() @ Applied General Topology c© Universidad Politécnica de Valencia Volume 12, no. 2, 2011 pp. 81-94 Extensions defined using bornologies Alessandro Caterino and M. Cristina Vipera Abstract Many extensions of a space X such that the remainder Y is closed can be constructed as B-extensions, that is, by defining a topology on the disjoint union X ∪ Y , provided there exists a map, satisfying some conditions, from a basis of Y into the family of the subsets of X which are “unbounded” with respect to a given bornology in X. We give a first example of a (nonregular) extension with closed remain- der which cannot be obtained as B-extension. Extensions with closed discrete remainders and extensions whose remainders are retract are mostly considered. We answer some open questions about separation properties and metrizability of B-extensions. 2010 MSC: 54D35, 54D10, 54D20. Keywords: Boundedness, bornology, topological extension, B-extension. 1. Introduction. The construction of the one-point compactification of a locally compact space can be generalized by taking as open neighborhoods of the new point the complements of the closed members of any boundedness. A nonempty family FX of subsets of a space X is said to be a boundedness if it is closed with respect to subsets and finite unions. FX is said to be closed (open) if every bounded set, that is, every member of FX, is contained in a closed (resp. open) bounded set ([10]). The one-point extension naturally associated to a boundedness FX in X is denoted by o(FX). In order to obtain T1 one-point extensions of a T1 space X we need that all singletons of X are bounded. In this case the boundedness is also called bornology. It turns out that all possible T1 one-point extensions of X can be defined in this way. In fact, every extension aX of X determines a closed bornology HX(aX) 82 A. Caterino and M. C. Vipera in X, namely the family of sets whose closure in X is also closed in aX. If aX is a one-point extension, then aX is equivalent to o(HX(aX)). If X is Hausdorff, then o(FX) will be Hausdorff provided every point has a bounded closed neighborhood. It is known that every Hausdorff n-point compactification X ∪{x1. . . . , xn} can be obtained associating to every xi an open non-relatively compact subset Ui of X, where Ui∩Uj = ∅ and X \( ⋃ Ui) is compact. Using a suitable bornology, an analogous result was proved for Hausdorff n-point extensions ([5]). It is natural to try a generalization of this kind of construction to other Haus- dorff extensions. The question is whether it is possible to obtain every extension aX of X, in which X is open, using a bornology in X and some kind of cor- respondence between a basis of the remainder Y = aX \ X and the family of unbounded open subsets of X. This idea inspired the construction of the so-called ESH-compactifications, first defined in [4]. The authors used an essential semilattice homomorphism π from a basis of a compact space K into the family of nonrelatively compact open subsets of a locally compact space X to obtain a compactification of X whose remainder is homeomorphic to K. Large families of compactifications (in most cases even the Stone-Čech compactification) can be obtained in this way. The word “essential” stands for “up to relatively compact sets”. This construction can also be generalized using different (closed) bornologies. An extension which can be obtained in this way is said to be a B-extension. In ([5]) B-extensions were first introduced and mostly used in order to construct Lindelöf extensions of locally Lindelöf spaces. Two questions naturally arise: - which extensions are B-extensions - which conditions a bornology FX must satisfy to produce a B-extension aX = X ∪Y preserving some specific topological property of X and Y . In this paper we give a first example of a (non-regular) extension which is not a B-extension. We do not know if every regular extension can be obtained as B-extension. The problem is still open even for compactifications. In section 5 we will show that every regular extension such that the remainder is a retract is a B-extension. Some results are known about the second question. If FX is a closed bornology in a regular (normal) space X, then the extension o(FX) is regular (resp. normal) if and only if FX is open ([12]). It was proved in ([6]) that o(FX) is Tychonoff if and only if X is Tychonoff and FX is functionally open. This means that for every (closed) F ∈ FX there is an open W ∈ FX such that F and X \ W are completely separated. This concept was used in [3] to obtain a topological and bornological immersion of a Tychonoff space into a cube. It is also known that o(FX) is metrizable if and only X is metrizable and FX is induced by a metric ([2], [6]). In [6] the results about regularity, normality and metrizability of one-point extensions were extended to B-extensions with compact remainder. It was also observed that the Moore-Niemytzki plane and the Mrówka space can be Extensions defined using bornologies 83 naturally obtained as B-extensions. These examples was used to prove that the results about normality and metrizability do not hold if we remove the compactness hypothesis. The case of a regular extension with regular non- compact remainder was not solved there. In Section 3 of this paper we mostly study B-extensions with closed discrete remainders. We prove that all regular extensions of this kind are B-extensions. We also show that the condition that the given bornology is open is not in general either necessary or sufficient to obtain a regular B-extension. We give conditions on the bornology FX which are equivalent to the B-extension being regular (Tychonoff) when the remainder is discrete and X is regular (resp. Tychonoff). In section 4 we discuss the weight of a B-extension. We prove by an example (the so-called butterfly space) that the weight of a B-extension aX can be greater than max{w(X), w(Y ), χ(aX)}. In section 5 we study a particular case of B-extensions, when the map from the open subsets of the remainder Y and the unbounded open subsets of X is induced by a map from X to Y . In this case almost all results about separation and metrizability properties of one-point extensions can be generalized. 2. Basic definitions. We recall that a boundedness (or bornology) FX is said to be local if every x ∈ X has a neighborhood in FX. In this case one can also say that X is locally bounded with respect to FX. An open bornology is obviously local. If aX is a regular extension of X then HX(aX) is local. A boundedness FX is nontrivial if X is not bounded. A basis for a boundedness is a cofinal subfamily. An open and closed bornol- ogy with a countable basis is said to be of metric type (or M-boundedness), since the usual boundedness induced by a metric has these properties. It was proved in [11] that every M-boundedness in a metrizable space is induced by a compatible metric. Let X, Y be Hausdorff spaces, FX a nontrivial closed local bornology on X and B an open basis of Y , closed with respect to finite unions. Let us denote by TX the topology of X. A map π = π(B, FX) : B → (TX \ FX) ∪ {∅}, is said to be a B-map provided that B0) U 6= ∅ implies π(U) 6= ∅, for every U ∈ B; B1) If {Ui}i∈A ⊂ B is a cover of Y , then X \ [ ⋃ i∈Aπ(Ui) ] ∈ FX; B2) If U, V ∈ B then π(U ∪ V ) ∆ [π(U) ∪ π(V )] ∈ FX; B3) If U, V ∈ B and ClY (U) ∩ ClY (V ) = ∅ then π(U) ∩ π(V ) ∈ FX. Putting on the disjoint union X ∪ Y the topology generated by TX ∪ {U ∪ (π(U) \ F) | U ∈ B, F = ClX(F) ∈ FX}, 84 A. Caterino and M. C. Vipera we obtain a dense extension of X, denoted by X ∪π Y . If Y is T3 the extension is Hausdorff. If Y is T2 not T3 we can obtain a Hausdorff extension if we replace B3) by the stronger condition • If U, V ∈ B and U ∩ V = ∅ then π(U) ∩ π(V ) ∈ FX. A B-extension of X is any extension of X which can be constructed in this way. As we have already mentioned, the family of B-extensions includes ESH-compactific- ations and Hausdorff extensions aX with finite remainder. Moreover normal extensions with 0-dimensional remainder are B-extensions ([5]). If aX = X ∪π Y is a B-extension, where π = π(B, FX), then π is also a B-map with respect to HX(aX) and X ∪π(B,HX (aX)) Y is equivalent to aX (see [5]). 3. Extensions with discrete remainders. All spaces will be Hausdorff and the word “extension” will always mean “dense extension”. Theorem 3.1. Let aX be a regular extension of X such that Y = aX \ X is closed in aX. Suppose there is a basis B of Y consisting of open and closed sets and for every B ∈ B there are disjoint open subsets U and V of aX such that B ⊂ U and Y \ B ⊂ V . Then aX is a B-extension (with respect to HX(aX)). The proof is essentially the same as the one of Theorem 1.5 in [5]. In fact, the hypothesis that aX is normal was used only to find disjoint open neighborhood of B and Y \ B for every B in the clopen basis. For a discrete space Y we denote by B0 the basis consisting of all finite subsets of Y . By the above theorem we easily obtain the following result. Theorem 3.2. Every regular extension aX such that Y = aX \ X is closed and discrete is a B-extension. Proof. It suffices to take as B, in the above theorem, the family B0. � If aX = X ∪π Y is a B-extension, where π = π(B, FX) and Y is discrete, then B contains B0, hence B can be replaced by B0 and π by its restriction (see [6], Lemma 4.7). In the previous theorem, the hypothesis that aX is regular cannot be re- moved. In fact we can give an example of a nonregular Hausdorff extension aX, such that Y = aX \ X is closed and discrete but aX cannot be obtained as a B-extension. Example 3.3. Let Y = M∪{0} ⊂ R, where M = { 1 n ∣ ∣ n ∈ N } . Put X = R\Y and let aX = (R, T ), where T is generated by the union of the usual topology and the family {(−a, a) \ M | a ∈ R+} ((R, T ) is often used as example of Hausdorff nonregular space). T induces the Euclidean topology on X and the discrete topology on Y . Suppose aX is a B-extension, that is aX = X ∪π Y . We can suppose π = π(B0, HX(aX)) (see above). By definition, {0} ∪ π({0}) Extensions defined using bornologies 85 is open in aX so it must contain a set of the form (−a, a) \ M with a ∈ R+. Let m ∈ N such that 1 m < a. The open set { 1 m } ∪ π ({ 1 m }) must contain an interval (b, c), with 0 < b < 1 m < c. The intersection (b, c) ∩ (−a, a) \ M ⊂ X cannot be in HX(aX) because 1 m belongs to its closure. This means that π does not satisfy B3), a contradiction. For any subset A of a space X we denote by FrX(A) the boundary of A in X. Lemma 3.4. Let aX = X ∪π Y be a B-extension, where Y is T3 and π = π(B, FX). Then, for every B ∈ B we have (i) ClaX(π(B)) = ClY (B) ∪ ClX(π(B)); (ii) FraX(B ∪ π(B)) = FrY (B) ∪ FrX(π(B)). Proof. (i) We only have to prove that if for y ∈ Y , y /∈ ClY (B) then y /∈ ClaX(π(B)). Let B ′ ∈ B such that y ∈ B′ ⊂ ClY (B ′) ⊂ Y \ ClY (B). Then, by B3, π(B) ∩ π(B′) = K ∈ FX. Therefore y ∈ B ′ ∪ (π(B′) \ K) which is disjoint from π(B). Then the conclusion follows. (ii) Let B ∈ B. Clearly we have FraX(B ∪π(B)) = (ClaX(B ∪π(B)))\(B ∪ π(B)). We note that ClaX(B)\(B ∪π(B)) ⊂ ClaX(π(B))\(B ∪π(B)). In fact if y ∈ ClaX(B)\(B∪π(B)) then y ∈ Y and, for every B ′ ∈ B and F ∈ FX such that y ∈ B′, we have (B′ ∪(π(B′)\F))∩π(B) 6= ∅ otherwise, for some B′ ∈ B, (B′ ∪ π(B′) \ F) ∩ (B ∪ π(B)) = B ∩ B′ would be a non-empty open subset of aX = X ∪π Y contained in Y . It follows that (ClaX(B ∪π(B)))\ (B ∪π(B)) = (ClaX(π(B)))\(B∪π(B)). By (i) we get (ClaX(π(B)))\(B∪π(B)) = (ClY (B)∪ ClX(π(B))) \ (B ∪ π(B)) and finally the conclusion follows by the obvious equalities (ClY (B)∪ClX (π(B)))\ (B ∪π(B)) = (ClY (B)\ B)∪((ClX (π(B))\ π(B)) = FrY (B) ∪ FrX(π(B)). � Remark 3.5. If B consists of open and closed sets then (ii) becomes (ii′) FraX(B ∪ π(B)) = FrX(π(B)), hence FrX(π(B)) is closed in aX. From now on, for a B-extension X ∪π Y , where π = π(B, FX) and Y is discrete, we will always put B = B0 and Uy = π({y}) for every y ∈ Y . Corollary 3.6. Let aX = X∪π Y be a B-extension, where Y is discrete. Then, for every y ∈ Y we have (i) ClaX(Uy) = {y} ∪ ClX(Uy); (ii) FrX(Uy) = FraX({y} ∪ Uy), hence FrX(Uy) is closed in aX. Proposition 3.7. Let aX = X ∪π Y be a B-extension of a regular space X, where π = π(B, FX) and B is a basis of Y consisting of open and closed subsets of Y . Then (i) if aX is regular and B is compact then FrX(π(B)) belongs to FX; (ii) if FX is open and, for each B ∈ B, FrX(π(B)) belongs to FX, then aX is regular. Proof. (i) Let aX be regular and let B ∈ B, B compact. By the Remark 3.5 A = FrX(π(B)) is closed in aX. Then for every y ∈ B there exists 86 A. Caterino and M. C. Vipera an open subset Wy of aX that contains A and is disjoint from a basic open neighborhood By ∪ (π(By) \ Ky) of y. We can suppose Wy ⊂ X. From the compactness of B it follows that B ⊂ B′ = ⋃n i=1 Byi for some y1, . . . , yn ∈ B. Hence π(B) \ π(B′) = π(B ∪ B′)∆[π(B) ∪ π(B′)] ∈ FX by B2) and clearly π(B) \ ( ⋃n i=1 π(Byi)) also belongs to FX. If we put W = ⋂n i=1 Wyi then W ∩ ( ⋃n i=1 π(Byi)) ⊂ ⋃n i=1 Kyi and W ∩ π(B) is bounded. Now let x ∈ A and let V be any neighborhood of x in X. Then V ∩W ∩π(B) 6= ∅ and this means that x ∈ ClX(W ∩ π(B)). Therefore A ⊂ ClX(W ∩ π(B)), which is bounded. (ii) Let x ∈ X. The hypotheses imply that x has a local basis consisting of bounded closed subset of X, which are also closed neighborhoods in aX. Now let y ∈ Y and let F be a closed subset of aX such that y /∈ F . Then there are B ∈ B and a closed member K of FX such that y ∈ B ∪ (π(B) \ K), which is disjoint from F . Put V = aX\[ClaX(B∪π(B))] = aX\[B∪ClX(π(B))] = aX\[B∪π(B)∪FrX(π(B))]. Then we have F \ V = (F ∩ (B ∪ π(B))) ∪ (F ∩ FrX(π(B))) ⊂ K ∪ FrX(π(B)). Since K ∪FrX(π(B)) is bounded, it is contained in an open member W of FX. Then V ∪ W is an open subset of aX which contains F and is disjoint from B ∪ [π(B) \ ClX(W)], which is a basic neighborhood of y. � Proposition 3.8. Let aX = X∪π Y be a B-extension of X, where Y is discrete and π = π(B0, FX). Suppose X is regular. Then (i) if aX is regular, then for each y ∈ Y , FrX(Uy) belongs to FX; (ii) if FX is open and, for each y ∈ Y , FrX(Uy) belongs to FX, then aX is regular. Proof. It easily follows by Proposition 3.7, since FrX(Uy) ∈ FX for every y ∈ Y implies that FrX(π(B)) ∈ FX for every B ∈ B0. � The following example shows that, for a B-extension aX = X ∪π Y , the conditions that X, Y are regular and FX is open do not ensure that aX is regular, even if Y is discrete. Example 3.9. Let X be the upper half plane, defined by {(x, y) ∈ R2 | y > 0}, with the usual topology, and Y be the x-axis with the discrete topology. Let FX = {A ⊂ X | d(A, Y ) > 0}, where d is the Euclidean metric. Clearly FX is an open boundedness and X is locally bounded. For every z = (a, 0) ∈ Y , put Uz = {(x, y) ∈ X | |x − a| < y < 1}. Uz is clearly unbounded. Let π : B0 → TX \ FX be defined by B 7→ ⋃ z∈B Uz. It is easy to see that π is a B-map. Then we can define the Hausdorff B- extension aX = X ∪π Y . Let z be any point of Y . Clearly E = FrX(Uz) is unbounded, so by Proposition 3.7, aX is not regular. In fact, E is closed Extensions defined using bornologies 87 in aX and z /∈ E, but no open set containing E can be disjoint from a basic neighborhood {z} ∪ (Uz \ K), with K ∈ FX. Remark 3.10. In the above example, although FX is open, HX(aX) is not open. By Lemma 3.4(ii) and Proposition 3.8(ii), if aX = X ∪π(B,FX) Y , where X is regular, Y is discrete and HX(aX) is open, then aX is regular. However, as we will see below, the condition that HX(aX) is open is not necessary to obtain a regular B-extension. We notice that, for any extension aX with closed remainder Y , HX(aX) is open if and only if Y is separated, by disjoint open subsets of aX, from every closed subset of aX which is contained in X. There exists a Tychonoff B-extension aX of a space X, with discrete remainder, which cannot be obtained as B-extension with respect to any open boundedness FX. In particular HX(aX) is not open. Example 3.11. The Tychonoff plank T can be seen as an extension aX of X = ω1 × (ω + 1) with closed discrete remainder Y = {ω1} × ω. By Theorem 3.2, aX is a B-extension. Put F = ω1 × {ω}. Clearly HX(aX) is not open, otherwise, F and Y would be contained in disjoint open subsets of T . Now suppose that FX is a closed boundedness on X, and π = π(B, FX) a B-map such that T = X ∪π Y . Put yn = (ω1, n) for every n, so that Y = {yn}n∈N. For sake of simplicity, for every y ∈ Y , we will write π(y) instead of π({y}). Since {yn} ∪ π(yn) is open in T , π(yn) must contain a set of the form (βn, ω1) × {n}, where βn < ω1. Property B1) implies that G = F \ ( ⋃ n∈N π(yn) ) belongs to FX. Now suppose FX is open. Then G and Y are contained in disjoint open subsets of T . This implies G ⊂ [0, α] × {ω}, where α < ω1, hence (α, ω1) × {ω} ⊂ ⋃ n∈N π(yn). Let m ∈ N such that M = {γ > α | (γ, ω) ∈ π(ym)} is uncountable. For every γ ∈ M let Vγ = (η1(γ), η2(γ)) × (nγ, ω] be a basic open neighborhood of (γ, ω) contained in π(ym). Let h ∈ N such that H = {γ ∈ M | nγ = h} is uncountable and let k ∈ N, k > max(h, m). Then (βk, ω1) × {k}, which is contained in π(yk), contains all the points (γ, k) with γ ∈ H. But every (γ, k), with γ ∈ H belongs to Vγ ⊂ π(ym). Then the uncountable set {(γ, k) | γ ∈ H} is contained in both π(yk) and π(ym). This means that yk ∈ ClT (π(yk) ∩ π(ym)). Since k 6= m, by B3) π(yk) ∩ π(ym) belongs to FX, so no point of Y can be in its closure, contradiction. The following proposition provides a condition which is equivalent to the regularity of X ∪π Y where Y is discrete. Proposition 3.12. Let aX = X ∪π Y be a B-extension of X, where Y is discrete and π = π(B0, FX). Suppose X is regular. Then aX is regular if and only if, for every y ∈ Y and for every closed subset F of X such that F ∩ Uy is bounded, there is an open subset W of X containing F ∩ ClX(Uy) such that W ∩ Uy is bounded. Proof. Suppose aX is regular. Let y ∈ Y and F ∩ Uy ∈ FX, where F is closed in X. This is clearly equivalent to y /∈ ClaX(F). Then there exists an 88 A. Caterino and M. C. Vipera open subset W1 of aX that contains ClaX(F) and is disjoint from a basic open neighborhood {y} ∪ (Uy \ K) of y. Put W = W1 ∩ X. Then F ∩ ClX(Uy) ⊂ W and W ∩ Uy ⊂ K, so that W ∩ Uy is bounded. Conversely, let y ∈ aX and let G be a closed subset of aX such that y /∈ G. We can suppose y ∈ Y (see the proof of Proposition 3.7(ii)). Put F = G ∩ X. Then y /∈ ClaX(F), hence F ∩ Uy is bounded. By hypothesis, there exists an open subset W of X such that F ∩ ClX(Uy) ⊂ W and W ∩ Uy is bounded. We know that V = aX \ ClaX(Uy) contains all points of Y except y (see Lemma 3.4). Since y /∈ G, we have G \ V = F \ V ⊂ W . Therefore we have G ⊂ W ∪V. Clearly [{y}∪(Uy \W)]∩(V ∪W) = ∅. We have {y}∪(Uy \W) = {y} ∪ (Uy \ (W ∩ Uy)) ⊃ {y} ∪ (Uy \ ClX(W ∩ Uy)), which is a basic open neighborhood of y disjoint from V ∪ W . � We recall that every one-point extension aX of X can be obtained as o(FX) for a suitable closed bornology FX; moreover we have HX(aX) = FX ([6]). Then, by Theorem 3.3 in [6], a one-point extension aX is Tychonoff if and only if HX(aX) is functionally open. The following proposition provides a sufficient condition for the complete reg- ularity of a B-extension with discrete remainder. Proposition 3.13. Let aX = X ∪π Y be a B-extension of X, where Y is discrete and π = π(B0, FX). Suppose X is Tychonoff. If FX is functionally open and, for each y ∈ Y , FrX(Uy) belongs to FX, then aX is Tychonoff Proof. Let z ∈ aX and let F be a closed subset of aX such that z /∈ F . First suppose z ∈ X. Let H be a bounded open neighborhood of z in X which is disjoint from F . Since X is Tychonoff, there exists a continuous function f : X → I such that f(z) = 1 and f(X \ H) = 0. The map f̂ : aX → I defined by f̂ |X= f, f̂(Y ) = 0, is continuous because no y ∈ Y belongs to ClaX(H). Clearly f̂ separates z from F . Let now z ∈ Y . Put A = FrX(Uz), T = ClX(Uz) = Uz ∪ A and T ∗ = ClaX(T ) = {z} ∪ T . T ∗, with the topology induced by aX, is a one-point extension of T . We want to prove that HT (T ∗) is functionally open. Let G be a subset of T which is closed in T ∗. There exists a neighborhood {z}∪(Uz \K) which is disjoint from G, where K is a closed member of FX. Since G is closed in T and is contained in K ∪ A, G is a closed member of FX. Then there is an open W ∈ FX such that G and X \ W are completely separated. Put V = W ∩ T . Since T is unbounded, T \ V = T \ W is nonempty and completely separated from G. Moreover, ClT ∗(V ) ⊂ ClaX(W) which does not meet Y . Then V ∈ HT (T ∗). We have proved that HT (T ∗) is functionally open, that is, T ∗ is Tychonoff. Let f : T ∗ → I be a continuous function such that f(z) = 1, f((F ∩ T ) ∪ A) = 0. We define an extension f̂ : aX → I of f putting f̂(aX \ T ∗) = 0. Since f is equal to 0 on A, which is the boundary of T ∗ in aX, ĥ is continuous. Moreover f̂ separates z from F . � In the above proposition the hypothesis FrX(Uy) ∈ FX for every y is clearly necessary (see proposition 3.7(i)). The Examples 3.9 and 3.11 show that the Extensions defined using bornologies 89 condition that FX is functionally open is a neither sufficient nor necessary con- dition. In view of Corollary 3.6(ii), the condition that HX(aX) is functionally open is sufficient, but it is not necessary (see Example 3.11 again). It is easy to see the following Proposition 3.14. For every normal extension aX, HX(aX) is functionally open. Corollary 3.15. Every normal extension aX such that Y = aX \ X is closed and discrete is a B-extension with respect to a functionally open boundedness. Proof. By the proof of Theorem 3.1, aX is a B-extension with respect to HX(aX). � However, there exist nonnormal B-extensions aX with discrete remainder where X is normal and HX(aX) is functionally open. Example 3.16. Let Ψ be the Mrówka space. It is known that Ψ is a nonnormal Tychonoff space (see for instance [9]), and is a B-extension of N with respect to the boundedness FN of the finite subsets of N (see [6], Example 4.10). Clearly HN(M) = FX is functionally open, since its members are clopen. 4. Weight of B-extensions. For a boundedness FX on X we put µ(FX) = min{|C| : C is a basis of FX}. Proposition 4.1. Let aX = X ∪π Y be any B-extension of a space X,where π = π(B, FX). Then we have w(aX) ≤ max{w(X), w(Y ), µ(FX)}. Proof. By [6] Lemma 4.7, we can suppose |B| = w(Y ). Let B1 be a basis of X with |B1| = w(X) and C a basis of FX with |C| = µ(FX). It is easy to see that B1 ∪ {U ∪ (π(U) \ F) | U ∈ B, F ∈ C} is a basis for the topology of aX whose cardinality is max{w(X), w(Y ), µ(FX)}. � The weight of a B-extension aX = X ∪π Y can be greater than max{w(X), w(Y ), χ(aX)}. Example 4.2. Let us consider the so called butterfly space, that is the space Z = (R × R, T ) where T can be described as follows. Let us denote the x- axis by Y and put X = (R × R) \ Y . The points of X have the same open neighborhoods as in the ordinary topology. Let pa = (a, 0) ∈ Y and let r ∈ R +. We put Rr(a) = (a− r, a+ r)× (− 1 r , 1 r ) and we denote by C1r (a) and C 2 r (a) the 90 A. Caterino and M. C. Vipera (closed) circles of radious 1 r and center (a, 1 r ), (a, −1 r ), respectively (so that the circles are tangent to the x-axis in pa). We also put Br(a) = [Rr(a) \ (C 1 r (a) ∪ C 2 r (a))] ∪ {pa}. A local basis for pa will be the family {Br(a)}r∈R+. If we consider the sub- family {Br(a)}r∈ Q, we obtain a countable local basis, hence χ(Z) = ω. It is known that w(Z) = c. The topology induced on both X and Y are the usual ones. We also observe that the sets of the form X ∩ Cir(a) = C i r(a) \ {pa} are closed in Z. The butterfly space can be considered a dense extension aX of X such that Y = aX \ X is naturally homeomorphic to the real line R. We want to prove that aX can be obtained as a B-extension X ∪π R with respect to the bound- edness HX(aX). Let B be the family of finite unions of bounded open intervals in R. Given an open interval (a − r, a + r), with a ∈ R, r ∈ R+, we put π(a − r, a + r) = Br(a)∩X, which is clearly an open unbounded subset of X. For a finite disjoint union U = ⋃ i (ai − ri, ai + ri) we put π(U) = ⋃ i π(ai − ri, ai + ri). We want to prove that π is a B-map. The property B3) is obviously satisfied, since for U1, U2 ∈ B such that U1∩U2 = ∅, we have π(U1) ∩ π(U2) = ∅. Let {Uj}j∈J ⊂ B be a cover of R. For Uj = ⋃n i=1(ai − ri, ai + ri), where the union is disjoint, we put Wj = ⋃n i=1 Bri(ai), so that one has Wj ∩ X = π(Uj). Therefore X \ ( ⋃ j π(Uj) ) = X \ ( ⋃ j Wj ) = aX \ ( ⋃ j Wj ) ∈ HX(aX). We have proved that B1) is satisfied. Now we prove B2) in case U, V are intervals, U = (a−r, a+r), V = (b−s, b+s). The general case will easily follow. If U and V are disjoint the proof is trivial, hence we can suppose that U ∪ V is an interval (c − t, c + t). We want to show that E = π(c − t, c + t)△[π(a − r, a + r) ∪ π(b − s, b + s)] is bounded. We have E ⊂ (c− t, c+ t)× (R\ {0}). Moreover every x ∈ E must be in X \ π(c − t, c + t) = X \ Bt(c) or in X \ [π(a − r, a + r) ∪ π(b − s, b + s)] = (X \ Br(a)) ∩ (X \ Bs(b)). Suppose x ∈ E, and x /∈ Bt(c). Then either x ∈ (c−t, c+t)×[R\(−1 t , 1 t )], which is clearly bounded, or x ∈ (C1t (c)∪C 2 t (c))∩X which is also bounded. Similarly we can prove that, if x ∈ E and x /∈ Br(a), x /∈ Bs(b) then x belongs to a bounded set. Therefore E is contained in a finite union of bounded sets and so it is bounded. We have proved that π is a B-map. If we identify R with the x-axis Y in Z = (R × R, T ), then the topology of X ∪π R is equal to T . In fact, for a ∈ R, r ∈ R +, we have Br(a) = U ∪ π(U), where U = (a − r, a + r), identified with (a − r, a + r) × {0}. Conversely, every set of the form U ∪ (π(U) \ F), where U ∈ B and F is a closed member of Extensions defined using bornologies 91 HX(aX), is open in Z, because U ∪ π(U) is a union of sets of the form Br(a) and F is closed in Z by the definition of HX(aX). 5. Extensions whose remainders are retracts. Given a nontrivial local closed bornology on X, a continuous mapping f from X to any space Y is said to be B-singular with respect to FX if f −1(U) /∈ FX for every nonempty open subset U of Y . If f is B-singular, then the map π : TY → (TX \ FX) ∪ {∅}, π(U) = f −1(U), is clearly a B-map. The B-extension induced by π, denoted by X ∪f Y , is said to be B-singular ([5], [7]). Regular extensions are B-singular if and only if the remainder is a retract. In fact we have Theorem 5.1. Let (aX, TaX) be a regular extension of X such that there exists a retraction g : aX → Y = aX \ X. Then f = g|X is B-singular with respect to HX(aX) and (aX, TaX) = X ∪f Y . Conversely, for every B-singular extension aX = X ∪f Y , the map f̂ : aX → Y defined by f̂|X = f and f̂|Y = 1Y is a continuous extension of f, hence a retraction. Proof. Let g : aX → Y be a retraction and let U be a nonempty open subset of Y . It is easy to see that every point of U belongs to ClaX(f −1(U)), hence f−1(U) /∈ HX(aX). Since aX is T3, X is locally bounded with respect to HX(aX). Now we will prove that the topology of X ∪f Y coincides with TaX. First we observe that TX is contained in both topologies. Let U1 = U ∪ [f −1(U) \ F ] where U is open in Y and F is a closed member of HX(aX). Then U1 = g−1(U) \ F ∈ TaX. This proves the first inclusion. Let us choose any W ∈ TaX. We only need to prove that, for every y ∈ W ∩ Y , there is a basic open set U1 in X ∪f Y such that y ∈ U1 ⊂ W . Let U be an open neighborhood of y in Y such that y ∈ U ⊂ ClY (U) ⊂ W ∩ Y . Put A = g−1(U) \ W which is a subset of X. We want to prove that no point of Y belongs to ClaX(A). This is obvious for z ∈ W . If z ∈ Y \ W then z /∈ ClY (U). Let V be an open neighborhood of z in Y such that U ∩ V = ∅. Then g−1(V )∩g−1(U) = ∅, that is, g−1(V ) is a neighborhood of z in (aX, TaX) which is disjoint from A. We have proved that ClX(A) is a (closed) member of HX(aX). Then U1 = g −1(U) \ ClX(A) = U ∪ [f −1(U) \ ClX(A)] is a basic neighborhood of y in X ∪f Y and, by the definition of A, we have U1 ⊂ W . We have proved (aX, TaX) = X ∪f Y . Let now aX = X ∪f Y a B-singular extension and let U be open in Y . Then f̂−1(U) = U ∪ f−1(U) is a basic open set of X ∪f Y . � In [6], Theorem 4.11, it was proved that a B-singular extension X ∪f Y of X, with respect to an open (and closed) bornology FX, is regular provided X and 92 A. Caterino and M. C. Vipera Y are both regular. If we replace “open” by “functionally open”, we obtain an analogous result for the Tychonoff property. Theorem 5.2. Let X, Y be Tychonoff spaces and let f : X → Y be a B- singular map with respect to a functionally open bornology FX. Then aX = X ∪f Y is Tychonoff. Proof. First suppose that Y is compact. Let q : X ∪f Y → o(FX) = X ∪ {p} be the natural mapping which takes every point of Y to the point p. By [5], Proposition 1.1, q is a quotient map. Let now z ∈ aX and let A be a closed subset of aX with z /∈ A. First suppose z ∈ X and put B = A ∪ Y . Then q(B) is a closed subset of o(FX) which does not contain z. Since o(FX) is Tychonoff ([6], Theorem 3.3), z and q(B) are separated by a continuous function g from o(FX) to I, where I is the unit interval. Clearly g ◦ q separates z and A. Now, let z ∈ Y and A ⊂ X. Then A = q(A) is closed in o(FX). Let g : o(FX) → I be a function such that g(p) = 1 and g(A) = 0. Then g ◦ q separates z from A. Note that g ◦ q maps all of Y onto 1. Finally let C = A ∩ Y 6= ∅ and z ∈ Y . Take a map v : Y → I such that v(C) = 0 and v(z) = 1. Put h = v ◦ f̂ : aX → I and U = h−1([0, 1/2)). Then A \ U is a closed subset of aX contained in X. We can take, as before, a function u : aX → I such that u(Y ) = 1 and u(A \ U) = 0. Then the map h ∧ u is less than 1/2 in U ∪ A and maps z to 1. We have proved that aX is Tychonoff in case Y is compact. Let now Y be any Tychonoff space and let (K, k) be any compactification of Y , where k : Y → K is the embedding. Then f1 = k ◦ f : X → K is B-singular and X ∪f1 K is Tychonoff. It is easy to see that aX = X ∪f Y is a subspace of X ∪f1 K, hence aX is also Tychonoff. This completes the proof. � Lemma 5.3. Let aX = X ∪f Y be a B-singular extension of X and suppose V be a locally finite family in Y . Then the family V1 = {V ∪ f −1(V ) | V ∈ V} is locally finite in aX. Proof. Let y ∈ Y . There is an open neighborhood Ny of y in Y such that Ny ∩ V 6= ∅ for only finitely many V ∈ V. This implies that Ny ∪ f −1(Ny) meets only finitely many members of V1. Let x ∈ X and y = f(x). Then f−1(Ny) is a neighborhood of x which meets only finitely many members of V1. � Theorem 5.4. Let aX = X ∪f Y be a B-singular extension of X with respect to an M-boundedness FX. Suppose X and Y are metrizable. Then aX is metrizable. Proof. Since X ∪f Y is T3 by [6], Theorem 4.11, we need only to prove that it admits a σ-locally finite basis. By hypothesis, FX has a countable basis {Mk}k∈N, where Mk is open and ClX(Mk) is bounded. We have ⋃ k∈N Mk = Extensions defined using bornologies 93 X. Let C = ⋃ n∈N Cn be a basis for TX, where every Cn is a locally finite family. Put Ckn = {C ∩ Mk | C ∈ Cn}, n, k ∈ N. Similarly, let U = ⋃ n∈N Un be a basis of TY , where every Un is locally finite. For every n, k ∈ N, let Ukn = {U ∪ [f −1(U) \ ClX(Mk)] | U ∈ Un}. We claim that S = ( ⋃ n,k∈N Ckn ) ∪ ( ⋃ n,k∈N Ukn ) is a σ-locally finite basis for aX. Let W be an open subset of aX and let x be in W . If x ∈ X, then x ∈ Mk for some k and there is C ∈ Cn, for some n, such that x ∈ C ⊂ W . Then x ∈ C ∩ Mk ⊂ W , where C ∩ Mk ∈ C k n. If x ∈ Y , then there is U ∈ Un for some n and F = ClX(F) ∈ FX such that x ∈ U ∪ (f−1(U) \ F) ⊂ W . We have F ⊂ Mk for some k, hence x ∈ U ∪ [f−1(U) \ ClX(Mk)] ⊂ (U ∪ (f −1(U) \ F)) ⊂ W, where U ∪ [f−1(U) \ ClX(Mk)] ∈ U k n. Every Ckn is locally finite. In fact, for every x ∈ X, there is a neighborhood that meets only finitely many members of Cn, hence of C k n. If x ∈ Y , any basic neighborhood of x of the form V ∪ [f−1(V ) \ ClX(Mk)] meets no member of Ckn. By Lemma 5.3, Ukn is locally finite for every n, k. This completes the proof. � References [1] G. Beer, On metric boundedness structures, Set-Valued Anal. 7 (1999), 195–208. [2] G. Beer, On convergence to infinity, Monathsh. Math. 129 (2000), 267–280. [3] G. Beer, Embeddings of bornological universes, Set-Valued Anal. 16, no. 4 (2008), 477– 488. [4] A. Caterino, G. D. Faulkner, M. C. Vipera, Construction of compactifications using essential semilattice homomorphisms, Proc. Amer. Math. Soc. 116 (1992), 851–860. [5] A. Caterino and S. Guazzone, Extensions of unbounded topological spaces, Rend. Sem. Mat. Univ. Padova 100 (1998), 123–135. [6] A. Caterino, F. Panduri and M. C. Vipera, Boundedness, one-point extensions and B-extensions, Math. Slovaca. 58, no. 1 (2008), 101–114. [7] R. E. Chandler and G. D. Faulkner, Singular compactifications: the order structure, Proc. Amer. Math. Soc. 100 (1987), 377–382. [8] R. Engelking, General Topology, Heldermann, Berlin, 1989. [9] L. Gillman and M. Jerison, Rings of continuous functions, Van Nostrand, Princeton, 1960. [10] S. T. Hu, Boundedness in a topological space, J. Math. Pures Appl. 28 (1949), 287–320. [11] S. T. Hu, Introduction to general topology, Holden-Day inc., San Francisco, 1969. [12] M. C. Vipera, Some results on sequentially compact extensions, Comment. Math. Univ. Carolinae 39 (1998), 819–831. (Received November 2008 – Accepted August 2009) 94 A. Caterino and M. C. Vipera Alessandro Caterino (caterino@dipmat.unipg.it) Dipartimento di Matematica e Informatica, Università degli Studi di Perugia, Via Vanvitelli 1, 06123, Perugia, Italy. (Corresponding Author) M. Cristina Vipera (vipera@dipmat.unipg.it) Dipartimento di Matematica e Informatica, Università degli Studi di Perugia, Via Vanvitelli 1, 06123, Perugia, Italy. Extensions defined using bornologies. By A. Caterino and M. C. Vipera