this theorem is not true for spaces having finite Hausdorff number which are not T 1. Indeed the space X = D {p} in Remark 9 is first countable and compact (hence Lindelöf). Question 33. Is the previous theorem true for spaces having uncountable weak Hausdorff number? In particular, does there exists a Lindelöf, first countable, T 1 space X with H(X) = ω and large cardinality? In case H(X) = 2 the following corollary gives Arhangel skii s theorem. Corollary 34. Let X be a T 1 space such that H(X) is finite. 2 l(x)χ(x). Then X Theorem 31 was generalized in the following way: Proposition 35. [8] If X is a Hausdorff space, then X 2 l(x)t(x)ψ(x). We will prove the following generalization of Corollary 34 for 3-Hausdorff spaces. Theorem 36. For every T 1 3-Hausdorff space X, X 2 l(x)t(x)ψ(x). We will need some propositions and lemmas. Proposition 37. Let (X, T ) be a 3-Hausdorff space. There exists a family B T having cardinality d(x) t(x) such that for every x X, denoting B x = {B B : x B}, one of the following holds:

12 1024 MADDALENA BONANZINGA (1) B x = {x}; (2) B x = 2 and there exists a unique y such that B x = {x, y} = B y ; (3) B x 2 and for every y B x \ {x}, B y = {y}. Proof. Let d(x) = σ and t(x) = τ. Let S be a dense subspace of X such that S = σ. Put A = {A : A [S] τ }. Then A σ τ. Put B = {X \ H : H A}. Then B σ τ. Claim 1: Let x, y X distinct. If y B x, then B y B x. (Follows from definition of B x ) Claim 2: For every x X, B x \ {x} is relatively Hausdorff in X. Proof of Claim 2: Suppose x, y, z X are distinct points such that y, z B x. Since X is 3-Hausdorff, there are open sets V x x, V y y and V z z such that V x V y V z =. Now we prove that y / V y V z. By contradiction, assume y V y V z. Then y V y V z S. Since t(x) τ, there is A V y V z S such that A τ and y A. Put B = X \ A. Then y / B and B B. Moreover, x B. Indeed, since V x V y V z = we have V x A = and then x / A. So, since y / B, it follows that y / B x ; a contradiction. Similarly, z V y V z. Put W y = V y \ V y V z and W z = V z \ V y V z. Then W y T and W z T are disjoint neighborhoods of y and z respectively. Claim 3: For every x, y X, y x, if y B x, then B y {y, x}. Proof of Claim 3: Let z B x distinct from x and y. By Claim 2, there are open in X disjoint sets W y x and W z z. Then z W z S. Since t(x) τ, there is A W z S such that A τ and z A. Put B = X \ A. Then z / B and B B. Moreover y B because A W z X \ W y and then, since X \ W y is closed, A W y =. So z / B y. On the other hand, if z / B x, then by Claim 1, z / B y. Claim 4: Let x, y X be distinct. If y B x, then either B y = {y} or B y = B x = {x, y} (Follows directly from Claim 3.) Then for every x X, case (1), (2) or (3) holds. Let (X, T ) and B be like in Proposition 37. Denote X i = {x X : x satisfies condition (i) in Proposition 37}, i = 1, 2, 3. Then X = X 1 X 2 X 3. Note that if x X 1 X 3, then for every y x, B y B x ; for every x X 2, there exists exactly one point y x such that B y = B x. Then we may introduce the following equivalence relation on X: x, y X, x y B x = B y. For each x X, we will denote [x] = {y X : y x}. Of course, if X is 3-Hausdorff (or n-hausdorff for any finite n) then all classes of equivalence are finite and thus if X is infinite then X/ = X. Now define the following technical notions:

13 ON THE HAUSDORFF NUMBER OF A TOPOLOGICAL SPACE 1025 Definition 38. Let (X, T ) be a topological space. A -pseudobase for X is a family B T such that for every x X, {B B : x B} = {U T : x U}. Put pw (X) = min{ B : B is a -pseudobase forx}. Definition 39. Let (X, T ) be a 3-Hausdorff space and let B be a -pseudobase for X. A -network for X is a family N P(X) such that for every x X and for every U T such that [x] U, there exists V N such that [x] V U. Put nw (X) = min{n : N is a -network for X}. Proposition 40. Let (X, T ) be a 3-Hausdorff space and let N be a -network for X. For every x X, denote N x = {N N : [x] N}. Then x, y X, x y N x = N y. Proof. Let B be a -pseudobase for X. Then B satisfies conditions (1)-(3) of Proposition 37. Let x, y X. If x y, we have [x] = [y] and then for every N N, N [x] N [y]. Therefore N x = N y. Now assume that N x = N y. If x y, then B x B y. This implies N x N y ; a contradiction. Lemma 41. For a 3-Hausdorff space X, X nw (X) ψ(x) Proof. Let S be a -network for X such that S = nw (X). Put ψ(x) = κ. For every x X, denote by U x a family of open sets of X such that U x κ and Ux = {x}. For every x X, put V x = U x if [x] = {x} and V x = {U x U y : U x U x, U y U y } if [x] = {x, y}, with y x. Since S is a -network, for every V V x there exists S V S such that [x] S V V. Of course, {S U : V V x } κ. Also [x] {S V : V V x } {V : V V x } = [x]; then [x] = {S V : V V x }. Define the function g : X / [S] κ : [x] {S V : V V x }. g is a 1-1 mapping, then X [S] κ S κ = nw (X) ψ(x). Note that the previous result is similar to the following Proposition 42. [8] For a T 1 space X, X nw(x) ψ(x) Lemma 43. For a 3-Hausdorff space X, nw (X) pw (X) l(x). Proof. Let B be a -pseudobase with B = κ. Then [B] l(x) B l(x) = κ l(x). Put N = {X \ G : G [B] l(x) }. Then N = [B] l(x) κ l(x). Now we prove that N is a -network. Let x X and U be an open set such that [x] U. Then for every y X \ U, y x. So B y B x and then there exists B y B y such that x / B y. Since X \ U is closed in X, l(x) l(x \ U) and then there exists A {B y : y X \ U} such that A l(x) and A X \ U. By arbitrarity of y, we have [x] ( A) =. Then X \ A N.

14 1026 MADDALENA BONANZINGA Note that the previous result is similar to the following: Proposition 44. [12] For a T 1 space X, nw(x) pw(x) l(x). Lemma 45. If X is a T 1 3-Hausdorff space, then pw (X) d(x) t(x). Proof. It follows from Proposition 37. The previous result is similar to the following: Proposition 46. [12] If X is Hausdorff, then pw(x) d(x) t(x). Lemma 47. If X is T 1 3-Hausdorff space, then X d(x) l(x)t(x)ψ(x) Proof. By Lemmas 41, 43 and 45, we have X nw (X) ψ(x) pw (X) l(x)ψ(x) d(x) t(x)l(x)ψ(x). Proof of Theorem 36. Put κ = l(x) = t(x) = ψ(x). In the proof of Theorem 32 replace local base B x, x X with local pseudobase B x, x X and inequality from Proposition 28 with inequality from Lemma 47. Question 48. Is it true that if X is a T 1 space such that H (X) ω, then X 2 l(x)t(x)ψ(x)? 4.3. Others inequalities. Combinatorial set theory plays an important role in the theory of cardinal function. In [9] the author explores the close connection between theorems of combinatorial set theory and cardinal functions. In particular the following result is considered. Theorem 49. ([7],[8]) For a Hausdorff space X, X 2 c(x)χ(x). Note that the space in Example 13 shows that there exists a space X with H(X) = ω such that X 2 c(x)χ(x) is not true. There are easy proofs of Theorem 49 (see [7] or [8]) using Erdös-Rado Partition Theorem. Recall that: Theorem 50. [5] (Erdös-Rado Partition Theorem) Let κ be an infinite cardinal, let E be a set with E > 2 κ and suppose [E] 2 = α<κ P α. Then there exists α < κ and a subset A of E with A > κ such that [A] 2 P α. In Theorem 49 character can be replaced with Hausdorff pseudocharacter in the way similar to the proof of the same theorem (see [7], [8] or [9]). Theorem 51. For a Hausdorff space X, X 2 c(x)hψ(x).

15 ON THE HAUSDORFF NUMBER OF A TOPOLOGICAL SPACE 1027 Proof. Let Hψ(X)c(X) κ. For every p X, enumerate the collection {V p,α : α < κ} of open sets such that for every pair of distinct points p, q, there exist disjoint sets V p,α, V q,α. Put {p, q} in R α if V p,α V q,α =. This defines a partition of [X] 2 into κ subsets. Assume that X > 2 κ. By Theorem 50, there exists A X with A > κ and [A] 2 is contained in one of R α. Consider the family {V α,p : p A}. It is a cellular family of cardinality > κ. This contradicts c(x) κ. Recall the following result: Theorem 52. [5] Let κ be an infinite cardinal, let E be a set with E > 2 2κ and suppose [E] 3 = α<κ P α. Then there exists α < κ and a subset A of E with A > κ such that [A] 3 P α. Theorem 53. For a 3-Hausdorff space X, X 2 2c(X)3-Hψ(X). Proof. Let 3-Hψ(X)c(X) κ. For every p X, enumerate a collection {V p,α : α < κ} of open sets such that if p, q, t are distinct points from X, there exist disjoint sets V p,α, V q,α, V t,α. Put {p, q, t} in R α if V p,α V q,α V t,α =. This defines a partition of [X] 3 into κ subsets. Assume that X > 2 2κ. By Theorem 52, there exists A X such that A > κ and [A] 3 is contained in one of R α. Consider the family {V α,p : p A}. This is a family of cardinality > κ witnessing that p 2 (X) > κ and equivalently, by Proposition 2, c(x) > κ; a contradiction because c(x) κ. Corollary 54. For every 3-Hausdorff space X, X 2 2c(X)χ(X). Question 55. Is X 2 c(x)χ(x) true for every X such that H(X) is finite? Recall that the star-lindelöf number of a space X is st-l(x) = min{κ : open cover U there exists a subset F such that F κ and st(f, U) = X}, where st(f, U) = {U U : U F } [15]. In [1] the following was proved: Proposition 56. [1] For a Hausdorff space X, e(x) 2 st-l(x)χ(x). In the previous proposition character can be replaced with Hausdorff pseudocharacter as the following proposition shows. Proposition 57. For every Hausdorff space X, e(x) 2 Hψ(X)st-l(X). Proof. Let Hψ(X)st-l(X) κ. For every p X, enumerate a collection {V p,α : α < κ} of open sets such that for every pair of distinct points p, q, there exist disjoint sets V p,α, V q,α. Assume that e(x) > 2 κ and let Y be a closed discrete

16 1028 MADDALENA BONANZINGA subspace of X such that Y > 2 κ. For each {p, q} [Y ] 2, put {p, q} in R α if V p,α V q,α =. This defines a partition of [Y ] 2 into κ subsets. By Theorem 50, there exists Z Y such that Z > κ and [Z] 2 is contained in one of R α, say R α. Put U = {V p,α : p Z} {X \ Z}. Since Z is closed, U is an open cover of X. Moreover, for every z Z, V p,α is the only element of z Z that contains z. Let A X be such that st(a, U) = X. Since st(a, U) Z, A must intersect each V p,α. Since the sets V p,α are pairwise disjoint, we conclude that A Z > κ; a contradiction with the assumption st-l(x) κ. Theorem 58. For a 3-Hausdorff space X, e(x) 2 23-Hψ(X)st-l(X). Proof. Let 3-Hψ(X)st-l(X) κ. For every p X, enumerate a collection {V p,α : α < κ} of open sets such that if p, q, t are distinct points from X, there exist disjoint sets V p,α, V q,α, V t,α. Assume that e(x) > 2 2κ and let Y be a closed discrete subspace of X such that Y > 2 2κ. For each {p, q, t} [Y ] 3, put {p, q, t} in R α if V p,α V q,α V t,α =. This defines a partition of [Y ] 3 into κ subsets. By Theorem 52, there exists Z Y such that Z > κ and [Z] 3 is contained in one of R α, say R α. Put U = {V p,α : p Z} {X \ Z}. Since Z is closed, U is an open cover of X. Moreover, for every z Z, V p,α is the only element of z Z that contains z. Let A X be such that st(a, U) = X. Since st(a, U) Z, A must intersect each V p,α. Since each point from A can not be contained in more than two sets V p,α, we conclude that A Z > κ; a contradiction with the assumption st-l(x) κ. Corollary 59. For a 3-Hausdorff space X, e(x) 2 2χ(X)st-l(X). Recall the following result: Theorem 60. [8] For a T 1 space X, X 2 s(x)ψ(x). Then we have the following application of Proposition 20: Theorem 61. For every space X, X 2 s(x)ψ(x)h(x). Proof. By Proposition 20, if H(X) < X, there exists a T 1 subset Y of X such that Y = X. Then by Theorem 60, we have X = Y 2 s(y )ψ(y ) 2 s(x)ψ(x), hence X 2 s(x)ψ(x)h(x). Of course, if H(X) = X, X 2 s(x)ψ(x)h(x). Recall the following theorem: Theorem 62. [6] For a T 1 space X, X 2 e(x) (X). Then we have another application of Proposition 20:

17 ON THE HAUSDORFF NUMBER OF A TOPOLOGICAL SPACE 1029 Theorem 63. For every space X, X 2 e(x) (X)H(X). Problem 64. To find other applications of Proposition 20. Acknowledgements. The author expresses gratitude to Mikhail Matveev and Angelo Bella for useful communications. References [1] M. Bonanzinga, Star-Lindelöf and absolutely star-lindelöf spaces, Quest. Answ. Gen. Topol., 16 (1998) [2] M. Bonanzinga, F. Cammaroto, M. Matveev, On the Urysohn number of a topological space, Quaest. Math. 34 (2011), [3] M. Bell, J. Ginsburg, G. Woods, Cardinal inequalities for topological spaces involving the weak Lindelöf number, Pacific J. Math. 79 (1978) [4] Engelking R. Engelking, General Topology, Ed. Helderman Verlag, Berlin (1989). [5] P. Erdös, R. Rado, A partition calculus in set theory Bull. Amer. Math. Soc. 62 (1956) [6] J. Ginsburg, G. Woods, A cardinal inequalities for topological spaces involving closed discrete sets, Proc. Amer. Math. Soc. 64 (1977), [7] A. Hajnal, I.Juhasz, Discrete subsets of topological spaces, Indag. Math. 29 (1967) [8] R.E. Hodel, Cardinal functions I, Handbook of Set-Theoretic Topology, K. Kunen and J.E. Vaughan, Eds., Elsevier (1984) [9] R.E. Hodel, Combinatorial set theory and cardinal function inequalities, Proc. Americ. Math. Soc. 111 no. 2 (1991) [10] R.E. Hodel, Arhangelskii s solution to Alexandroff problem: A survey, Topol. And Appl. 153 (2006) [11] I. Juhàsz, Cardinal Functions in Topology, Math. Centre Tracts, Vol. 34, Math. Centrum, Amsterdam (1971). [12] I. Juhàsz, Cardinal Functions in Topology - Ten Years Later, Math. Centre Tracts, Vol. 123, Math. Centrum, Amsterdam (1980). [13] D. Làzàr, On a problem in the theory of aggregates, Composio Math. 3 (1936), 304. Zbl 14, 396. [14] Jech, Set theory, the third millennium edition, Springer (Erdos-Rado, p.111, Theorem 9.6). [15] M. V. Matveev, Pseudocompact and related spaces, Thesis, Moscow State University (1985). [16] V.V. Tkachuck, On cardinal invariants of Suslin number type, Soviet Math. Dokl. 27 (1983) no. 3, Received March 14, 2011 Revised version received June 8, 2011

18 1030 MADDALENA BONANZINGA Second revision received June 9, 2011 Dipartimento di Matematica e Informatica, Università di Messina, Via F. Stagno d Alcontres N.31, Messina (Italy) address: mbonanzinga@unime.it