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Scattered spaces and selections
Topology and its Applications 231 (2017) 306–315
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Topology and its Applications www.elsevier.com/locate/topol
Virtual Special Issue – In honor of Professor Yukihiro Kodama on his 85th birthday
Scattered spaces and selections Valentin Gutev Department of Mathematics, Faculty of Science, University of Malta, Msida MSD 2080, Malta
a r t i c l e
i n f o
Article history: Received 24 May 2016 Received in revised form 24 August 2016 Accepted 7 September 2016 Available online 2 October 2017 MSC: 54B20 54C65 54F05 54F65 54G12
a b s t r a c t If the Vietoris hyperspace F (X) of the nonempty closed subsets of a regular space X has a continuous zero-selection, then so does F (Z) for every nonempty Z ⊂ X. The present paper deals with the inverse problem showing that X is a scattered space provided F (Z) has a continuous selection for every nonempty countable Z ⊂ X. This is obtained by showing that a crowded regular space X contains a copy of the rational numbers provided its Vietoris hyperspace F (X) has a continuous selection. Some related problems and applications are discussed as well. © 2017 Elsevier B.V. All rights reserved.
Keywords: Vietoris topology Continuous selection Scattered space
1. Introduction All spaces in this paper are assumed T1 topological spaces. Let X be a space, and F (X) be the set of all nonempty closed subsets of X. A map f : F (X) → X is a selection for F (X) if f (S) ∈ S for every S ∈ F (X), and f is continuous if it is continuous with respect to the Vietoris topology τV on F (X). Recall that τV is generated by all collections of the form V = S ∈ F (X) : S ⊂ V and S ∩ V = ∅, whenever V ∈ V , where V runs over the finite families of open subsets of X. A space X is scattered if every nonempty subset A ⊂ X has an element p ∈ A which is isolated relative to A, i.e. p is an isolated point of A. Since each x ∈ A \ A is a non-isolated point of A, a space X is scattered if every nonempty closed subset of X has an isolated point. A selection f : F (X) → X is called E-mail address: [email protected]. https://doi.org/10.1016/j.topol.2017.09.023 0166-8641/© 2017 Elsevier B.V. All rights reserved.
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a zero-selection if f (S) is an isolated point of S for every S ∈ F (X). Evidently, every scattered space has a zero-selection, and every space which has a zero-selection is scattered. Continuous zero-selections imply some interesting properties. For instance, it was shown in [7] that a compact space X is an ordinal space if and only if F (X) has a continuous zero-selection. Here, X is an ordinal space if it is an ordinal equipped with the open interval topology. For some related results in the non-compact case, the interested reader is referred to [1,2,6,10,14]. In the present paper, we are interested in another aspect of zero-selections. Following Michael [18], let A (X) be the collection of all nonempty subsets of a set X. For a space X, one can endow A (X) with the Vietoris topology in precisely the same way as this was done for F (X) [18], see also the next section. However, the resulting hyperspace (A (X), τV ) is not so interesting. For instance, it is a folklore result that (A (X), τV ) is a T1 -space if and only if X is discrete. Despite of this, (A (X), τV ) plays an interesting role for zero-selections. Indeed, in the next section we will show that F (X) has a continuous zero-selection if and only if A (X) has a continuous selection, Theorem 2.1. The idea behind this result is simple. Namely, if f : F (X) → X is a zero-selection and S ∈ A (X), then f S ∈ S. The converse is also evident, and for a continuous selection g : A (X) → X, the value g(S) must be an isolated point of S ∈ A (X), Corollary 2.5. These considerations reveal a natural relationship between zero-selections and the closure operator c(S) = S ∈ F (X), for S ∈ A (X), see Theorem 2.1. An interesting aspect of this relationship is that to each zero-selection f for F (X) and Z ∈ A (X), one can associate the zero-selection f ◦ c F (Z) for F (Z), Corollary 2.3. In this construction, the continuity of the selection is preserved as far as X is regular, see Corollary 2.6. Thus, for a regular space X with a continuous zero-selection for F (X), each hyperspace F (Z) for Z ∈ A (X), has a continuous selection, Theorem 2.1. The substantial part of this paper deals with the inverse problem. Briefly, in Section 4 we show that a regular space X with a continuous selection for F (X) is scattered if and only if F (Z) has a continuous selection for each countable Z ∈ A (X), Theorem 4.1. This is based on an interesting result that each crowded regular space X with a continuous selection for F (X), contains a copy of the rational numbers, Theorem 3.1. In the last Section 5 of the paper, various disconnectedness-like properties are related to continuous selections for hyperspaces F (Z), for special subsets Z ⊂ X, see e.g. Corollary 5.3 and Theorem 5.4. 2. Continuous zero-selections For a space X, as in [18], we will consider A (X) endowed with the Vietoris topology τV , i.e. with the topology generated by the basic τV -open sets of the form V = S ∈ A (X) : S ⊂ V and S ∩ V = ∅, V ∈ V , for V being a finite family of open subsets of X. In this section, we are interested in the closure operator c
A (X) A −−→ c(A) = A ∈ F (X)
as a map from the hyperspace (A (X), τV ) to the hyperspace (F (X), τV ). This operator is naturally related to the existence of continuous zero-selections as the following theorem asserts. Theorem 2.1. Let X be a regular space. Then (a) F (X) has a continuous zero-selection if and only if A (X) has a continuous selection. (b) If f : F (X) → X is a continuous zero-selection, then f ◦ c F (Z) is a continuous zero-selection for F (Z), for every Z ∈ A (X).
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It should be mentioned that (b) of Theorem 2.1 was shown in [2, Proposition 2.10] under the condition that X is normal. In fact, the authors asked if normality can be dropped, so this question of [2] is now settled. To prepare for the proof of Theorem 2.1, we proceed with several simple observations regarding the closure operation c : A (X) → F (X). Proposition 2.2. A map f : F (X) → X is a zero-selection if and only if the composite map f ◦ c : A (X) → X is a selection for A (X). Proof. The map f ◦ c is a selection for A (X) if and only if f (c(S)) ∈ S for every S ∈ A (X). If f ◦ c is a selection for A (X) and p ∈ c(S \ {p}) for some p ∈ X and S ∈ A (X), then f (c(S)) = f (c(S \ {p})) = p. Hence, f ◦ c is a selection for A (X) precisely when f is a zero-selection for F (X). 2 If Z ⊂ X is a nonempty subset, then F (Z) ⊂ A (X). Hence, Proposition 2.2 implies the following immediate consequence. Corollary 2.3. Let f : F (X) → X be a zero-selection for a space X, and Z ∈ A (X). Then f ◦ c F (Z) is a zero-selection for F (Z). The closure operator also has the following natural property, it is essentially what we need to handle continuity of zero-selections. Proposition 2.4. If S ∈ A (X), then c(S) belongs to the τV -closure {S}
τV
of {S} in (A (X), τV ).
Proof. Indeed, if V is a basic τV -open set in (A (X), τV ) with c(S) ∈ V , then S ∈ V because V ∩S = ∅ for every V ∈ V . 2 The proof of Theorem 2.1 is now easily completed by combining the following two applications of Proposition 2.4. Corollary 2.5. If g : (A (X), τV ) → X is continuous, then g(c(S)) = g(S) for every S ∈ A (X). In particular, g F (X) is a continuous zero-selection for F (X) provided g : A (X) → X is a continuous selection for A (X). τV Proof. Take S ∈ A (X). By Proposition 2.4, c(S) ∈ {S} ⊂ g −1 g(S) because X is T1 . Accordingly, g(c(S)) = g(S). The second part now follows from Proposition 2.2. 2 Corollary 2.6. Let Y be a regular space and f : F (X) → Y be a continuous map. Then the map f ◦ c : A (X) → Y is also continuous. Proof. Let S ∈ A (X) and U ⊂ Y be an open set with f (c(S)) ∈ U . Since f is continuous, there exists a finite collection V of open subsets of X such that c(S) ∈ V and f (V ) ⊂ U . Consider V in A (X). If τV τV T ∈ V , then c(T ) ∈ {T } ⊂ V ⊂ f −1 U , by Proposition 2.4. Hence, f (c(T )) ∈ U which implies the continuity of f ◦ c because Y is regular. 2 It was shown in [18, Theorem 5.3] that the closure operator c : A (X) → F (X) is continuous provided X is normal. This now follows from Corollary 2.6. Namely, if X is normal, then (F (X), τV ) is regular [18, Theorem 4.9]. Then according to Corollary 2.6, c = idF (X) ◦ c : A (X) → F (X) is continuous, where idF (X) is the identity of F (X). In fact, normality is also a necessary condition.
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Proposition 2.7. A space X is normal if and only if c : A (X) → F (X) is continuous. Proof. Suppose that c is continuous, and take disjoint closed sets A, B ∈ F (X). For U = X \ B, we have that A ∈ U and, by the continuity of c, there exists a basic τV -open set V such that A ∈ V and c(T ) : T ∈ V ⊂ U . Since W = V ∈ V , it follows that c(W ) ∈ U . Then W is an open set with A ⊂ W ⊂ c(W ) ⊂ U = X \ B. Thus, X is normal. 2 Related to this, it was claimed in [18, Corollary 5.4] that for a space X and a nonempty subset Z ⊂ X, the map c F (Z) is an injective continuous map from F (Z) to F (X). However, this statement doesn’t seem to be true in general. Proposition 2.8. Let X be a first countable space such that c F (Z) is a continuous map from F (Z) to F (X), for every Z ∈ A (X). Then X is a normal space. Proof. Take disjoint closed sets A, B ∈ F (X) and set Z = X \B. Since c F (Z) is continuous and A ∈ Z, there exists a finite family V of open subsets of Z such that A ∈ V and c(S) ⊂ Z for every S ∈ V . Then V ∩ B = ∅. On the contrary, assume that there is x ∈ V ∩ B. Hence, there also exists a sequence {xn } ⊂ V with x = limn→∞ xn . Since x ∈ / Z, the set A ∪ {xn } is closed in Z and clearly A ∪ {xn } ∈ V . Accordingly, c(A ∪ {xn }) ⊂ Z which is impossible because x ∈ c(A ∪ {xn }) and x ∈ / Z. Thus, X must be normal. 2 3. Copies of the rationals A space X is dense in itself, often called also crowded, if it has no isolated points. Here, we are going to establish the following result. Theorem 3.1. If X is a crowded regular space with a continuous selection for F (X), then X contains a copy of the rational numbers. To prepare for the proof, let us recall that a sequence {An } ⊂ A (X) is convergent to a point p ∈ X if every neighbourhood W of p contains all but finitely many members of {An }. The following property represents the main idea of [11, Lemma 4.4]. Lemma 3.2. Let X be a regular space with a continuous selection f : F (X) → X. If f (X) is a non-isolated point, then X has a pairwise disjoint sequence of nonempty open sets convergent to some point q ∈ X. Proof. Since p = f (X) is a non-isolated point, by [11, Proposition 4.3], there exists a nonempty closed subset T0 ⊂ X with p ∈ / T0 and f (T0 ∪ {y}) = y for some y ∈ / T0 . Since f is continuous and X is regular, we may find a nonempty open set U0 ⊂ X such that U0 ⊂ X \ (T0 ∪ {p}) and f (T0 ∪ {x}) = x for every x ∈ U0 . Set H0 = T0 ∪ U0 , and use [11, Proposition 4.3] once more to find a closed subset T1 ⊂ X with H0 ⊂ T1 ⊂ X \ {p} and f (T1 ∪ {y}) = y for some y ∈ / T1 . Just like before, there exists a nonempty open set U1 ⊂ X such that U1 ⊂ X \ (T1 ∪ {p}) and f (T1 ∪ {x}) = x for every x ∈ U1 . Setting H1 = T1 ∪ U1 , the arguments can be extended by induction. Thus, there exists an increasing sequence {Tn : n < ω} of closed subsets, and a pairwise disjoint sequence {Un : n < ω} of nonempty open sets such that for every n < ω, (a) Tn ∪ Un ⊂ Tn+1 ⊂ X \ (Un+1 ∪ {p}), (b) f (Tn ∪ {x}) = x, for every x ∈ Un .
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Let Y = n<ω Tn , and q = f (Y ). We are going to show that {Un } is convergent to q. To this end, according to (a), we have that q = f (Y ) = f
{Tn : n < ω}
= lim f (Tn ). n→∞
Whenever yn ∈ Un , n < ω, it follows from (a) that Tn ⊂ Tn ∪ {yn } ⊂ Tn+1 , and so, by (b), q = lim f (Tn+1 ) = lim f (Tn ∪ {yn }) = lim yn . n→∞
n→∞
n→∞
(3.1)
This implies that the sequence Un , n < ω, is convergent to q. Indeed, if this fails, then q is contained in an open set W ⊂ X such that Un \ W = ∅ for infinitely many n < ω. Hence, there is a strictly increasing sequence {nk : k < ω} ⊂ ω with Unk \ W = ∅, for each k < ω. So, taking yn ∈ Un with ynk ∈ Unk \ W , for k < ω, it follows from (3.1) that q = limk→∞ ynk ∈ X \ W , which is clearly impossible because q ∈ W . The proof is complete. 2 To adapt Lemma 3.2 to the situation of Theorem 3.1, let us explicitly state the following consequence. Corollary 3.3. Let X be a regular crowded space with a continuous selection for F (X). Then every nonempty open set U ⊂ X contains a point q and a pairwise disjoint sequence {Un } of nonempty open sets which is convergent to q. Proof. Take a nonempty open set V ⊂ X with Y = V ⊂ U . Then Y remains regular and crowded; moreover, F (Y ) has a continuous selection being a subspace of F (X). Hence, by Lemma 3.2, there exists a pairwise disjoint sequence {Wn } of nonempty open subsets of Y convergent to some point q ∈ Y ⊂ U . Setting Un = V ∩ Wn , n < ω, we get a pairwise disjoint sequence {Un } of nonempty open subsets of U convergent to q. 2 Proof of Theorem 3.1. The proof follows an idea of Eric van Douwen [23]. Namely, we are going to construct a sequence {Un : n < ω} of countable pairwise disjoint open families in X and a sequence Qn = {qU ∈ U : U ∈ Un }, n < ω, of countable subsets of X such that for each n < ω, each U ∈ Un contains a sequence of elements of Un+1 convergent to qU ∈ U . If this is done, then Q = n<ω Qn is a countable crowded subset of X. Moreover, the collection n<ω Un defines a countable base for Q. Hence, by Urysohn’s metrization theorem [22], Q is metrizable being regular. Finally, by a result of Sierpiński [20], Q is homeomorphic to the rational numbers. Thus, it remains to construct these sequences. Since F (X) has a continuous selection and X is a regular crowded space, by Lemma 3.2, there exists a pairwise disjoint sequence U1 = {Un : n < ω} of nonempty open subsets of X convergent to some q ∈ X. Set Q0 = {q} and U0 = {X}. Whenever n < ω, by Corollary 3.3, there exists a pairwise disjoint sequence U2n = {Ukn : k < ω} of nonempty open subsets of Un convergent to some point qn ∈ Un . Take U2 = n<ω U2n and Q1 = {qn : n < ω}, and extend the construction in an obvious manner. 2 4. Many selections and scattered spaces Theorem 3.1 implies the following interesting result. Theorem 4.1. Let X be a regular space with a continuous selection for F (X). Then X is scattered if and only if F (Z) has a continuous selection for every countable Z ∈ A (X).
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Proof. Suppose that X is not a scattered space. Then it contains a closed crowded subset Y ⊂ X. Since F (Y ) has a continuous selection, by Theorem 3.1, it contains a copy Q of the rational numbers. However, F (Q) has no continuous selection [5, Theorem 6.1]. This shows the one direction. Conversely, suppose that X is scattered and Z ⊂ X is a nonempty countable set. Since Z is Lindelöf (being countable), it is paracompact. For the same reason, each point of Z is Gδ . Hence, by a result of Kolesnikov [17, Theorem 1], F (Z) has a continuous selection. 2 It should be remarked that in [17, Theorem 1], Kolesnikov proved that a paracompact scattered space Z, all points of which are Gδ , has a continuous selection for F (Z). The same result was also obtained in [8, Theorem 2.3], and was used in [8, Theorem 2.4] to show that a countable regular space Z has a continuous selection for F (Z) if and only if it is scattered. Earlier, a more general result was obtained by Kolesnikov [16, Theorem 2] that an Fσ -discrete paracompact space Z has a continuous selection for F (Z) if and only if Z is scattered. Finally, for countable spaces, a slight generalisation was obtained in [9, Theorem 3.2] that each countable space Z with a continuous selection for F (Z) is scattered. In this regard, let us remark that there exists a countable, first countable scattered Hausdorff space Z which has no continuous selection for F (Z) [8, Example 2.5]. In the rest of this section, we discuss the subtle difference between Theorems 2.1 and 4.1. To this end, recall that a selection f : F (X) → X is p-maximal for a point p ∈ X if f (S) = p for every S ∈ F (X) with p ∈ S. Similarly, f is called p-minimal if f (S) = p for every S ∈ F (X) with S = {p}. Finally, for spaces X and Y , we will use X ∨p=q Y to denote the wedge sum obtained by identifying the points p ∈ X and q ∈ Y to a single point in the disjoint union X Y . The following property was obtained in [9, Lemma 6.4]. Proposition 4.2 ([9]). Let X be a space which has a continuous p-maximal selection for some p ∈ X, and Y be a space which has a continuous q-minimal selection for some q ∈ Y . Then F (X ∨p=q Y ) also has a continuous selection. A point p ∈ X is called selection extreme if F (X) has either a continuous p-maximal selection, or a continuous p-minimal one. Since the ordinal space ω + 1 has both continuous ω-maximal and continuous ω-minimal selections for F (ω + 1), it follows from Proposition 4.2 that F (ω + 1) ∨ω=p X has a continuous selection for every space X which has a selection extreme point p ∈ X. For limit ordinals λ and μ, the wedge sum (λ + 1) ∨λ=μ (μ + 1) is often denoted by L(λ, μ), see [7]. This space is compact, scattered and orderable, but is an ordinal space if and only if both λ and μ are of a countable cofinality [7, Theorem 3]. We now have the following example. Example 4.1. The space L(ω, ω1 ) has no continuous zero-selection, but F (Z) has a continuous selection for every nonempty Z ⊂ L(ω, ω1 ). Proof. The space L(ω, ω1 ) is not an ordinal space because the cofinality of ω1 is not countable. Hence, it has no continuous zero-selection [7, Theorem 1]. Take a nonempty Z ⊂ L(ω, ω1 ), and let X = Z ∩ (ω + 1) and Y = Z ∩ (ω1 + 1). If Z is the topological sum of X and Y , then F (Z) has a continuous selection because both ω + 1 and ω1 + 1 have continuous zero-selections being ordinal spaces. Suppose that Z = X ∨p=q Y , where p = ω ∈ X \ {ω} and q = ω1 ∈ Y \ {ω1 }. Then F (X) has a continuous p-maximal selection and F (Y ) has a continuous q-minimal selection. By Proposition 4.2, F (Z) has a continuous selection. 2 In contrast, it was shown in [8, Example 4.1] that there exists a nonempty subset Z ⊂ L(ω1 , ω1 ) such that F (Z) has no continuous selection. Here is a similar example dealing with L(λ, μ) for uncountable regular cardinals λ < μ.
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Example 4.2. Let λ and μ be uncountable regular cardinals with λ < μ. Then some nonempty subset Z ⊂ L(λ, μ) has no continuous selection for F (Z). Proof. For convenience, let Δλ ⊂ λ and Δμ ⊂ μ be the isolated points of λ and, respectively, μ. Also, let p = λ = μ. Take Z = Δλ ∪ {p} ∪ Δμ ⊂ L(λ, μ), so that Z is a space with only one non-isolated point p ∈ Z. Suppose that F (Z) has a continuous selection f . Then f is a zero-selection. Indeed, if f (T ) is a non-isolated point for some T ∈ F (Z), then f (T ) = p and T contains a nontrivial convergent sequence [9, Proposition 4.4]. Evidently, this sequence is convergent to p, which is impossible because both λ and μ are with uncountable cofinality. Thus, f is a continuous zero-selection. We will show that this is impossible. To this end, observe that Z is a suborderable space when Δλ is equipped with the order of λ and Δμ — with the reverse one on μ. Thus, α ≺ p ≺ β for every α ∈ Δλ and β ∈ Δμ . Whenever α ≺ β, let [α, β] = {γ ∈ Z : α γ β} be the corresponding closed interval in Z. Then for each α ∈ Δλ we have that f ([α, p]) ≺ p. Hence, by the continuity of f , there exists βα ∈ Δμ such that f ([α, β]) ≺ p, for every β ∈ Δμ with β βα . Since λ < μ, there exists β ∈ Δμ with β βα for each α ∈ Δλ . Accordingly, we have f ([p, β]) = p because α f ([α, β]) ≺ p, for every α ∈ Δλ . Clearly, this is impossible. 2 These examples bring the natural question to characterise those scattered compact orderable spaces X with the property that each nonempty Z ⊂ X admits a continuous selection for F (Z). Here are some more specific questions based on the following simple observations. Proposition 4.3. Let X be an ordinal space and p ∈ X. Then F (X) has both a continuous p-maximal selection and a continuous p-minimal one. Proof. Suppose that X = μ for some ordinal μ, and p = λ < μ. Then X is the topological sum of λ + 1 = [0, λ] and [λ + 1, μ). Since F (λ + 1) has both λ-maximal and λ-minimal continuous selections, the property follows. 2 Proposition 4.4. Let X = L(λ, μ) for some limit ordinals λ and μ, and p ∈ X be the point at which λ and μ are identified. Then F (X) has a continuous p-maximal selection if and only if both λ and μ are of countable cofinality. Proof. Since X = (λ + 1) ∨λ=μ (μ + 1) and λ and μ are limit, U = λ and V = μ are disjoint open subsets of X with U ∩ V = {p}. For such special points, it was shown in [15, Theorem 3.1] that F (X) has a continuous p-maximal selection if and only if X is zero-dimensional and first countable at p. 2 Thus, we have the following particular question. Question 1. Let X be a compact scattered space such that F (X) has a continuous p-maximal selection for each p ∈ X. Then, is it true that X is an ordinal space? To emphasise on the distinction, let us remark that X = L(ω, ω1 ) has a continuous selection f for F (X) with f (X) = p, where p ∈ X is the point of identifying ω and ω1 [13, Lemma 2.3]. However, L(ω, ω1 ) has no continuous p-maximal selection because the cofinality of ω1 is not countable, Proposition 4.4. This suggests the following further question. Question 2. Let X = L(λ, μ) for some limit ordinals λ and μ, and p ∈ X be the point at which λ and μ are identified. If F (X) has a continuous selection f with f (X) = p, then is it true that one of the ordinals λ or μ is of countable cofinality?
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The question makes sense also in a more general setting. Question 3. Let X be a space and p ∈ X be such that X \ {p} is the union of two open sets U, V ⊂ X with U ∩ V = {p}. If F (X) has a continuous selection f with f (X) = p, then is it true that p is Gδ in one of the sets U or V ? 5. Disconnectedness-like properties If X is a compact space with the property that F (Z) has a continuous selection for each nonempty Z ⊂ X, then X is both scattered and ordered. Thus, X must be zero-dimensional. However, it is well known that there are completely regular scattered spaces which are not zero-dimensional, see [21]. This brings the following natural question. Question 4. Let X be a (completely) regular space with the property that F (Z) has a continuous selection for every Z ∈ A (X). Then, is it true that X is zero-dimensional? We conclude this paper discussing the role of special subsets Z ∈ A (X) in the setting of Question 4. To this end, let us recall that each selection f for a collection D ⊂ F (X) with F2 (X) = S ∈ F (X) : |S| ≤ 2 ⊂ D , generates an order like relation f on X defined by x f y if f ({x, y}) = x [18, Definition 7.1]. The relation f is emulating a linear order being both total and antisymmetric, but is not necessarily transitive. Motivated by this, we often write x ≺f y if x f y and x = y. Finally, recall that a point p ∈ X of a connected space X is cut if X \ {p} is not connected; otherwise, if X \ {p} is connected, the point is called noncut. Proposition 5.1. Let X be a connected space such that F (Z) has a continuous selection for every connected Z ∈ A (X) with |X \ Z| ≤ 1. Then X is compact. Proof. Taking Z = X, there exists a continuous selection f for F (X). Since X is connected, f is p-maximal for p = f (X), see [18, Lemma 7.3]. Moreover, f is a linear order on X and the order topology on X is coarser than the topology of X [18, Lemma 7.2], i.e. X is weakly orderable with respect to f . Therefore, p is a noncut point of X because p f x for every x ∈ X, see [12, Corollary 2.7]. So, Z = X \ {p} is connected and, by condition, F (Z) has a continuous selection g. Hence, just like before, q = g(Z) is a noncut point of Z and Z \ {q} is connected. We now have that p f x f q for every x ∈ X, see [12, Proposition 2.6]. Accordingly, the linear orders f and g are different on Z. However, Z is weakly orderable with respect to both f and g . Thus, by a result of Eilenberg [4], the order f is inverse to g on Z. Keeping this in mind, define another selection h : F (X) → X by h(S) = g(S \ {p}) for S ∈ F (X) with S = {p}, and h({p}) = p. By [18, Lemma 7.3], h(S) is the first (g )-element of S \ {p}, for each S ∈ F (X) with S = {p}. Since p is the last (f )-element of X and f is inverse to g on Z, we get that h(S) is the last (f )-element of S, for each S ∈ F (X). Since X is weakly orderable with respect to f , the selection h is continuous [18, Lemma 7.5.1]. Since h = f , the space X is compact [19, Theorem 1]. 2 Here are two natural consequences. Corollary 5.2. Let X be a space such that F (Z) has a continuous selection for every Z ∈ A (X) with |X \ Z| ≤ 1. Then each connected component of X is compact. Proof. Let Y be a connected component of X. Then Y is closed in X and, in particular, F (Y \ {y}) ⊂ F (X \ {y}) for every y ∈ Y . Hence, F (Z) has a continuous selection for every nonempty Z ⊂ Y with |Y \ Z| ≤ 1. By Proposition 5.1, Y must be compact. 2
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Corollary 5.3. Let X be a space such that F (Z) has a continuous selection for every Z ∈ A (X) with |X \ Z| ≤ 2. Then X is totally disconnected. Proof. According to [13, Theorem 4.1], each quasi-component of X is connected because F (X) has a continuous selection. Hence, it suffices to show that each connected component of X is a singleton. Suppose that Y ⊂ X is a nontrivial component. Then Y is compact by Corollary 5.2, hence it is also orderable being weakly orderable [18, Lemma 7.2]. Thus, Y has precisely 2 noncut points p, q ∈ Y . By condition, F (X \ {p, q}) has a continuous selection, hence so does F (Y \ {p, q}) because Y is closed in X. Let f be a continuous selection for F (Y \ {p, q}). Since Y \ {p, q} is connected, it has a first (f )-element r ∈ Y \ {p, q} [18, Lemma 7.3]. Hence, r is a noncut point of Y \ {p, q}, see e.g. [12, Corollary 2.7]. However, r is a cut point of Y \ {p, q} because Y \ {p, q} is open in Y and r is a cut point of Y . A contradiction! Thus, Y must be a singleton. 2 Recall that a family P of open subsets of X is a π-base for X, or a pseudobase, if each nonempty open subset of X contains a nonempty element of P . We now have also the following property. Theorem 5.4. Let X be a regular space such that F (Z) has a continuous selection for every nonempty open Z ⊂ X. Then X has a clopen π-base. Proof. Take a nonempty open set U ⊂ X. We are going to show that U contains a nonempty clopen set. By condition, F (U ) has a continuous selection f . Let p = f (U ) and V ⊂ X be an open set with p ∈ V ⊂ V ⊂ U . If p is isolated in U or U = V , then clearly U contains a nonempty clopen subset of X. So, suppose that p is a non-isolated point, and set F = U \ V = ∅. Since p = f (U ) ∈ / F , by [11, Proposition 4.3], there exists a closed in U subset T ⊂ U such that F ⊂ T and f (T ∪ {x}) = x for some x ∈ U \ T . Let W = z ∈ U : f (T ∪ {z}) = z . Then W is closed in U because f is continuous. Moreover, W \ T = ∅ is open in U because y ∈ W \ T implies f (T ∪ {y}) = y. Indeed, by the continuity of f , the point y has a neighbourhood O ⊂ U \ T with f (T ∪ {z}) = z for every z ∈ O, hence y ∈ O ⊂ W . If W ∩ T = ∅, then W is clopen in U and, evidently, W ⊂ U \ T ⊂ U \ F ⊂ V . Hence, W is also clopen in X. Suppose finally that W ∩ T = ∅. If q ∈ W ∩ T , then f (T ) = f (T ∪ {q}) = q and, therefore, W ∩ T is the singleton {q}. Since W \ T = ∅ and X is totally disconnected (by Corollary 5.3), q has a clopen neighbourhood G with H = W \ G = ∅. Just like before, H is a clopen in X. 2 If X is a completely metrizable strongly zero-dimensional space, then F (X) has a continuous selection [3,5]. However, complete metrizability is inherited on Gδ -sets; moreover, such sets remain strongly zerodimensional as well. Thus, F (Z) has a continuous selection for every nonempty Gδ -subset Z ⊂ X. This suggests the following natural question. Question 5. Let X be a (completely) metrizable space with the property that F (Z) has a continuous selection for every nonempty Gδ -subset Z ⊂ X. Then, is it true that X is strongly zero-dimensional? References [1] G. Artico, U. Marconi, Selections and topologically well-ordered spaces, Topol. Appl. 115 (2001) 299–303. [2] G. Artico, U. Marconi, J. Pelant, L. Rotter, M. Tkachenko, Selections and suborderability, Fundam. Math. 175 (2002) 1–33. [3] M. Choban, Many-valued mappings and Borel sets. I, Trans. Mosc. Math. Soc. 22 (1970) 258–280. [4] S. Eilenberg, Ordered topological spaces, Am. J. Math. 63 (1941) 39–45. [5] R. Engelking, R.W. Heath, E. Michael, Topological well-ordering and continuous selections, Invent. Math. 6 (1968) 150–158. [6] S. Fujii, Characterizations of ordinal spaces via continuous selections, Topol. Appl. 122 (2002) 143–150. [7] S. Fujii, T. Nogura, Characterizations of compact ordinal spaces via continuous selections, Topol. Appl. 91 (1999) 65–69. [8] S. Fujii, K. Miyazaki, T. Nogura, Vietoris continuous selections on scattered spaces, J. Math. Soc. Jpn. 54 (2) (2002) 273–281.
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[9] S. García-Ferreira, V. Gutev, T. Nogura, M. Sanchis, A. Tomita, Extreme selections for hyperspaces of topological spaces, Topol. Appl. 122 (2002) 157–181. [10] V. Gutev, Fell continuous selections and topologically well-orderable spaces II, in: Proceedings of the Ninth Prague Topological Symposium, 2001, Topology Atlas, Toronto, 2002, pp. 157–163 (electronic). [11] V. Gutev, Approaching points by continuous selections, J. Math. Soc. Jpn. (4) (2006) 1203–1210. [12] V. Gutev, Weak orderability of second countable spaces, Fundam. Math. 196 (3) (2007) 275–287. [13] V. Gutev, T. Nogura, Vietoris continuous selections and disconnectedness-like properties, Proc. Am. Math. Soc. 129 (2001) 2809–2815. [14] V. Gutev, T. Nogura, Fell continuous selections and topologically well-orderable spaces, Mathematika 51 (2004) 163–169. [15] V. Gutev, T. Nogura, Selection pointwise-maximal spaces, Topol. Appl. 146–147 (2005) 397–408. [16] O.N. Kolesnikov, Continuous selections, in: Continuous Functions on Topological Spaces, Latv. Gos. Univ., Riga, 1986, pp. 78–84 (Russian), iii–iv, xi. [17] O.N. Kolesnikov, Selections of multivalued mappings with values in scattered spaces, Sib. Mat. Zh. 27 (1) (1986) 70–78, 198. [18] E.A. Michael, Topologies on spaces of subsets, Trans. Am. Math. Soc. 71 (1951) 152–182. [19] T. Nogura, D. Shakhmatov, Characterizations of intervals via continuous selections, Rend. Circ. Mat. Palermo, Ser. II 46 (1997) 317–328. [20] W. Sierpiński, Sur un propriété topologique des ensembles dénombrables denses en soi, Fundam. Math. 1 (1920) 11–16. [21] R.C. Solomon, A scattered space that is not zero-dimensional, Bull. Lond. Math. Soc. 8 (3) (1976) 239–240. [22] P. Urysohn, Zum metrisationsproblem, Math. Ann. 94 (1925) 309–315. [23] E.K. van Douwen, Closed copies of the rationals, Comment. Math. Univ. Carol. 28 (1) (1987) 137–139.
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
Virtual Special Issue – In honor of Professor Yukihiro Kodama on his 85th birthday
Scattered spaces and selections Valentin Gutev Department of Mathematics, Faculty of Science, University of Malta, Msida MSD 2080, Malta
a r t i c l e
i n f o
Article history: Received 24 May 2016 Received in revised form 24 August 2016 Accepted 7 September 2016 Available online 2 October 2017 MSC: 54B20 54C65 54F05 54F65 54G12
a b s t r a c t If the Vietoris hyperspace F (X) of the nonempty closed subsets of a regular space X has a continuous zero-selection, then so does F (Z) for every nonempty Z ⊂ X. The present paper deals with the inverse problem showing that X is a scattered space provided F (Z) has a continuous selection for every nonempty countable Z ⊂ X. This is obtained by showing that a crowded regular space X contains a copy of the rational numbers provided its Vietoris hyperspace F (X) has a continuous selection. Some related problems and applications are discussed as well. © 2017 Elsevier B.V. All rights reserved.
Keywords: Vietoris topology Continuous selection Scattered space
1. Introduction All spaces in this paper are assumed T1 topological spaces. Let X be a space, and F (X) be the set of all nonempty closed subsets of X. A map f : F (X) → X is a selection for F (X) if f (S) ∈ S for every S ∈ F (X), and f is continuous if it is continuous with respect to the Vietoris topology τV on F (X). Recall that τV is generated by all collections of the form V = S ∈ F (X) : S ⊂ V and S ∩ V = ∅, whenever V ∈ V , where V runs over the finite families of open subsets of X. A space X is scattered if every nonempty subset A ⊂ X has an element p ∈ A which is isolated relative to A, i.e. p is an isolated point of A. Since each x ∈ A \ A is a non-isolated point of A, a space X is scattered if every nonempty closed subset of X has an isolated point. A selection f : F (X) → X is called E-mail address: [email protected]. https://doi.org/10.1016/j.topol.2017.09.023 0166-8641/© 2017 Elsevier B.V. All rights reserved.
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a zero-selection if f (S) is an isolated point of S for every S ∈ F (X). Evidently, every scattered space has a zero-selection, and every space which has a zero-selection is scattered. Continuous zero-selections imply some interesting properties. For instance, it was shown in [7] that a compact space X is an ordinal space if and only if F (X) has a continuous zero-selection. Here, X is an ordinal space if it is an ordinal equipped with the open interval topology. For some related results in the non-compact case, the interested reader is referred to [1,2,6,10,14]. In the present paper, we are interested in another aspect of zero-selections. Following Michael [18], let A (X) be the collection of all nonempty subsets of a set X. For a space X, one can endow A (X) with the Vietoris topology in precisely the same way as this was done for F (X) [18], see also the next section. However, the resulting hyperspace (A (X), τV ) is not so interesting. For instance, it is a folklore result that (A (X), τV ) is a T1 -space if and only if X is discrete. Despite of this, (A (X), τV ) plays an interesting role for zero-selections. Indeed, in the next section we will show that F (X) has a continuous zero-selection if and only if A (X) has a continuous selection, Theorem 2.1. The idea behind this result is simple. Namely, if f : F (X) → X is a zero-selection and S ∈ A (X), then f S ∈ S. The converse is also evident, and for a continuous selection g : A (X) → X, the value g(S) must be an isolated point of S ∈ A (X), Corollary 2.5. These considerations reveal a natural relationship between zero-selections and the closure operator c(S) = S ∈ F (X), for S ∈ A (X), see Theorem 2.1. An interesting aspect of this relationship is that to each zero-selection f for F (X) and Z ∈ A (X), one can associate the zero-selection f ◦ c F (Z) for F (Z), Corollary 2.3. In this construction, the continuity of the selection is preserved as far as X is regular, see Corollary 2.6. Thus, for a regular space X with a continuous zero-selection for F (X), each hyperspace F (Z) for Z ∈ A (X), has a continuous selection, Theorem 2.1. The substantial part of this paper deals with the inverse problem. Briefly, in Section 4 we show that a regular space X with a continuous selection for F (X) is scattered if and only if F (Z) has a continuous selection for each countable Z ∈ A (X), Theorem 4.1. This is based on an interesting result that each crowded regular space X with a continuous selection for F (X), contains a copy of the rational numbers, Theorem 3.1. In the last Section 5 of the paper, various disconnectedness-like properties are related to continuous selections for hyperspaces F (Z), for special subsets Z ⊂ X, see e.g. Corollary 5.3 and Theorem 5.4. 2. Continuous zero-selections For a space X, as in [18], we will consider A (X) endowed with the Vietoris topology τV , i.e. with the topology generated by the basic τV -open sets of the form V = S ∈ A (X) : S ⊂ V and S ∩ V = ∅, V ∈ V , for V being a finite family of open subsets of X. In this section, we are interested in the closure operator c
A (X) A −−→ c(A) = A ∈ F (X)
as a map from the hyperspace (A (X), τV ) to the hyperspace (F (X), τV ). This operator is naturally related to the existence of continuous zero-selections as the following theorem asserts. Theorem 2.1. Let X be a regular space. Then (a) F (X) has a continuous zero-selection if and only if A (X) has a continuous selection. (b) If f : F (X) → X is a continuous zero-selection, then f ◦ c F (Z) is a continuous zero-selection for F (Z), for every Z ∈ A (X).
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It should be mentioned that (b) of Theorem 2.1 was shown in [2, Proposition 2.10] under the condition that X is normal. In fact, the authors asked if normality can be dropped, so this question of [2] is now settled. To prepare for the proof of Theorem 2.1, we proceed with several simple observations regarding the closure operation c : A (X) → F (X). Proposition 2.2. A map f : F (X) → X is a zero-selection if and only if the composite map f ◦ c : A (X) → X is a selection for A (X). Proof. The map f ◦ c is a selection for A (X) if and only if f (c(S)) ∈ S for every S ∈ A (X). If f ◦ c is a selection for A (X) and p ∈ c(S \ {p}) for some p ∈ X and S ∈ A (X), then f (c(S)) = f (c(S \ {p})) = p. Hence, f ◦ c is a selection for A (X) precisely when f is a zero-selection for F (X). 2 If Z ⊂ X is a nonempty subset, then F (Z) ⊂ A (X). Hence, Proposition 2.2 implies the following immediate consequence. Corollary 2.3. Let f : F (X) → X be a zero-selection for a space X, and Z ∈ A (X). Then f ◦ c F (Z) is a zero-selection for F (Z). The closure operator also has the following natural property, it is essentially what we need to handle continuity of zero-selections. Proposition 2.4. If S ∈ A (X), then c(S) belongs to the τV -closure {S}
τV
of {S} in (A (X), τV ).
Proof. Indeed, if V is a basic τV -open set in (A (X), τV ) with c(S) ∈ V , then S ∈ V because V ∩S = ∅ for every V ∈ V . 2 The proof of Theorem 2.1 is now easily completed by combining the following two applications of Proposition 2.4. Corollary 2.5. If g : (A (X), τV ) → X is continuous, then g(c(S)) = g(S) for every S ∈ A (X). In particular, g F (X) is a continuous zero-selection for F (X) provided g : A (X) → X is a continuous selection for A (X). τV Proof. Take S ∈ A (X). By Proposition 2.4, c(S) ∈ {S} ⊂ g −1 g(S) because X is T1 . Accordingly, g(c(S)) = g(S). The second part now follows from Proposition 2.2. 2 Corollary 2.6. Let Y be a regular space and f : F (X) → Y be a continuous map. Then the map f ◦ c : A (X) → Y is also continuous. Proof. Let S ∈ A (X) and U ⊂ Y be an open set with f (c(S)) ∈ U . Since f is continuous, there exists a finite collection V of open subsets of X such that c(S) ∈ V and f (V ) ⊂ U . Consider V in A (X). If τV τV T ∈ V , then c(T ) ∈ {T } ⊂ V ⊂ f −1 U , by Proposition 2.4. Hence, f (c(T )) ∈ U which implies the continuity of f ◦ c because Y is regular. 2 It was shown in [18, Theorem 5.3] that the closure operator c : A (X) → F (X) is continuous provided X is normal. This now follows from Corollary 2.6. Namely, if X is normal, then (F (X), τV ) is regular [18, Theorem 4.9]. Then according to Corollary 2.6, c = idF (X) ◦ c : A (X) → F (X) is continuous, where idF (X) is the identity of F (X). In fact, normality is also a necessary condition.
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Proposition 2.7. A space X is normal if and only if c : A (X) → F (X) is continuous. Proof. Suppose that c is continuous, and take disjoint closed sets A, B ∈ F (X). For U = X \ B, we have that A ∈ U and, by the continuity of c, there exists a basic τV -open set V such that A ∈ V and c(T ) : T ∈ V ⊂ U . Since W = V ∈ V , it follows that c(W ) ∈ U . Then W is an open set with A ⊂ W ⊂ c(W ) ⊂ U = X \ B. Thus, X is normal. 2 Related to this, it was claimed in [18, Corollary 5.4] that for a space X and a nonempty subset Z ⊂ X, the map c F (Z) is an injective continuous map from F (Z) to F (X). However, this statement doesn’t seem to be true in general. Proposition 2.8. Let X be a first countable space such that c F (Z) is a continuous map from F (Z) to F (X), for every Z ∈ A (X). Then X is a normal space. Proof. Take disjoint closed sets A, B ∈ F (X) and set Z = X \B. Since c F (Z) is continuous and A ∈ Z, there exists a finite family V of open subsets of Z such that A ∈ V and c(S) ⊂ Z for every S ∈ V . Then V ∩ B = ∅. On the contrary, assume that there is x ∈ V ∩ B. Hence, there also exists a sequence {xn } ⊂ V with x = limn→∞ xn . Since x ∈ / Z, the set A ∪ {xn } is closed in Z and clearly A ∪ {xn } ∈ V . Accordingly, c(A ∪ {xn }) ⊂ Z which is impossible because x ∈ c(A ∪ {xn }) and x ∈ / Z. Thus, X must be normal. 2 3. Copies of the rationals A space X is dense in itself, often called also crowded, if it has no isolated points. Here, we are going to establish the following result. Theorem 3.1. If X is a crowded regular space with a continuous selection for F (X), then X contains a copy of the rational numbers. To prepare for the proof, let us recall that a sequence {An } ⊂ A (X) is convergent to a point p ∈ X if every neighbourhood W of p contains all but finitely many members of {An }. The following property represents the main idea of [11, Lemma 4.4]. Lemma 3.2. Let X be a regular space with a continuous selection f : F (X) → X. If f (X) is a non-isolated point, then X has a pairwise disjoint sequence of nonempty open sets convergent to some point q ∈ X. Proof. Since p = f (X) is a non-isolated point, by [11, Proposition 4.3], there exists a nonempty closed subset T0 ⊂ X with p ∈ / T0 and f (T0 ∪ {y}) = y for some y ∈ / T0 . Since f is continuous and X is regular, we may find a nonempty open set U0 ⊂ X such that U0 ⊂ X \ (T0 ∪ {p}) and f (T0 ∪ {x}) = x for every x ∈ U0 . Set H0 = T0 ∪ U0 , and use [11, Proposition 4.3] once more to find a closed subset T1 ⊂ X with H0 ⊂ T1 ⊂ X \ {p} and f (T1 ∪ {y}) = y for some y ∈ / T1 . Just like before, there exists a nonempty open set U1 ⊂ X such that U1 ⊂ X \ (T1 ∪ {p}) and f (T1 ∪ {x}) = x for every x ∈ U1 . Setting H1 = T1 ∪ U1 , the arguments can be extended by induction. Thus, there exists an increasing sequence {Tn : n < ω} of closed subsets, and a pairwise disjoint sequence {Un : n < ω} of nonempty open sets such that for every n < ω, (a) Tn ∪ Un ⊂ Tn+1 ⊂ X \ (Un+1 ∪ {p}), (b) f (Tn ∪ {x}) = x, for every x ∈ Un .
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Let Y = n<ω Tn , and q = f (Y ). We are going to show that {Un } is convergent to q. To this end, according to (a), we have that q = f (Y ) = f
{Tn : n < ω}
= lim f (Tn ). n→∞
Whenever yn ∈ Un , n < ω, it follows from (a) that Tn ⊂ Tn ∪ {yn } ⊂ Tn+1 , and so, by (b), q = lim f (Tn+1 ) = lim f (Tn ∪ {yn }) = lim yn . n→∞
n→∞
n→∞
(3.1)
This implies that the sequence Un , n < ω, is convergent to q. Indeed, if this fails, then q is contained in an open set W ⊂ X such that Un \ W = ∅ for infinitely many n < ω. Hence, there is a strictly increasing sequence {nk : k < ω} ⊂ ω with Unk \ W = ∅, for each k < ω. So, taking yn ∈ Un with ynk ∈ Unk \ W , for k < ω, it follows from (3.1) that q = limk→∞ ynk ∈ X \ W , which is clearly impossible because q ∈ W . The proof is complete. 2 To adapt Lemma 3.2 to the situation of Theorem 3.1, let us explicitly state the following consequence. Corollary 3.3. Let X be a regular crowded space with a continuous selection for F (X). Then every nonempty open set U ⊂ X contains a point q and a pairwise disjoint sequence {Un } of nonempty open sets which is convergent to q. Proof. Take a nonempty open set V ⊂ X with Y = V ⊂ U . Then Y remains regular and crowded; moreover, F (Y ) has a continuous selection being a subspace of F (X). Hence, by Lemma 3.2, there exists a pairwise disjoint sequence {Wn } of nonempty open subsets of Y convergent to some point q ∈ Y ⊂ U . Setting Un = V ∩ Wn , n < ω, we get a pairwise disjoint sequence {Un } of nonempty open subsets of U convergent to q. 2 Proof of Theorem 3.1. The proof follows an idea of Eric van Douwen [23]. Namely, we are going to construct a sequence {Un : n < ω} of countable pairwise disjoint open families in X and a sequence Qn = {qU ∈ U : U ∈ Un }, n < ω, of countable subsets of X such that for each n < ω, each U ∈ Un contains a sequence of elements of Un+1 convergent to qU ∈ U . If this is done, then Q = n<ω Qn is a countable crowded subset of X. Moreover, the collection n<ω Un defines a countable base for Q. Hence, by Urysohn’s metrization theorem [22], Q is metrizable being regular. Finally, by a result of Sierpiński [20], Q is homeomorphic to the rational numbers. Thus, it remains to construct these sequences. Since F (X) has a continuous selection and X is a regular crowded space, by Lemma 3.2, there exists a pairwise disjoint sequence U1 = {Un : n < ω} of nonempty open subsets of X convergent to some q ∈ X. Set Q0 = {q} and U0 = {X}. Whenever n < ω, by Corollary 3.3, there exists a pairwise disjoint sequence U2n = {Ukn : k < ω} of nonempty open subsets of Un convergent to some point qn ∈ Un . Take U2 = n<ω U2n and Q1 = {qn : n < ω}, and extend the construction in an obvious manner. 2 4. Many selections and scattered spaces Theorem 3.1 implies the following interesting result. Theorem 4.1. Let X be a regular space with a continuous selection for F (X). Then X is scattered if and only if F (Z) has a continuous selection for every countable Z ∈ A (X).
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Proof. Suppose that X is not a scattered space. Then it contains a closed crowded subset Y ⊂ X. Since F (Y ) has a continuous selection, by Theorem 3.1, it contains a copy Q of the rational numbers. However, F (Q) has no continuous selection [5, Theorem 6.1]. This shows the one direction. Conversely, suppose that X is scattered and Z ⊂ X is a nonempty countable set. Since Z is Lindelöf (being countable), it is paracompact. For the same reason, each point of Z is Gδ . Hence, by a result of Kolesnikov [17, Theorem 1], F (Z) has a continuous selection. 2 It should be remarked that in [17, Theorem 1], Kolesnikov proved that a paracompact scattered space Z, all points of which are Gδ , has a continuous selection for F (Z). The same result was also obtained in [8, Theorem 2.3], and was used in [8, Theorem 2.4] to show that a countable regular space Z has a continuous selection for F (Z) if and only if it is scattered. Earlier, a more general result was obtained by Kolesnikov [16, Theorem 2] that an Fσ -discrete paracompact space Z has a continuous selection for F (Z) if and only if Z is scattered. Finally, for countable spaces, a slight generalisation was obtained in [9, Theorem 3.2] that each countable space Z with a continuous selection for F (Z) is scattered. In this regard, let us remark that there exists a countable, first countable scattered Hausdorff space Z which has no continuous selection for F (Z) [8, Example 2.5]. In the rest of this section, we discuss the subtle difference between Theorems 2.1 and 4.1. To this end, recall that a selection f : F (X) → X is p-maximal for a point p ∈ X if f (S) = p for every S ∈ F (X) with p ∈ S. Similarly, f is called p-minimal if f (S) = p for every S ∈ F (X) with S = {p}. Finally, for spaces X and Y , we will use X ∨p=q Y to denote the wedge sum obtained by identifying the points p ∈ X and q ∈ Y to a single point in the disjoint union X Y . The following property was obtained in [9, Lemma 6.4]. Proposition 4.2 ([9]). Let X be a space which has a continuous p-maximal selection for some p ∈ X, and Y be a space which has a continuous q-minimal selection for some q ∈ Y . Then F (X ∨p=q Y ) also has a continuous selection. A point p ∈ X is called selection extreme if F (X) has either a continuous p-maximal selection, or a continuous p-minimal one. Since the ordinal space ω + 1 has both continuous ω-maximal and continuous ω-minimal selections for F (ω + 1), it follows from Proposition 4.2 that F (ω + 1) ∨ω=p X has a continuous selection for every space X which has a selection extreme point p ∈ X. For limit ordinals λ and μ, the wedge sum (λ + 1) ∨λ=μ (μ + 1) is often denoted by L(λ, μ), see [7]. This space is compact, scattered and orderable, but is an ordinal space if and only if both λ and μ are of a countable cofinality [7, Theorem 3]. We now have the following example. Example 4.1. The space L(ω, ω1 ) has no continuous zero-selection, but F (Z) has a continuous selection for every nonempty Z ⊂ L(ω, ω1 ). Proof. The space L(ω, ω1 ) is not an ordinal space because the cofinality of ω1 is not countable. Hence, it has no continuous zero-selection [7, Theorem 1]. Take a nonempty Z ⊂ L(ω, ω1 ), and let X = Z ∩ (ω + 1) and Y = Z ∩ (ω1 + 1). If Z is the topological sum of X and Y , then F (Z) has a continuous selection because both ω + 1 and ω1 + 1 have continuous zero-selections being ordinal spaces. Suppose that Z = X ∨p=q Y , where p = ω ∈ X \ {ω} and q = ω1 ∈ Y \ {ω1 }. Then F (X) has a continuous p-maximal selection and F (Y ) has a continuous q-minimal selection. By Proposition 4.2, F (Z) has a continuous selection. 2 In contrast, it was shown in [8, Example 4.1] that there exists a nonempty subset Z ⊂ L(ω1 , ω1 ) such that F (Z) has no continuous selection. Here is a similar example dealing with L(λ, μ) for uncountable regular cardinals λ < μ.
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Example 4.2. Let λ and μ be uncountable regular cardinals with λ < μ. Then some nonempty subset Z ⊂ L(λ, μ) has no continuous selection for F (Z). Proof. For convenience, let Δλ ⊂ λ and Δμ ⊂ μ be the isolated points of λ and, respectively, μ. Also, let p = λ = μ. Take Z = Δλ ∪ {p} ∪ Δμ ⊂ L(λ, μ), so that Z is a space with only one non-isolated point p ∈ Z. Suppose that F (Z) has a continuous selection f . Then f is a zero-selection. Indeed, if f (T ) is a non-isolated point for some T ∈ F (Z), then f (T ) = p and T contains a nontrivial convergent sequence [9, Proposition 4.4]. Evidently, this sequence is convergent to p, which is impossible because both λ and μ are with uncountable cofinality. Thus, f is a continuous zero-selection. We will show that this is impossible. To this end, observe that Z is a suborderable space when Δλ is equipped with the order of λ and Δμ — with the reverse one on μ. Thus, α ≺ p ≺ β for every α ∈ Δλ and β ∈ Δμ . Whenever α ≺ β, let [α, β] = {γ ∈ Z : α γ β} be the corresponding closed interval in Z. Then for each α ∈ Δλ we have that f ([α, p]) ≺ p. Hence, by the continuity of f , there exists βα ∈ Δμ such that f ([α, β]) ≺ p, for every β ∈ Δμ with β βα . Since λ < μ, there exists β ∈ Δμ with β βα for each α ∈ Δλ . Accordingly, we have f ([p, β]) = p because α f ([α, β]) ≺ p, for every α ∈ Δλ . Clearly, this is impossible. 2 These examples bring the natural question to characterise those scattered compact orderable spaces X with the property that each nonempty Z ⊂ X admits a continuous selection for F (Z). Here are some more specific questions based on the following simple observations. Proposition 4.3. Let X be an ordinal space and p ∈ X. Then F (X) has both a continuous p-maximal selection and a continuous p-minimal one. Proof. Suppose that X = μ for some ordinal μ, and p = λ < μ. Then X is the topological sum of λ + 1 = [0, λ] and [λ + 1, μ). Since F (λ + 1) has both λ-maximal and λ-minimal continuous selections, the property follows. 2 Proposition 4.4. Let X = L(λ, μ) for some limit ordinals λ and μ, and p ∈ X be the point at which λ and μ are identified. Then F (X) has a continuous p-maximal selection if and only if both λ and μ are of countable cofinality. Proof. Since X = (λ + 1) ∨λ=μ (μ + 1) and λ and μ are limit, U = λ and V = μ are disjoint open subsets of X with U ∩ V = {p}. For such special points, it was shown in [15, Theorem 3.1] that F (X) has a continuous p-maximal selection if and only if X is zero-dimensional and first countable at p. 2 Thus, we have the following particular question. Question 1. Let X be a compact scattered space such that F (X) has a continuous p-maximal selection for each p ∈ X. Then, is it true that X is an ordinal space? To emphasise on the distinction, let us remark that X = L(ω, ω1 ) has a continuous selection f for F (X) with f (X) = p, where p ∈ X is the point of identifying ω and ω1 [13, Lemma 2.3]. However, L(ω, ω1 ) has no continuous p-maximal selection because the cofinality of ω1 is not countable, Proposition 4.4. This suggests the following further question. Question 2. Let X = L(λ, μ) for some limit ordinals λ and μ, and p ∈ X be the point at which λ and μ are identified. If F (X) has a continuous selection f with f (X) = p, then is it true that one of the ordinals λ or μ is of countable cofinality?
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The question makes sense also in a more general setting. Question 3. Let X be a space and p ∈ X be such that X \ {p} is the union of two open sets U, V ⊂ X with U ∩ V = {p}. If F (X) has a continuous selection f with f (X) = p, then is it true that p is Gδ in one of the sets U or V ? 5. Disconnectedness-like properties If X is a compact space with the property that F (Z) has a continuous selection for each nonempty Z ⊂ X, then X is both scattered and ordered. Thus, X must be zero-dimensional. However, it is well known that there are completely regular scattered spaces which are not zero-dimensional, see [21]. This brings the following natural question. Question 4. Let X be a (completely) regular space with the property that F (Z) has a continuous selection for every Z ∈ A (X). Then, is it true that X is zero-dimensional? We conclude this paper discussing the role of special subsets Z ∈ A (X) in the setting of Question 4. To this end, let us recall that each selection f for a collection D ⊂ F (X) with F2 (X) = S ∈ F (X) : |S| ≤ 2 ⊂ D , generates an order like relation f on X defined by x f y if f ({x, y}) = x [18, Definition 7.1]. The relation f is emulating a linear order being both total and antisymmetric, but is not necessarily transitive. Motivated by this, we often write x ≺f y if x f y and x = y. Finally, recall that a point p ∈ X of a connected space X is cut if X \ {p} is not connected; otherwise, if X \ {p} is connected, the point is called noncut. Proposition 5.1. Let X be a connected space such that F (Z) has a continuous selection for every connected Z ∈ A (X) with |X \ Z| ≤ 1. Then X is compact. Proof. Taking Z = X, there exists a continuous selection f for F (X). Since X is connected, f is p-maximal for p = f (X), see [18, Lemma 7.3]. Moreover, f is a linear order on X and the order topology on X is coarser than the topology of X [18, Lemma 7.2], i.e. X is weakly orderable with respect to f . Therefore, p is a noncut point of X because p f x for every x ∈ X, see [12, Corollary 2.7]. So, Z = X \ {p} is connected and, by condition, F (Z) has a continuous selection g. Hence, just like before, q = g(Z) is a noncut point of Z and Z \ {q} is connected. We now have that p f x f q for every x ∈ X, see [12, Proposition 2.6]. Accordingly, the linear orders f and g are different on Z. However, Z is weakly orderable with respect to both f and g . Thus, by a result of Eilenberg [4], the order f is inverse to g on Z. Keeping this in mind, define another selection h : F (X) → X by h(S) = g(S \ {p}) for S ∈ F (X) with S = {p}, and h({p}) = p. By [18, Lemma 7.3], h(S) is the first (g )-element of S \ {p}, for each S ∈ F (X) with S = {p}. Since p is the last (f )-element of X and f is inverse to g on Z, we get that h(S) is the last (f )-element of S, for each S ∈ F (X). Since X is weakly orderable with respect to f , the selection h is continuous [18, Lemma 7.5.1]. Since h = f , the space X is compact [19, Theorem 1]. 2 Here are two natural consequences. Corollary 5.2. Let X be a space such that F (Z) has a continuous selection for every Z ∈ A (X) with |X \ Z| ≤ 1. Then each connected component of X is compact. Proof. Let Y be a connected component of X. Then Y is closed in X and, in particular, F (Y \ {y}) ⊂ F (X \ {y}) for every y ∈ Y . Hence, F (Z) has a continuous selection for every nonempty Z ⊂ Y with |Y \ Z| ≤ 1. By Proposition 5.1, Y must be compact. 2
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Corollary 5.3. Let X be a space such that F (Z) has a continuous selection for every Z ∈ A (X) with |X \ Z| ≤ 2. Then X is totally disconnected. Proof. According to [13, Theorem 4.1], each quasi-component of X is connected because F (X) has a continuous selection. Hence, it suffices to show that each connected component of X is a singleton. Suppose that Y ⊂ X is a nontrivial component. Then Y is compact by Corollary 5.2, hence it is also orderable being weakly orderable [18, Lemma 7.2]. Thus, Y has precisely 2 noncut points p, q ∈ Y . By condition, F (X \ {p, q}) has a continuous selection, hence so does F (Y \ {p, q}) because Y is closed in X. Let f be a continuous selection for F (Y \ {p, q}). Since Y \ {p, q} is connected, it has a first (f )-element r ∈ Y \ {p, q} [18, Lemma 7.3]. Hence, r is a noncut point of Y \ {p, q}, see e.g. [12, Corollary 2.7]. However, r is a cut point of Y \ {p, q} because Y \ {p, q} is open in Y and r is a cut point of Y . A contradiction! Thus, Y must be a singleton. 2 Recall that a family P of open subsets of X is a π-base for X, or a pseudobase, if each nonempty open subset of X contains a nonempty element of P . We now have also the following property. Theorem 5.4. Let X be a regular space such that F (Z) has a continuous selection for every nonempty open Z ⊂ X. Then X has a clopen π-base. Proof. Take a nonempty open set U ⊂ X. We are going to show that U contains a nonempty clopen set. By condition, F (U ) has a continuous selection f . Let p = f (U ) and V ⊂ X be an open set with p ∈ V ⊂ V ⊂ U . If p is isolated in U or U = V , then clearly U contains a nonempty clopen subset of X. So, suppose that p is a non-isolated point, and set F = U \ V = ∅. Since p = f (U ) ∈ / F , by [11, Proposition 4.3], there exists a closed in U subset T ⊂ U such that F ⊂ T and f (T ∪ {x}) = x for some x ∈ U \ T . Let W = z ∈ U : f (T ∪ {z}) = z . Then W is closed in U because f is continuous. Moreover, W \ T = ∅ is open in U because y ∈ W \ T implies f (T ∪ {y}) = y. Indeed, by the continuity of f , the point y has a neighbourhood O ⊂ U \ T with f (T ∪ {z}) = z for every z ∈ O, hence y ∈ O ⊂ W . If W ∩ T = ∅, then W is clopen in U and, evidently, W ⊂ U \ T ⊂ U \ F ⊂ V . Hence, W is also clopen in X. Suppose finally that W ∩ T = ∅. If q ∈ W ∩ T , then f (T ) = f (T ∪ {q}) = q and, therefore, W ∩ T is the singleton {q}. Since W \ T = ∅ and X is totally disconnected (by Corollary 5.3), q has a clopen neighbourhood G with H = W \ G = ∅. Just like before, H is a clopen in X. 2 If X is a completely metrizable strongly zero-dimensional space, then F (X) has a continuous selection [3,5]. However, complete metrizability is inherited on Gδ -sets; moreover, such sets remain strongly zerodimensional as well. Thus, F (Z) has a continuous selection for every nonempty Gδ -subset Z ⊂ X. This suggests the following natural question. Question 5. Let X be a (completely) metrizable space with the property that F (Z) has a continuous selection for every nonempty Gδ -subset Z ⊂ X. Then, is it true that X is strongly zero-dimensional? References [1] G. Artico, U. Marconi, Selections and topologically well-ordered spaces, Topol. Appl. 115 (2001) 299–303. [2] G. Artico, U. Marconi, J. Pelant, L. Rotter, M. Tkachenko, Selections and suborderability, Fundam. Math. 175 (2002) 1–33. [3] M. Choban, Many-valued mappings and Borel sets. I, Trans. Mosc. Math. Soc. 22 (1970) 258–280. [4] S. Eilenberg, Ordered topological spaces, Am. J. Math. 63 (1941) 39–45. [5] R. Engelking, R.W. Heath, E. Michael, Topological well-ordering and continuous selections, Invent. Math. 6 (1968) 150–158. [6] S. Fujii, Characterizations of ordinal spaces via continuous selections, Topol. Appl. 122 (2002) 143–150. [7] S. Fujii, T. Nogura, Characterizations of compact ordinal spaces via continuous selections, Topol. Appl. 91 (1999) 65–69. [8] S. Fujii, K. Miyazaki, T. Nogura, Vietoris continuous selections on scattered spaces, J. Math. Soc. Jpn. 54 (2) (2002) 273–281.
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