THE COMPACTIFICABILITY CLASSES : THE BEHAVIOR AT INFINITY

We study the behavior of certain spaces and their compactificability classes at infinity. Among other results we show that every noncompact, locally compact, second countable Hausdorff space X such that each neighborhood of infinity (in the Alexandroff compactification) is uncountable, has (X) = (R). We also prove some criteria for (non-) comparability of the studied classes of mutual compactificability.

Throughout the paper we mostly use the standard topological notions as in [1] or [2], however with a few exceptions.Space always refers to topological space, usually considered without any additional separation axioms.Especially compactness is understood without the Hausdorff separation axiom.For the terminology related to θ-regularity, we refer the reader to [3,4].An ordinal is the set of smaller ordinals, and a cardinal is an initial ordinal.Let S be a set.The cardinality of S we denote by |S|.Let (X,τ) be a topological space.If we do not wish to specify its topology explicitly, we will sometimes, for our convenience, speak less precisely about the space X.Conversely, if we decide to specify the topology of a space X introduced in some previous steps, we will usually denote it by τ or τ X (in the case that we will work simultaneously with more topological spaces or more topologies on the same set).In a space X, a point x ∈ X is in the θclosure of a set A ⊆ X (x ∈ cl θ A) if every closed neighborhood of x intersects A. A filter base Φ in X has a θ-cluster point x ∈ X if x ∈ {cl θ F | F ∈ Φ}.We say that a space X is θ-regular if every filter base in X with a θ-cluster point has a cluster point.For more detailed characterization of θ-regularity, the reader is referred to [3][4][5].The points x, y in a space X are T 0 -separable if there is an open set containing only one of the points x, y.The points x, y are T 2 -separable if they have disjoint open neighborhoods.Let X be a space.Two disjoint sets A,B ⊆ X are said to be point-wise separated in X if every x ∈ A, y ∈ B are T 2 -separable in X.Several modifications of local compactness have been defined by various authors in the literature.In this paper, we say that a space is (strongly) locally compact if its every point has a compact (closed) neighborhood.One can easily check that a space is strongly locally compact if and only if it is θ-regular and locally compact.We will often use the following simple, but important property of strongly locally compact spaces.If X is strongly locally compact and γX is a compactification such that X and the remainder γX \ X are point-wise separated in γX, then X is open in γX.A filter in a space X is said to be ultra-closed if it is maximal among all filters in X having a base consisting of closed sets [1].By the Wallman compactification of X, we mean the set ωX = X ∪ {y | y is a nonconvergent ultra-closed filter in X}.The sets constitute an open base of ωX.For more detail, we refer the reader to [1].Some properties of the Wallman compactification of a θ-regular space were studied in [5].

Preliminaries and introduction
Let us recall some notions and results from the previous papers [6][7][8].Let (X,τ X ), (Y ,τ Y ) be spaces with X ∩ Y = ∅.We say that the space X is compactificable by the space Y or, in other words, X, Y are said to be mutually compactificable if there is a compact topology τ K on K = X ∪ Y such that the topologies on X, Y induced by τ K coincide with τ X , τ Y , respectively, and the sets X, Y are point-wise separated in (K,τ K ).Then we say that the topology τ K is Ꮿ-acceptable.Recall that mutually compactificable spaces are always θ-regular, and any two disjoint strongly locally compact spaces are always mutually compactificable [6].
Let Top be the class of all topological spaces.For any pair of two spaces X, Z, we define X ∼ Z if for every nonempty space Y disjoint from X, Z the space X is compactificable by Y if and only if Z is compactificable by Y .It can be easily seen that ∼ is reflexive, symmetric, transitive, and hence it is an equivalence relation.Let us denote by Ꮿ(X) the equivalence subclass of Top with respect to ∼ containing X and call it the compactificability class of X.For any spaces X, Z, we put Ꮿ(X) Ꮿ(Z) if for every nonempty space Y it holds: if the space X is compactificable by Y disjoint from X, Z, then Z is compactificable by Y .The relation is reflexive, antisymmetric, transitive, and hence it is an order relation between the compactificability classes.If for some spaces X, Z it holds Ꮿ(X) Ꮿ(Z) but Ꮿ(X) = Ꮿ(Z), we write Ꮿ(X) Ꮿ(Z).We proved in [7] that every compactificability class contains a T 1 representative, but there are compactificability classes with no Hausdorff representatives.
Let A = [1,∞), I = [0,1] be equipped with the Euclidean topology induced from R, and D = {0, 1} equipped with the discrete topology.By 0, we denote the constant function equal to 0. In [8], we proved that for any k,n ∈ N the spaces are of the same class of mutual compactificability.Also we proved that if X is a noncompact locally connected metrizable generalized continuum, then Ꮿ(X) = Ꮿ(R).All these spaces are uncountable but second countable, locally compact and Hausdorff.On the other hand, it can be proved that there exist spaces compactificable by R which are not compactificable by N, but not conversely.Hence, Ꮿ(N) Ꮿ(R).Note that we omit the proof now, because it will follow as a corollary from a theorem that will be presented in the next section.Further, it seems that connectedness does not affect the compactificability classes much because D ℵ0 is homeomorphic to a subspace of R, also known as the Cantor Discontinuum.Hence, there is a natural question whether it is true that every uncountable, second countable, or separable locally compact Hausdorff space must be of the same class of compactificability as R. Next we will give a counterexample for this conjecture.After some further investigation we will find out that the essential property which plays the most important role in determining the compactificability classes is the behavior at infinity.
Before we start, let us recapitulate a couple of theorems from the previous papers that we will need in our proofs in the next section.For the proofs of these result, we refer the reader to [7, Theorem 2.12] for Theorem 2.1, and to [8, Theorems 3.1 and 3.2, Corollary 3.2] for Theorems 2.2, 2.3, Corollary 2.1, respectively.Theorem 2.1.Let X be a T 3.5 space which is not locally compact and let Z be a strongly locally compact (not necessarily Hausdorff) space.Then Ꮿ(X) and Ꮿ(Z) are not comparable in the order .Theorem 2.2.Let (X,τ X ) be a closed subspace of a strongly locally compact space (Z,τ Z ).

Main results
Our first theorem studies what happens with the classes of mutual compactificability if a closed compact subspace is collapsed to a singleton.
Proof.Let (Z,τ Z ) be a space which is mutually compactificable by (X,τ), X ∩ Z = ∅.We put K = X ∪ Z and denote by τ K the Ꮿ-acceptable topology on K.We will show that We will extend the equivalence relation ∼ to K by setting Let f : K → K /∼ be the corresponding quotient mapping.We put L = K /∼ and consider L with its quotient topology τ L .For simplicity, we may identify the singleton equivalence classes with their elements, so the quotient mapping f : Then, W ∈ τ L because the quotient mapping is continuous.It follows that τ Y is weaker or equal to the topology on Y induced by τ L .Conversely, let W ∈ τ L and denote Let g : Y → X be the quotient mapping given by the original (not extended) equivalence ∼ on X.Then g is a restriction of f , so we have y ∈ P, z ∈ Q, and P ∩ Q = ∅.Let y = h.Since X, Z are point-wise separated in (K,τ K ) and H is compact, there exist U,V ∈ τ K such that H ⊆ U, z ∈ V , and Hence, Y and Z are point-wise separated in (L,τ L ).Finally, the compactness of (L,τ L ) follows from the continuity of the quotient mapping f : Proof.Clearly, [−1,0] is the closed compact subspace of X, so by the previous theorem we have Ꮿ(X) Ꮿ(N).The converse inequality follows from the fact that N is also a closed subspace of X, which is strongly locally compact, and from Theorem 2.2.
Martin Maria Kovár 5 Now we will study the behavior of spaces at infinity and its influence on the compactificability classes.For that purpose, we will need some auxiliary assertions.The following lemma is a variation on Cantor-Bendixson theorem (cf., [2, Problem 1.7.10-11,page 59]).Lemma 3.3.Let (X,τ) be a second countable space and let M ⊆ X be an uncountable set.Then there is a closed set C ⊆ X such that M \ C is (at most) countable and for every open set O ⊆ X, either Proof.Let σ ⊆ τ be a countable base for the topology τ.First, it is easy to see that if the condition stated in the theorem holds for every open basic set O ∈ σ, then it holds also for every open set from τ.Therefore, we may restrict our considerations to σ instead of τ.By transfinite induction, for some ordinal δ, we define a family {O α | α < δ} ⊆ σ as follows: ( and so (3.7) The set O 1 ∩ M is countable, from the previous steps (β) where β < α, we suppose O β ∩ M countable and from the current step (α) we know that Hence, by transfinite induction, O α ∩ M is countable as a countable union of countable sets.Then the set ∅ and from the last induction step (δ) it follows that O ∩ (M \ α<δ O α ) is uncountable, because otherwise we could set O δ = O, which would contradict to the fact that (δ) is the last step of the induction.But then also the set M ∩ O is uncountable and we can see that C has all the required properties.
6 The behavior at infinity The next lemma is a variation on Cantor's well-known result that D ℵ0 embeds into every nonempty perfect set of reals (cf., [2, Problem 3.12.11,page 230; Problem 4.5.5, page 290]).Lemma 3.4.Let (X,τ) be a locally compact, noncompact, second countable Hausdorff space, αX = X ∪ {∞} the Alexandroff compactification of X.If every neighborhood of ∞ is uncountable, then there exists an embedding e : D ℵ0 → αX which maps 0 to ∞.
Proof.By Lemma 3.3, there is a closed set C ⊆ αX such that every open set that intersects C is an uncountable set and such that X \ C is countable.Clearly, ∞ ∈ C. Thus, without loss of generality we may assume that C = αX.In other words, we may assume that every nonempty open subset of αX is uncountable.In particular, it means that no point in αX is isolated.
Since αX is second countable, it is metrizable.Let d : αX × αX → R be a metric on X, which induces the topology τ.For a point x ∈ X and a positive real number r > 0, we denote (3.9) Let D <ω = n∈ω D n be the set of all nonempty finite sequences whose members consist of 0's and 1's.Every sequence s ∈ D <ω of the length n may be extended by 0 or 1, respectively, to the sequence having the length n + 1.We denote the extended sequence by s 0 or s 1, respectively.For every s ∈ D <ω , we define inductively an open set O s ∈ τ as follows: (1) we put x 0 = ∞, and as x 1 we take any point from X.We also put r 0 = r 1 = (1/3)d(x 0 ,x 1 ), O 0 = B(x 0 ,r 0 ), O 1 = B(x 1 ,r 1 ).( 2) suppose that O s = B(x s ,r s ) is defined for every s ∈ D n .Let s ∈ D n .The point x s is not isolated, so there exists x s 1 ∈ B(x s ,2/3r s ), x s 1 = x s .Further, we put x s 0 = x s , r s 0 = r s 1 = 1/3d(x s 0 ,x s 1 ), O s 0 = B(x s 0 ,r s 0 ), O s 1 = B(x s 1 ,r s 1 ).Let p ∈ D ℵ0 .We denote p n the restriction of the infinite sequence p to the first n elements.From the construction it follows that Since the space αX is compact, the intersection n∈N cl αX O pn of the closed balls is nonempty and since lim n→∞ r pn = 0, it contains exactly one element, say e(p).In particular, e(0) = ∞.The mapping e : D ℵ0 → X is an injection.Indeed, let p, q ∈ D ℵ0 , p = q, and let n ∈ N be the least number for which p n = q n .We Finally, we can formulate and prove the main theorem.
So far, our effort was concentrated especially on the problem how to prove that the compactificability classes of two or more different spaces coincide.Perhaps it is just the right time to find some general conditions under which the compactificability classes of different spaces must differ.
Let T be some class of directed sets with ∅ / ∈ T. Define the following properties of a topological space X: (i) property α(T): there exists an injective net ϕ(A,≥) with We consider I with its natural, Euclidean topology.Let τ K be the product topology of the Wallman compactification ωX of (X,τ X ) with the Euclidean topology of I and let τ Y be the subspace topology on Y induced from (K,τ K ).Let a 1 ,a 2 ,..., b 1 ,b 2 ,... be two disjoint sequences in X having no cluster point in X.Let (Z,τ Z ) be a topological space disjoint from Y and let L = Y ∪ Z be equipped with a topology τ L such that (Y ,τ Y ), (Z,τ Z ) are subspaces of (L,τ L ).
Let c 1 ,c 2 ,... be a sequence in Z such that each c i is a cluster point of the nets α i (t) = (a i ,t), with their values in Y for t → 0. Then the topology τ L is not Ꮿ-acceptable.
Proof.Suppose that all the conditions stated in the lemma are satisfied and the topology τ L is Ꮿ-acceptable.We put where we identify the elements of ωX \ X with the nonconvergent ultra-closed filters in (X,τ X ).The set (V ) is open in ωX.Let h ∈ ωX \ X be a cluster point of the sequence a 1 ,a 2 ,.... Then F ∈ h, because otherwise, (X \ F) would be a neighborhood of h, which does not contain any element from a 1 ,a 2 ,....Then, Consequently, all cluster points of the sequence a 1 ,a 2 ,... are in (V ).Analogously, if g is a cluster point of the sequence b 1 ,b 2 ,..., then necessarily G ∈ g.Then V = X \ G / ∈ g, which gives g / ∈ (V ).Therefore, all cluster points of the sequence b 1 ,b 2 ,... are outside of (V ), that is, in ωX \ (V ).Then both sequences a 1 ,a 2 ,..., b 1 ,b 2 ,... have no common cluster point in ωX.
(3.12) Further, for every (k,t) ∈ M, we put The nets ϕ(M,≥), χ(M,≥) have their values in Y and they have no cluster point in X × {0}.Suppose that, for instance, ϕ(M,≥) has a cluster point z ∈ Z. Then there exists a net ϕ (M , ) finer than ϕ(M,≥), which converges to z.Since K is compact, ϕ (M , ) has a cluster point, say y ∈ K.But y is also a cluster point of ϕ(M, ), which has no cluster point in X × {0}, so y ∈ K \ (X × {0}) = Y .However, since τ L is Ꮿ-acceptable, Y and Z are point-wise separated in (L,τ L ).This is not possible since ϕ (M , ) converges to z ∈ Z and has a cluster point y ∈ Y , both in the topology τ L , which on Y coincides with the topology induced from (K,τ K ).Hence, the net ϕ(M,≥), and similarly also χ(M,≥), has no cluster point in Z.
Let w ∈ L be the cluster point of the sequence c 1 ,c 2 ,.... Take W ∈ τ L such that w ∈ W. Let (k 0 ,t 0 ) ∈ M.There exists k ≥ k 0 such that c k ∈ W.However, c k is a cluster point of the nets α k (t), β k (t) for t → 0. So, there exist t,s ∈ (0,1], We have (k,t) ≥ (k 0 ,t 0 ) and (k,s) ≥ (k 0 ,t 0 ), which mean that w ∈ L is a common cluster point of the nets ϕ(M,≥), χ(M,≥).

Because of the previous paragraph w /
Then w = (v,0), where v ∈ ωX \ X.But this implies that v is a common cluster point of the sequences a 1 ,a 2 ,..., b 1 ,b 2 ,..., which is a contradiction.
Thus our assumption that τ L is Ꮿ-acceptable is incorrect.
Proof.Suppose Ꮿ(X) Ꮿ(Z).We put I = [0,1] with the Euclidean topology, K = ωX × I, Y = K \ (X × {0}).We consider the product topology τ K on K and the topology τ Y induced from (K,τ K ) on Y .Clearly, the spaces (X,τ X ) and X × {0} with the topology induced from (K,τ K ) are homeomorphic.Since (X,τ X ) is θ-regular, ωX \ X and X are pairwise separated in ωX.Then, also X × {0} and Y are pairwise separated in (K,τ K ).Then Ꮿ(X × {0}) = Ꮿ(X) Ꮿ(Z).Hence, there exists a Ꮿ-acceptable topology τ L on L = Y ∩ Z, where Z is supposed to be disjoint from Y (if Y ∩ Z = ∅, we will replace Z by a homeomorphic copy, disjoint from Y ).Let ξ i (A i ,≥) be an injective net in P i such that A i ∈ T, having a limit l i ∈ P i .For any α ∈ A i and t ∈ (0,1], we define Suppose that ψ i is injective.Since (Z,τ Z ) has the property α(T), the net ψ i (A i ,≥) has no cluster point in Z, so it must have a cluster point in Y , say y ∈ Y .We denote M i = A i × (0,1] and for every (α,t),(β,s) ∈ M i , we define For every i ∈ N, we have a net ϕ i (M i , ) with values in Y .Let us show that ϕ i (M i , ) has the limit (l i ,0) ∈ X × {0} in (K,τ K ).Choose V ∈ τ X such that l i ∈ V and ε > 0. Then there exists α 0 ∈ A i such that ξ i (α) ∈ V for α α 0 .Then, for every (α,t) ∈ M i such that (α,t i ) (α 0 ,ε/2), we have But ψ i (β) is a cluster point of the net ζ i,β .Hence, there exists s ∈ (0,1], s ≤ t such that ϕ i (β,s) = ζ i,β (s) = (ξ i (β),s) ∈ U. Since the net ϕ i (M i , ) has its values in Y , we have ϕ i (β,s) ∈ W. Then y ∈ Y is a cluster point of ϕ i (M i , ), which is a contradiction, because in the previous step we proved that ϕ i (M i , ) has the limit (l i ,0) ∈ X × {0} and the sets X × {0}, Y are point-wise separated in (K,τ K ).Hence, the only possible conclusion is that ψ i is not injective.
By the previous paragraph, there exist α i ,β i ∈ A i such that α i = β i and ψ i (α i ) = ψ i (β i ).We put a i = ξ i (α i ), b i = ξ i (β i ), c i = ψ i (α i ) = ψ i (β i ).Since the net ξ i (A i ,≥) is injective, a i = b i .We have a i ,b i ∈ P i , so the sequences a 1 ,a 2 ,... and b 1 ,b 2 ,... are disjoint because of the discreteness of the collection {P 1 ,P 2 ,...}.For the same reason, they have no cluster point in X.Moreover, for every i ∈ N, c i = ψ i (α i ) = ψ i (β i ) ∈ Z is a cluster point of the nets ζ i,αi (t) = (ξ i (α i ),t) = (a i ,t), ζ i,βi (t) = (ξ i (β i ),t) = (b i ,t) for t → 0 in (L,τ L ).By Lemma 3.6, the topology τ L is not Ꮿ-acceptable, which is a contradiction.Therefore, it must be Ꮿ(X) Ꮿ(Z).
Modifying the technique of the proof of the previous theorem analogously, we can obtain the following result.Theorem 3.8.Let (X,τ X ) be a θ-regular T 1 space containing a discrete infinite sequence of subspaces P 1 ,P 2 ,... with the property |P i | > κ (κ is a cardinal number).Then for any space (Z,τ Z ) with the property |Z| ≤ κ it follows Ꮿ(X) Ꮿ(Z).
have e(p) ∈ cl αX O pn , e(q) ∈ cl αX O qn , but cl αX O pn ∩ cl αX O qn = ∅.Thus e(p) = e(q).Finally, we will show that e : D ℵ0 → X is continuous.Let O ∈ τ be an open set containing e(p).There is some m ∈ N such that e(p) ∈ O pm ⊆ O.The set U = {y | y ∈ D ℵ0 , y m = p m } is open in D ℵ0 since it is an intersection of m sub-basic sets of the product topology on D ℵ0 .But if y ∈ U, then e(y) ∈ cl αX O ym+1 ⊆ O ym = O pm ⊆ O. Hence, e : D ℵ0 → X is continuous.Since D ℵ0 is compact, e is a homeomorphism onto its image.
1) if for every O ∈ σ either M ∩ O is empty or uncountable, then C = cl X M has all the required properties and we are done.Otherwise, there exists O 1 ∈ σ such that O 1 ∩ M is nonempty and countable, (2) now, suppose that we have already chosen {O β | β < α} for some ordinal α.If there exists a basic open set O ∈ σ such that O ∩ (M \ β<α O β ) is nonempty and countable, we let O α be any such set O. Otherwise, we stop and put δ = α.Having the family {O α Y are pointwise separated in (K,τ K ), ζ i,α has no cluster point in Y .Since (L,τ L ) is compact, ζ i,α has a cluster point in Z, say ψ .14) If we consider (0,1] as a directed set, then ζ i,α is a net with values in Y and a limit (ξ i (α),0) ∈ X × {0} for t → 0. Since the net ζ i,α has a limit in X × {0} and the sets X × {0},