Topology and its Applications 168 (2014) 94–102
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
Sequential properties of lexicographic products F. Azarpanah, M. Etebar Department of Mathematics, Chamran University, Ahvaz, Iran
a r t i c l e
i n f o
MSC: 54F05 Keywords: Lexicographic product Almost P -point P + (P − )-point Sequential space Sequentially connected space Quasi F -space
a b s t r a c t In this article, using the characterization of almost P -points of a linearly ordered topological space (LOTS) in terms of sequences, we observe that in the category of linearly ordered topological spaces, quasi F -spaces and almost P -spaces coincide. This coincidence gives examples of quasi F -spaces with no F -points. We also use the characterization of sequentially connected LOTS in terms of almost P -points to show that whenever each LOTS Xn has first and last elements, the lexicographic product ∞ each n=1 Xn is sequentially connected if and only if each Xn is. Whenever X Xn is a LOTS without first and last elements, thenit is shown that ∞ n is n=1 always a sequential space. The lexicographic product α<ω1 Xα , where ω1 is the first uncountable ordinal, isalso investigated and it is shown that if each Xα contains at least two points, then α<ω1 Xα is always an almost P -space (a quasi F -space) but it is neither sequential nor sequentially connected. Using this lexicographic product, we give an example of a quasi F -space in which the set of F -points and the set of ω1 , does not have first and last non-F -points are dense. Whenever each Xα , α < elements, we show that the lexicographic product α<ω1 Xα is a P -space without isolated points. © 2014 Elsevier B.V. All rights reserved.
1. Introduction For each f in C(X), the ring of all real-valued continuous functions on a completely regular Hausdorff (Tychonoff) space X, Z(f ) = {x ∈ X: f (x) = 0} is called a zeroset and X \ Z(f ) is called a cozeroset. A point x in a completely regular Hausdorff space X is said to be a P -point (an almost P -point) if every Gδ -set or every zeroset containing x is a neighborhood of x (has a nonempty interior) and X is called a P -space (an almost P -space) if every point of X is a P -point (an almost P -point), see [1,6,8] for more details and properties of these spaces. A point x in a completely regular Hausdorff space X is said to be an F -point if the ideal Ox = {f ∈ C(X): x ∈ intβX clβX Z(f )} is prime, where βX is the Stone–Čech compactification of X. A space X is called an F -space if every point of βX is an F -point. It is well-known that a space X is an F -space if and only if every cozeroset is C ∗ -embedded in X, see Theorem 14.25 in [6]. A quasi F -space E-mail addresses:
[email protected] (F. Azarpanah),
[email protected] (M. Etebar). http://dx.doi.org/10.1016/j.topol.2014.02.011 0166-8641/© 2014 Elsevier B.V. All rights reserved.
F. Azarpanah, M. Etebar / Topology and its Applications 168 (2014) 94–102
95
is a space in which every dense cozeroset is C ∗ -embedded, see [3] for more properties and characterizations of quasi F -spaces. Throughout this paper, topological spaces under consideration are linearly ordered spaces and the reader is referred to [4] and [6] for undefined terms and notations. A linearly ordered topological space (LOTS) is a triple (X, τ, <), where < is a linear ordering of the set X and τ is the usual open interval topology defined by <. For every x in a LOTS X, x+ (x− ) denotes the immediate successor (predecessor) of x, if it exists and the set of all points y ∈ X satisfying y > x (y < x) is denoted by (x, →) ((←, x)). A subset S of a LOTS X is said to be cofinal if, for every x ∈ X, there exists s ∈ S such that s x. It is well-known that every LOTS is a normal Hausdorff space, see [4] and a LOTS is connected if and only if it is Dedekind-complete (i.e., every nonempty subset with an upper bound has a supremum or equivalently, every nonempty subset with a lower bound has an infimum) and it does not have consecutive elements, see [6, 3O]. Whenever (W, ) is a well-ordered set and for every α ∈ W , (Xα , α ) is a linearly ordered set, then the lexicographic product of the family {(Xα , α ): α ∈ W } is the set of all points x = (xα )α∈W with the order < defined as follows: if x = (xα )α∈W and y = (yα )α∈W have x = y, let σ be the first element of the set {α ∈ W : xα = yα } in the ordering ≺ and then define x < y provided xσ <σ yσ . In what follows, we shall omit putting subscripts on the various orderings because context will make clear which ordering is meant. The lexicographic product will be denoted by α∈W Xα . In this paper we consider two familiar well-ordered sets as index sets: ω1 , the set of all countable ordinals and N, the set of all natural numbers. We also assume that each factor space Xα in the lexicographic product α∈W Xα contains at least two elements. In a LOTS X, we call a point x a P + -point if every Gδ -set containing x contains an interval [x, y) for some x < y ∈ X. P − -points will be defined similarly. If we define O+ (x) = {f ∈ C(X): [x, y) ⊆ Z(f ), for some y ∈ X, x < y} and similarly O− (x) = {f ∈ C(X): (y, x] ⊆ Z(f ), for some y ∈ X, x > y}, then clearly x is a P + -point (P − -point) if and only if O+ (x) = Mx (O− (x) = Mx ), where Mx = {f ∈ C(X): f (x) = 0}. A LOTS X is said to be P + -space (P − -space) if every point of X is a P + -point (P − -point) and it is clear that a LOTS X is a P -space if and only if it is both a P + -space and a P − -space. The space of countable ordinals is an example of a P + -space which is not P − -space, whence it is not a P -space, see [4] and [6] for more details of the space of ordinals. A subset A of a topological space X is said to be sequentially open if whenever {xn } is a sequence in X that converges to a point in A, then {xn } is eventually in A. Similarly a subset B of X is called sequentially closed if whenever there is a sequence in B that converges to some point x ∈ X, then x ∈ B. We recall that a space X is a sequential space if every sequentially closed (open) subset of X is closed (open). It is well-known that a LOTS is sequential if and only if it is first countable, see [4] and [9]. A space X is said to be sequentially connected if X cannot be expressed as the union of two nonempty disjoint sequentially closed (open) sets. Our aim in this article is to reveal the importance of almost P -points and their role in characterizing of some sequential properties of lexicographic products. In Section 2, we give some preliminary results and cite some facts from [2]. By the characterization of almost P -points of a LOTS in terms of sequences, we observe that quasi F -spaces and almost P -spaces coincide in the category of linearly ordered spaces. Using the characterization of a sequential LOTS and a sequentially connected LOTS in terms of almost P -points, we observe that a LOTS is sequentially connected if and only if it is connected without any almost P -points. In Sections 3 and 4 we study the lexicographic product α∈W Xα , where the well-ordered set W is either the set of natural numbers N or the set of countable ordinals ω1 . We observe that the existence of almost P -points of the lexicographic product n∈N Xn depends on first and last elements of the factor spaces. Using this lexicographic product, we give examples of P + -spaces and P − -spaces in which the set of almost P -points is dense. Whenever the set of countable ordinals is considered as an index set, we will show that the lexicographic product is always an almost P -space and whenever each factor space does not have first and last elements, then this lexicographic product will be a P -space without isolated points. Sequential
F. Azarpanah, M. Etebar / Topology and its Applications 168 (2014) 94–102
96
connectedness of the lexicographic countable products is also investigated and it is shown that if each Xn ∞ has first and last elements, then n=1 Xn is sequentially connected if and only if each Xn is. Whenever ∞ each Xn is a LOTS without first and last elements, we show that the lexicographic product n=1 Xn is always a sequential space. Finally, using lexicographic products and the coincidence of quasi F -spaces and almost P -spaces in the category of linearly ordered spaces, we give an example of a quasi F -space (an almost P -space) without F -points (P -points) and an example of a quasi F -space (an almost P -space) in which the set of F -points (P -points) and the set of non-F -points (non-P -points) are dense. 2. Preliminaries In this section, using the characterization of almost P -points in terms of sequences, we observe that a LOTS is an almost P -space if and only if it is a quasi F -space. By the characterization of almost P -points in terms of sequences and the characterization of a sequentially connected LOTS in terms of almost P -points in this section together with the results in Section 3, enable us to prove the main results of the article in Section 4. In [6, 5O], it is shown that a point x of a LOTS is a P -point if and only if it is neither the limit of an increasing sequence nor the limit of a decreasing sequence. By the following lemma, we also characterize almost P -points of a LOTS in terms of sequences. We note that whenever x is not the first (last) point of a LOTS X and x is not the limit of an increasing (a decreasing) sequence in X and moreover x− (x+ ) does not exist, then every Gδ -set containing x, contains a nonempty interval (α, x] ([x, α)) for some α < x (α > x). On the other hand, for a point x ∈ X, whenever x− or x+ exists and x is both the limit of an increasing sequence and the limit of a decreasing sequence in X, then either x is an isolated point or the singleton {x} is a Gδ -set with empty interior. Using this argument the proof of the following result is evident. Lemma 2.1. A non-isolated point x in a LOTS X is an almost P -point if and only if one of the following statements holds. (a) x− does not exist and x is not the limit of an increasing sequence. (b) x+ does not exist and x is not the limit of a decreasing sequence. We note that whenever x is the first (last) element of X, then the condition (b) ( (a)) holds. We also cite the following lemma from [3] which is needed in the proof of the next theorem. Lemma 2.2. Let X be a completely regular Hausdorff quasi F -space and x ∈ X be a Gδ -point. Then x is not the limit of a distinct sequence in X. In particular, if X is a quasi F -space satisfying the first axiom of countability, then it is discrete. Now we prove the coincidence of quasi F -spaces and almost P -spaces in the category of linearly ordered spaces; for a different proof, see [2]. Theorem 2.3. A LOTS is an almost P -space if and only if it is a quasi F -space. Proof. Since the only dense cozeroset in an almost P -space X is X itself, clearly every almost P -space is a quasi F -space. Now let X be a quasi F -space and x ∈ X. If x+ and x− exist, then x is an isolated point and hence it is an almost P -point. Therefore suppose that one of x+ and x− does not exist, say x+ . If x is the limit of a decreasing sequence, then x is a Gδ -point and by Lemma 2.2, x is again an isolated point. Whenever x is not the limit of a decreasing sequence, then x is an almost P -point, by Lemma 2.1. Now
F. Azarpanah, M. Etebar / Topology and its Applications 168 (2014) 94–102
97
suppose that both x+ and x− do not exist. If x is the limit of a decreasing sequence and it is also the limit of an increasing sequence, then x is again a Gδ -point and hence x is an isolated point, by Lemma 2.2. Whenever one of these cases does not happen, x should be an almost P -point, by Lemma 2.1. Therefore X is an almost P -space and we are through. 2 Now using Theorem 2.3, examples of linearly ordered almost P -spaces with no P -points (see Example 2.9 in [2] and the example given in the first paragraph of Section 4 in [8]) are also examples of quasi F -spaces with no F -points. It is well-known that every first countable space is a Fréchet space (a space X is called a Fréchet space if for every A ⊆ X and every x ∈ clX A, there exists a sequence of points of A converging to x) and every Fréchet space is a sequential space, see [4, Theorem 1.6.14]. The converse is also true for any LOTS, i.e., every sequential LOTS is first countable, see [9]. It is also easy to see that a LOTS X is sequential if and only if every point of X is a Gδ -point. Now using these well-known facts and Lemma 2.1, the following proposition is evident. Proposition 2.4. A LOTS X is a sequential space if and only if every almost P -point of X is an isolated point. Clearly every disconnected space is sequentially disconnected, but not conversely. In fact a connected space may be sequentially disconnected. The following lemma characterizes connected sequentially disconnected linearly ordered spaces in terms of almost P -points, see [2] for the proof. Corollary 2.5. A LOTS X is sequentially connected if and only if X is connected without any almost P -point. Using Proposition 2.4 and Corollary 2.5, the following corollary is evident. Corollary 2.6. If a LOTS is sequentially connected, then it is sequential. 3. P -points and almost P -points in lexicographic products In this section we investigate almost P -points and P -points of the lexicographic products. We note that n the existence of almost P -points and P -points of the lexicographic finite product i=1 Xi only depends on n the last factor space Xn . In fact if x = (x1 , x2 , . . . , xn ) ∈ i=1 Xi , then x− (x+ ) exists if and only if x− n (x+ P -point (P -point) if and only if xn is an almost P -point (P -point). n ) exists in Xn and x is an almost ∞ But for a lexicographic product n=1 Xn , we will observe that the existence of almost P -points depends on first and last elements of the factor spaces. For lexicographic products α<ω1 Xα , we show that its every point is an almost P -point. First we need the following lemma. Lemma 3.1. Suppose that for every α < ω1 , Xα is a LOTS with at least two points. Let X be the lexicographic product α<ω1 Xα and for each x ∈ X, define F (x) = {α < ω1 : xα is not the first element of Xα } and L(x) = {α < ω1 : xα is not the last element of Xα }. (a) If F (x) is cofinal in ω1 , then x is not the first element of X, x− does not exist and x is not the supremum of any countable subset of (←, x). (b) If L(x) is cofinal in ω1 , then x is not the last element of X, x+ does not exist and x is not the infimum of any countable subset of (x, →). Proof. (a) Clearly x cannot be the first element of X, for F (x) is cofinal. Now suppose that C is a countable subset of (←, x). For each y ∈ C, there is δy < ω1 such that for every α < δy , yα = xα and yδy < xδy . Since
98
F. Azarpanah, M. Etebar / Topology and its Applications 168 (2014) 94–102
F (x) is cofinal and C is countable, there is some β ∈ F (x) with β > sup{δy : y ∈ C}. Since β ∈ F (x), we may choose zβ ∈ Xβ with zβ < xβ . Define zα = xα for each α ∈ ω1 \ {β}. Then y < z < x for every y ∈ C, so that x cannot be the supremum of the countable set C. A similar proof shows that x cannot have an immediate predecessor in X. (b) The proof is analogous. 2 We also need the following lemma in the sequel. Lemma 3.2. Suppose that for every n ∈ N, Xn is a LOTS with at least two points. Let X be the lexicographic ∞ product n=1 Xn and for each x ∈ X, define F (x) = {n ∈ N: xn is not the first element of Xn } and L(x) = {n ∈ N: xn is not the last element of Xn }. If F (x) (L(x)) is an infinite subset of N, then x is the limit of an increasing (a decreasing) sequence. Proof. Let L(x) be an infinite subset of N. Regard L(x) as an increasing subsequence {nk }∞ k=1 of integers ∞ and define a sequence f : N → n=1 Xn as follows: (f (k))n = xn , if n = nk and (f (k))n = ynk , if n = nk , where ynk ∈ Xnk and ynk > xnk . It is easy to see that f is a decreasing sequence and f → x. Whenever F (x) is infinite, the proof is similar. 2 Corollary 3.3. For each α < ω1 , let Xα be a LOTS with at least two points. Then every point of the lexicographic product α<ω1 Xα is an almost P -point. Proof. For any x ∈ X, at least one of the sets F (x) and L(x) defined in Lemma 3.1, is cofinal in ω1 . Hence each x ∈ X is an almost P -point by Lemmas 2.1 and 3.1. 2 Since linearly ordered quasi F -spaces and linearly ordered almost P -spaces coincide, and a point in a LOTS is an F -point if and only if it is a P -point, the following remark also gives an example of a quasi F -space in which both the set of F -points and the set of non-F -points are dense. Remark 3.4. If Xα is a LOTS, ∀α < ω1 , then the lexicographic product α<ω1 Xα is not necessarily a P -space. For example α<ω1 [0, 1] is an almost P -space by Corollary 3.3, but not every point of ω1 [0, 1] is a P -point. In fact both the set of P -points and the set of non-P -points of α<ω1 [0, 1] are dense. First we show that the set of P -points of α<ω1 [0, 1] is dense in α<ω1 [0, 1]. Let S=
(xα )α<ω1 ∈
[0, 1]: ∀β < ω1 , ∃δ > β such that xδ = 0, 1 .
α<ω1
For each x ∈ S, both F (x) and L(x) defined in Lemma 3.1 are cofinal. Hence by this lemma, x is neither the limit of an increasing nor the limit of a decreasing sequence. Now by 5O in [6], x is a P -point. On the other hand S is dense in α<ω1 [0, 1], for if a, b ∈ α<ω1 [0, 1], a < b, then there is smallest σ such that aσ = bσ , i.e., aσ < bσ . Now set x ∈ S such that aσ < xσ < bσ , xα = aα , ∀α < σ and xα ∈ (0, 1), ∀α > σ. Clearly x ∈ S ∩ (a, b), i.e., S is dense in α<ω1 [0, 1]. Next we consider the set T = (xα )α<ω1 ∈ [0, 1]: xδ ∈ (0, 1), and xβ = 0, ∀β > δ, for some δ < ω1 . α<ω1
We observe that every element of T is a non-P -point and T is dense in α<ω1 [0, 1]. It is easy to see that if x ∈ T , where xδ ∈ (0, 1) and xα = 0, ∀α > δ, then the sequence {xn } in α<ω1 [0, 1] such that xnδ = n−1 n xδ n and xα = xα , ∀α = δ converges to x. This means that x is not a P -point by 5O in [6]. Moreover, T is
F. Azarpanah, M. Etebar / Topology and its Applications 168 (2014) 94–102
99
dense in α<ω1 [0, 1], for if x < y and β is the smallest index such that xβ = yβ (xβ < yβ ), then we choose z ∈ α<ω1 [0, 1], where zα = xα , ∀α < β, xβ < zβ < yβ and zα = 0, ∀α > β. Clearly z ∈ T ∩ (x, y), i.e., T is dense in α<ω1 [0, 1]. The following proposition shows that
α<ω1
Xα may be a P -space.
Proposition 3.5. Let Xα be a LOTS with at least two points, ∀α < ω1 and F (L) be the set of all α < ω1 such that Xα does not have first (last) element. If F and L are cofinal, then α<ω1 Xα is a P -space without isolated points. Proof. Clearly F ⊆ F (x) and L ⊆ L(x), ∀x ∈ X. Thus by Lemma 3.1 and 5O in [6], X is a P -space. Now it remains to show that α<ω1 Xα has no consecutive elements. In fact if x, y ∈ α<ω1 Xα and x < y, then there exists a smallest index β < ω1 such that xβ = yβ and hence xβ < yβ . Since F is cofinal, there is λ ∈ F such that λ > β and yλ is not the first element of Xλ . Consider z ∈ α<ω1 Xα , so that zλ < yλ and zα = yα , ∀α < λ. Clearly x < z < y and this implies that α<ω1 Xα has no consecutive elements, whence it has no any isolated point. 2 The following corollary shows that
α<ω1
R is a P -space without isolated points.
Corollary 3.6. Let Xα be a LOTS, ∀α < ω1 and each Xα does not have first and last elements. Then α<ω1 Xα is a P -space without isolated points. Using a proof similar to that of Proposition 3.5, we obtain a more general result as follows: Corollary 3.7. Let Xα be a LOTS, ∀α < ω1 and F (L) be the set of all α < ω1 such that Xα does not have first (last) element. (a) If F is cofinal, then α<ω1 Xα is a P − -space without isolated points. (b) If L is cofinal, then α<ω1 Xα is a P + -space without isolated points. It is well-known that a finite product of (Tychonoff) P -spaces is a P -space. But no infinite product of (Tychonoff) spaces of more than one point is a P -space, see [6, 4K]. In Proposition 3.5, we observed that the lexicographic product α<ω1 Xα may be a P -space. But in the following proposition, we show that ∞ the lexicographic product n=1 Xn of linearly ordered spaces of more than one point does not have any P -point. Proposition 3.8. If Xn is a LOTS with at least two elements, ∀n ∈ N, then the lexicographic product ∞ X = n=1 Xn never has an almost P -point (a P -point). Proof. For each x ∈ X, clearly one of the sets F (x) and L(x) defined in Lemma 3.2 is infinite. Now by Lemma 3.2, x is the limit of a sequence and this means that x is not an almost P -point (a P -point). 2 ∞ Proposition 3.8 states that the lexicographic product n=1 Xn does not have any P -point even each Xn ∞ is a P -space (discrete). By part (a) of the following proposition, the lexicographic product n=1 Xn may not contain any almost P -point. Parts (b) and (c) of the following proposition show that whenever each factor space Xn is a P -space with first (last) element and without consecutive elements, then the set of ∞ almost P -points of the lexicographic product n=1 Xn is dense. Part (a) of this proposition also shows that the existence of first and last elements of the factor spaces is essential for existence of almost P -points of ∞ n=1 Xn .
100
F. Azarpanah, M. Etebar / Topology and its Applications 168 (2014) 94–102
Proposition 3.9. Let Xn be a LOTS with at least two points, ∀n ∈ N. (a) If F = {n ∈ N: Xn does not have first element} and L = {n ∈ N: Xn does not have last element} are ∞ infinite sets, then the lexicographic product n=1 Xn does not have any almost P -point. (b) If each Xn is a P + -space with last element and no element of Xn has immediate successor, then the ∞ set of almost P -points of the lexicographic product n=1 Xn is dense. (c) If each Xn is a P − -space with first element and no element of Xn has immediate predecessor, then the ∞ set of almost P -points of the lexicographic product n=1 Xn is dense. ∞ Proof. (a) For each x ∈ n=1 Xn , we have F ⊆ F (x) and L ⊆ L(x), hence F (x) and L(x) are infinite sets. Now by Lemma 3.2, x is the limit of increasing and decreasing sequences. This implies that x is not almost P -point by Lemma 2.1. ∞ (b) Let ln be the last element of Xn , ∀n ∈ N and consider A = {x ∈ n=1 Xn : xk = lk for some k ∈ ∞ N and xn = ln , ∀n > k}. We show that every point of A is an almost P -point and A is dense in n=1 Xn . + Let a ∈ A, where an = ln , ∀n > k and ak = lk . Since a+ k does not exist, a does not exist. On the other hand a is not the limit of a decreasing sequence. To see this let {y n } be a decreasing sequence and a < y n , n n ∀n ∈ N. Suppose mn is the smallest integer such that amn = ym (so, amn < ym ). Clearly mn k, n n ∀n ∈ N and this implies that there exists t k which is the smallest index such that at < ytnk for an infinite nk + sequence {nk }∞ at and hence there exists yt ∈ Xt such that for each k=1 . Since Xt is a P -space, yt ∞ nk n0 ∈ N, nk > n0 exists so that yt > yt > at . If we consider z ∈ n=1 Xn such that zt = yt and zn = an , ∀n = t, then a < z and for every n0 ∈ N, y nk > z for some nk > n0 , i.e., y n a. Now it remains to prove ∞ ∞ that A is dense in n=1 Xn . Let (b, c) be an interval in n=1 Xn and k be the smallest index such that ∞ bk = ck . So bk < ck and bi = ci , ∀i < k. We take x ∈ n=1 Xn , where xn = bn , ∀n < k, bk < xk < ck (note that no element of Xk has immediate successor) and xn = ln , ∀n > k. Thus x ∈ (b, c) ∩ A, i.e., A is dense ∞ in n=1 Xn . (c) The proof is similar. 2 4. Sequential properties of lexicographic products Theorem 2.2 in [7] shows that the countable Tychonoff product of sequentially connected spaces is sequentially connected. Theorem 4.1.1 in [5] also investigates the connectedness of lexicographic products. In this section, using characterization of a sequentially connected LOTS in terms of almost P -points in Corollary 2.5, we investigate the sequential connectedness of the lexicographic product spaces. First we need the following proposition. Although this proposition may be proved by Theorem 4.1.1 in [5], we give a direct proof. Proposition 4.1. Let Xn be a LOTS with first and last elements for each n ∈ N. Then the lexicographic ∞ product n=1 Xn is connected if and only if each Xn is. ∞ Proof. First suppose that n=1 Xn is connected and An is a nonempty subset of Xn . Let A be a subset of ∞ n=1 Xn containing the elements of the form (l1 , . . . , ln−1 , an , ln+1 , . . .), where an ∈ An and lm is the last element of Xm , ∀m = n. Since A is bounded above, it has a supremum. Let α = sup A, since α a, ∀a ∈ A, αi = li , ∀i < n and an αn , ∀an ∈ An . This implies that αn = sup An , i.e., Xn is Dedekind complete. Next, each Xn has no consecutive elements, for if an , bn ∈ Xn are consecutive elements of Xn and an < bn , ∞ then we may take x, y ∈ n=1 Xn such that xn = an , yn = bn , xm = ym , ∀m < n and xm = lm , ym = fm , ∞ ∀m > n, where fm is the first element of Xm . Then x and y are two consecutive elements of n=1 Xn which is a contradiction.
F. Azarpanah, M. Etebar / Topology and its Applications 168 (2014) 94–102
101
∞ Conversely, let each Xn be connected and A ⊆ n=1 Xn . Suppose that A1 is the set of all a1 ∈ X1 such that a1 is the first component of an element of A. Since A1 is bounded above and X1 is Dedekind complete, α1 = sup A1 exists. By induction, let Ak be the set of ak ∈ Xk such that α1 , α2 , . . . , αk−1 and ak be the first k components of an element of A and take αk = sup Ak . If Ak = ∅, ∀k ∈ N, then α = (α1 , . . . , αk , . . .) = sup A. In fact if β < α and n is the smallest index such that βn < αn (so βi = αi , ∀i < n), then we take a ∈ A such that α1 , . . . , αn are the first n components of a. Therefore β < a α, i.e., α = sup A. Now let Ak = ∅ for some integer k and n be the smallest integer such that An = ∅ (so, Am = ∅, ∀m < n). Consider α = (α1 , . . . , αn−1 , fn , fn+1 , . . .), where fm is the first element of Xm , ∀m n. We show that α = sup A. First, if a ∈ A, then a α, for otherwise, a > α implies that there is k < n (note that An = ∅) such that ak > αk and ai = αi , ∀i < k which contradicts αk = sup Ak . Next, let β < α and k be the smallest index such that βk = αk , so βi = αi , ∀i < k and βk < αk . Now either k = n − 1 or k < n − 1 for βm ≮ fm , ∀m n. If k = n − 1, then for all an−1 ∈ An−1 , we have an−1 < αn−1 and since αn−1 = sup An−1 , there exists a∗n−1 ∈ An−1 such that βn−1 < a∗n−1 < αn−1 (note that αn−1 ∈ / An−1 for An = ∅). Since An−1 = ∅, we may take a ∈ A whose first n − 1 components are α1 , . . . , αn−2 and a∗n−1 . Clearly β < a α, i.e., α = sup A. If k < n − 1, we can consider a ∈ A such that am = αm , ∀m < n − 1, for An−1 = ∅. Clearly β < a < α, whence we have α = sup A. 2 Now we prove our main theorem in this section. Theorem 4.2. Suppose that Xn is a LOTS with first and last elements for each n ∈ N. Then the lexicographic ∞ product n=1 Xn is sequentially connected if and only if each Xn is. ∞ Proof. By Proposition 4.1 and Corollary 2.5, it is enough to show that n=1 Xn contains a non-isolated almost P -point if and only if for some k ∈ N, Xk contains a non-isolated almost P -point. First suppose that for some k ∈ N, Xk contains an almost P -point xk ∈ Xk which is not isolated point. Then one of the conditions (a) and (b) of Lemma 2.1 holds, say (a). Hence x− k does not exist and xk is not the limit of an increasing sequence. Whenever fm is the first element of Xm , ∀m ∈ N, then x = (f1 , . . . , fk−1 , xk , fk+1 , . . .) ∞ is an almost P -point of n=1 Xn . In fact if y < x, and m is the smallest integer for which ym = xm , i.e., ym < xm , then m should be k. Hence yk < xk and this implies that x− does not exist for x− k does not exist. Now if {xn } is an increasing sequence which converges to x, then xn < x, ∀n ∈ N implies that xnk < xk , ∀n ∈ N and xnm = fm , ∀m < k. Thus {xnk } is an increasing sequence in Xk which converges to xk , a contradiction. In case part (b) of Lemma 2.1 holds for xk , then we take x = (l1 , . . . , lk−1 , xk , lk+1 , . . .), where lm is the last element of Xm , ∀m = k and the remainder of the proof is similar. ∞ Next, suppose that n=1 Xn has an almost P -point x = (x1 , . . . , xk , . . .) and let part (a) of Lemma 2.1 holds. Since x is not the limit of an increasing sequence, there exists k ∈ N such that xm = fm , ∀m > k. For otherwise if there is a subsequence {nk } of integers such that xnk is not the first element of Xnk , then ∞ we may define a sequence h : N → n=1 Xn , with (h(k))nk = fnk and (h(k))m = xm , ∀m = nk . Clearly h is increasing and h → x. To see h → x, let y < x and m be the smallest index such that ym = xm (i.e., ym < xm and yn = xn , ∀n < m). Then for nk > m, we have h(k) ∈ (y, x]. This contradicts our hypothesis and hence x is of the form x = (x1 , . . . , xk , fk+1 , . . .), where fm is the first element of Xm , ∀m > k. Now it is not hard to see that xk is an almost P -point of Xk . In case part (b) of Lemma 2.1 holds for x, then x will be of the form x = (x1 , . . . , xk , lk+1 , . . .) for some k ∈ N, where lm is the last element of Xm , ∀m > k and similarly xk will be an almost P -point of Xk . 2 In the proof of Theorem 4.2, we have shown that whenever each Xn is a LOTS with first and last elements, ∞ then the lexicographic product n=1 Xn has a non-isolated almost P -point if and only if some factor space ∞ Xk has. This indicates that the lexicographic product n=1 Xn is not sequential if and only if some factor ∞ space Xk is not sequential, by Proposition 2.4. Equivalently, the lexicographic product n=1 Xn is not first
102
F. Azarpanah, M. Etebar / Topology and its Applications 168 (2014) 94–102
countable if and only if some factor space is not first countable. Now as an immediate consequence of this fact, we have the following corollary. Corollary 4.3. Let Xn be a LOTS with first and last elements. Then the lexicographic product a sequential space if and only if each factor space Xn is.
∞ n=1
Xn is
∞ By Corollary 2.6, if the lexicographic product n=1 Xn is sequentially connected, then it is sequential. Whenever each factor space Xn is connected, the converse is also true. In fact, if the lexicographic product ∞ ∞ will be an isolated n=1 Xn is sequential, then every almost P -point of the lexicographic product n=1 Xn ∞ point by Proposition 2.4. But Proposition 3.8 states that the lexicographic product n=1 Xn does not ∞ contain any isolated point and this means that the lexicographic product n=1 Xn is sequentially connected by Corollary 2.5. Corollary 4.4. If each factor space Xn is a connected LOTS, then the lexicographic product sequential if and only if it is sequentially connected.
∞ n=1
Xn is
By Proposition 3.9(a) and Proposition 2.4, the following corollary is evident. Corollary 4.5. Let each Xn be a LOTS without first and last elements. Then the lexicographic product ∞ n=1 Xn is a sequential space. Corollary 4.6. The lexicographic product
α<ω1
Xα is never a sequential space.
Proof. By Corollary 3.3, every point of α<ω1 Xα is an almost P -point. Now consider x ∈ α<ω1 Xα such that for some λ < ω1 , xα is not the last element of Xα , ∀α > λ. If y ∈ α<ω1 Xα and x < y, then there is β < ω1 so that xβ < yβ and xα = yα , ∀α < β. Take σ < ω1 as the smallest element of the set {α < ω1 : α > β and xα is not the last element of Xα }. Now we consider z ∈ α<ω1 Xα such that zα = xα , ∀α < σ and zσ > xσ . Clearly x < z < y and this shows that x is not an isolated point of α<ω1 Xα which means by Proposition 2.4 that α<ω1 Xα is not sequential. 2 Finally, by Corollaries 2.6 and 4.6, the following result is evident. Corollary 4.7. The lexicographic product
α<ω1
Xα is never sequentially connected.
Acknowledgement We would like to thank the referee. His/her careful reading of the manuscript combined with some very useful comments have hopefully improved the quality of the paper. References [1] F. Azarpanah, O.A.S. Karamzadeh, A. Rezai Aliabad, z ◦ -ideals in C(X), Fundam. Math. 160 (1999) 15–25. [2] F. Azarpanah, M. Etebar, Linearly ordered P ± -spaces, JP J. Geom. Topol. 10 (2) (2010) 99–112. [3] F. Dashiell, A. Hager, M. Henriksen, Order-Cauchy completion of rings and vector lattices of continuous functions, Can. J. Math. XXXII (3) (1980) 657–685. [4] R. Engelking, General Topology, Heldermann Verlag, Berlin, 1989. [5] M.J. Faber, Metrizability in generalized ordered spaces, PhD dissertation, Vrije University te Amsterdam, Tract 53, Math. Centre Tracts, Amsterdam, 1974. [6] L. Gillman, M. Jerison, Rings of Continuous Functions, Springer, 1976. [7] Q. Huang, S. Lin, Notes on sequentially connected spaces, Acta Math. Hung. 110 (1–2) (2006) 159–164. [8] R. Levy, Almost P -spaces, Can. J. Math. 2 (1977) 284–288. [9] P.R. Meyer, Sequential properties of ordered topological spaces, Compos. Math. 21 (1969) 102–106.