A short note on hit-and-miss hyperspaces

@ Applied General Topology c Universidad Polit´ecnica de Valencia Volume 4, No. 2, 2003 pp. 281–288 A short note on hit–and–miss hyperspaces Ren´ e Bartsch and Harry Poppe Dedicated to Professor S. Naimpally on the occasion of his 70th birthday. Abstract. Based on some set-theoretical observations, compactness results are given for general hit-and-miss hyperspaces. Compactness here is sometimes viewed splitting into “κ-Lindel¨ofness” and “κcompactness” for cardinals κ. To focus only hit-and-miss structures, could look quite old-fashioned, but some importance, at least for the techniques, is given by a recent result, [8], of Som Naimpally, to who this article is hearty dedicated. 2000 AMS Classification: 54B20, 54D30, 54F99. Keywords: hit-and-miss topology, compactness, relative completeness, relative compact unions, upper Vietoris topology. 1. Introduction. Let (X, τ) be a topological space. By P(X), P0(X), Cl(X) and K(X) respectively we denote the power set, the power set without the empty set ∅, the family of all closed subsets and the set of all compact subsets of X. For B∈P(X) and A⊆P(X) we define B−A:= {A∈A|A∩B6=∅}(hit–set) and B+A:= {A∈ A|A∩B=∅}(miss–set). Specializing A:= Cl(X), we get the usual symbols B−, B+. By τl,Awe denote the topology for A, generated by the subbase of all G−A, G ∈τ. Now consider ∅6=α⊆P(X); by τα,Awe denote the topology for A which is generated from the subbase of all B+A, B ∈αand G−A, G ∈τ. Of course, for every possible αwe have τl,A⊆τα,A; for α=Cl(X) we get the Vietoris topology and for α=K(X) we get the Fell topology for A. If α= ∆ ⊆Cl(X), τα,Ais called ∆–topology by Beer and Tamaki [2], and was first introduced by Poppe [10]. By F(X) and F0(X) we denote the set of all filters and ultrafilters, respectively, on a set X(a filter is not allowed to contain the empty set ∅); the symbol F(ϕ) (resp. F0(ϕ)) means the set of all filters (resp. ultrafilters) which contain a given filter ϕ;. xis the filter generated by a singleton {x}, x ∈X. The symbol qτdenotes the convergence structure induced by a topology τ, i.e. qτ:= {(ϕ, x)∈F(X)×X|ϕ⊇ . x∩τ}, so qτis a relation between filters and points of a set X. If Xis a set, τ, Aare subsets of P(X), then we call Aweakly complementary w.r.t. τ, iff for every subset σ⊆τthere exist a subset B⊆A, s.t. SB∈BB=X\SS∈σS. 282 R. Bartsch and H. Poppe Lemma 1.1. Let Xbe a set, τ, A⊆P(X)and K⊆X. Then holds [ i∈I Gi⊇K=⇒[ i∈I G−A i⊇K−A for every collection Gi, i ∈I, Gi∈τ. If Ais weakly complementary w.r.t. τ, then for every collection Gi, i ∈I, Gi∈τ the implication [ i∈I Gi⊇K⇐=[ i∈I G−A i⊇K−A holds, too. Proof. Let Si∈IGi⊇K.A∈K−A⇒A∩K6=∅⇒∅6=A∩Si∈IGi⇒ ∃i0∈I: A∩Gi06=∅⇒A∈G−A i0⇒A∈Si∈IG−A i. Conversely, let Abe weakly complementary w.r.t. τand Si∈IG−A i⊇K−A. Assume Si∈IGi6⊇ K. Then X\Si∈IGi⊇K\Si∈IGi6=∅holds, so there is an A∈A, A ⊆ X\Si∈IGiwith A∩K\Si∈IGi6=∅. Thus A∈K−A, implying A∈Si∈IG−A i. This yields ∃i0∈I:A∩Gi06=∅in contradiction to the construction of A. Corollary 1.2. Let Xbe a set, τ, A⊆P(X)and K⊆X. Then holds (1.1) [ i∈I Gi⊇K⇐⇒ [ i∈I G−A i⊇K−A for every collection Gi, i ∈I, Gi∈τif and only if Ais weakly complementary w.r.t. τ. Proof. We only have to show, that Ais weakly complementary w.r.t. τ, if (1.1) holds. Assume, Ais not weakly complementary w.r.t. τ. Then there must be a collection {Gi|i∈I} ⊆ τ, such that S{A|A∈P(X\Si∈IGi)∩A} 6⊇ X\Si∈IGi. Now, we chose K:= X\Si∈IGi\S{A|A∈P(X\Si∈IGi)∩A} 6=∅. Then no element of A, which meets K, can be contained in X\Si∈IGi, i.e. every element of K−Ameets Si∈IGi, too. So, it must meet a Gi0, i0∈Iand consequently it is contained in Si∈IG−A i. But, by construction, the collection {Gi|i∈I}doesn’t cover K, so (1.1) would fail.  Obviously, if for every collection {Gi|i∈I} ⊆ τthe complement X\Si∈IGi itself belongs to A, or if all singletons {x}, x ∈Xare elements of A, then Ais weakly complementary w.r.t. τ. So, if τis a topology on X,Cl(X) and K(X) are weakly complementary w.r.t. τ. Corollary 1.3. Let (X, τ)be a topological space, K⊆Xand ∀i∈I:Gi∈τ. Then holds [ i∈I Gi⊇K⇐⇒ [ i∈I G− i⊇K− We have yet another easy, but useful set-theoretical lemma: Lemma 1.4. Let Xbe a set, A⊆P(X)and ϕ∈F(X). Assume, Ais closed under finite unions of its elements. Then holds ϕ∩A6=∅⇐⇒ ∀ψ∈F0(ϕ) : ψ∩A6=∅, i.e. a filter contains an A–set, iff each refining ultrafilter contains an A–set. A short note on hit-and-miss hyperspaces 283 Proof. Suppose ∀ψ∈F0(ϕ) : ∃Aψ∈A:Aψ∈ψ. Now, assume ϕ∩A=∅. From this automatically follows X6∈ A. Consider B:= {X\A|A∈A}. Because of the closedness of Aunder finite unions, Bis closed under finite intersection of its elements, and ∅6∈ B, because X6∈ A. For any F∈ϕ, B ∈Bwe have F∩B6=∅, because F∩B=∅would imply F⊆X\B∈Aand therefore ϕ∩A6=∅. So, ϕ∪Bis a subbase of a filter and consequently, there exists an ultrafilter ψ, containing ϕ∪B, therefore containing ϕ and the complement of every A–set - in contradiction to ∀ψ∈F0(ϕ) : ψ∩A6=∅. The other direction of the statement of the lemma is obvious.  Definition 1.5. Let κbe a cardinal. Then a topological space (X, τ) is called κ-compact, iff every open cover of Xwith cardinality at most κadmits a finite subcover. (X, τ) is called κ-Lindel¨of, iff every open cover of Xadmits a subcover of cardinality at most κ. A filter is called κ-generated, iff it has a base of cardinality at most κ. A filter ϕis called κ-completable, iff every subset B⊆ϕwith card(B) at most κfulfills TB∈BB6=∅. It is called κ-complete, iff TB∈BB∈ϕholds under this condition. Proposition 1.6. A topological space (X, τ)is κ-compact, if and only if every κ-generated filter on Xhas a convergent refining ultrafilter. Proof. Let (X, τ) be κ-compact and ϕa filter on Xwith a base Bof cardinality at most κ. Assume, all refining ultrafilters of ϕwould fail to converge in X. Then for each element x∈X, all refining ultrafilters of ϕcontain the complement of an open neighbourhood of x. But the set of complements of open neighbourhoods of a point xis closed w.r.t. finite unions, thus by Lemma 1.4, ϕcontains the complement of an open neighbourhood of x. So, for each x∈Xthere must exist Ox∈τ∩. xand Bx∈B, s.t. Bx⊆X\Ox, implying Bx⊆X\Oxand thus X\Bx⊇Ox. Now, for each B∈Bwe define OB:= X\Band find, that {OB|B∈B}is an open cover of X, because of the preceeding facts. So, there must exist a finite subcover OB1∪ · · · ∪ OBn=X, implying Sn i=1(X\Bi) = X, just meaning Tn i=1 Bi=∅, which is impossible, because all Bibelong to the filter ϕ. So, the assumption must be false; there must exist convergent refining ultrafilters of ϕ. Otherwise, let all κ-generated filters on Xhave a convergent refining ultrafilter. Assume, there would exist an open cover C:= {Oi∈τ|i∈I},Si∈IOi=X, card(I)≤ κsuch that all finite subcollections fail to cover X(implying κto be infinite). But the set of all finite subcollections of the infinite collection Cof cardinality at most κhas cardinality at most κ, too. So, B:= {X\Sn k=1 Oik|n∈IN, ik∈I}is a filterbasis of cardinality at most κ, thus there must exist an ultrafilter ϕ⊇B, which converges in Xleading to the usual contradiction, because every x∈Xis contained in an open Ox∈Cand X\Oxbelongs to B⊆ϕ. Analogously we get a characterization of κ-Lindel¨of-spaces. Proposition 1.7. If (X, τ)is κ-Lindel¨of, then every κ-completable filter on Xhas a convergent refining ultrafilter. If κis an infinite cardinal and every κ-complete filter on a topological space (X, τ) has a convergent refining ultrafilter, then (X, τ)is κ-Lindel¨of. 284 R. Bartsch and H. Poppe Of course, every κ-complete filter is κ-completable, so we may say, that a topological space (X, τ) is κ-Lindel¨of, if and only if each κ-complete filter on Xhas a convergent refinement. 2. Compactness Properties for Hyperspaces. Lemma 2.1. Let κbe a cardinal, (X, τ)a topological space and let A⊆P(X)be weakly complementary w.r.t. τ. If A0:= A\ {∅}is κ-Lindel¨of (resp. κ-compact) in τl,A0, then (X, τ)is κ-Lindel¨of (resp. κ-compact). Proof. If Ais weakly complementary w.r.t. τ, then A0is, too. So, Corollary 1.2 is applicable. Let {Gi|i∈I}be an open cover (resp. an open cover with cardinality at most κ) of X. By Corollary 1.2, then {G−A0 i|i∈I}is an open cover of X−A0=A0 (resp. of card. at most κ), so there exists a subset J⊆Iof cardinality at most κ (resp. a finite subset J), s.t. Sj∈JG−A0 j⊇A0=X−A0, implying Sj∈JGj⊇Xby Corollary 1.2.  Of course, the assumed topology τl,A0is not really hit-and-miss, because the miss-sets are missed. But every proper hit-and-miss topology would be stronger and therefore it would enforce the desired properties for (X, τ) as well. Lemma 2.2. Let (X, τ)be a κ-compact (resp. κ-Lindel¨of) topological space and assume Cl(X)⊆A⊆P(X). Then A0:= A\ {∅}is κ-compact (resp. κ-Lindel¨of) in τl,A0. Proof. Let ˆϕbe a κ-generated (resp. κ-complete) filter on A0. Then, for an arbitrary h∈ A := {g∈XP0(X)| ∀M∈P0(X) : g(M)∈M}the image h( ˆϕ) is a κ-generated (resp. κ-complete) filter on Xand consequently it has a τ-convergent refining ultrafilter ψh. Furthermore, there must exist an ultrafilter ˆ ψ⊇ˆϕ, s.t. h(ˆ ψ) = ψh. So, the set A:= {a∈X| ∃f∈ A : (f(ˆ ψ), a)∈qτ} is not empty and consequently the closure Abelongs to A0. Now, for any O∈τ with A∈O−A0(⇔A∩O6=∅) we get A∩O6=∅(because of the closureproperties). Now, the assumption O−A06∈ ˆ ψwould imply O+A0∈ˆ ψ, yielding ∀f∈ A :X\O∈f(ˆ ψ), thus ∀f∈ A :∀b∈A∩O: (f(ˆ ψ), b)6∈ qτin contradiction to the construction of A. Thus, O∈τ, A ∈O−A0always imply O−A0∈ˆ ψand consequently ˆ ψ τl,A0-converges to A. Definition 2.3. Let (X, τ) be a topological space. A subset A⊆Xis called weakly relatively complete in X, iff ∀ϕ∈F(A)∩q−1 τ(X) : F(ϕ)∩q−1 τ(A)6=∅, i.e. every filter ϕon A, which converges in X, has a refinement, converging in A. Proposition 2.4. Let (X, τ)be a topological space and A⊆X. Then holds: (a) Ais weakly relatively complete in X, iff F0(A)∩q−1 τ(X) = F0(A)∩q−1 τ(A), i.e. every ultrafilter on A, which converges in X, converges in A. (b) If Ais closed in X, then Ais weakly relatively complete in X. (c) If Ais compact, then Ais weakly relatively complete in X. A short note on hit-and-miss hyperspaces 285 (d) If (X, τ)is compact and Ais weakly relatively complete in X, then Ais compact. (e) If (X, τ)is Hausdorff, then every weakly relatively complete subset A⊆X is closed in (X, τ). (f) Ais compact iff Ais weakly relatively complete and relatively compact. (g) If (X, τ)is κ-compact and Ais weakly relatively complete in (X, τ), then A is κ-compact. (h) If (X, τ)is κ-Lindel¨of and Ais weakly relatively complete in (X, τ), then A is κ-Lindel¨of. (i) Weak relative completeness is transitive, i.e. for all A⊆B⊆Xwith Bweakly relatively complete in (X, τ)and Aweakly relatively complete in (B, τ|B), the subset Ais weakly relatively complete in (X, τ). There is also a useful description by coverings for weak relative completeness. Lemma 2.5. Let (X, τ)be a topological space and A⊆X. Then the following are equivalent: (1) Ais weakly relatively complete in X. (2) For every open cover Aof Aand every element xof X, there is an open neighbourhood Ux,Aof x, s.t. Ux,A∩Ais covered by finitely many members of A. (3) For every open cover Aof Athere exists an open cover A0⊇Aof X, such that the intersection of every member of A0with Acan be covered by finitely many members of A, i.e. ∀O∈A0:∃n∈IN, P1, ..., Pn∈A:Sn i=1 Pi⊇ O∩Aholds. Proof. (1)⇒(2): Let A⊆τwith SP∈AP⊇Abe given. For every x∈Awe can chose a single member of Aas open neighbourhood, whose intersection with Ais covered by itself. So, assume (2.2) ∃x∈X\A:∀Ux∈U(x)∩τ:∀n∈IN, P1, ..., Pn∈A:Ux∩A6⊆ n [ i=1 Pi Then B:= {(U∩A)\Sn i=1 Pi|U∈U(x)∩τ, n ∈IN, Pi∈A}would be closed under finite intersections and thus there would exist an ultrafilter ϕon Awith ϕ⊇B. By construction ϕ→xmust hold for this ultrafilter, and now by the weak relative completeness of Ait follows ∃a∈A:U(a)⊆ϕ. But Ais an open cover of A, so there is an open set P∈Awith a∈P, implying P∈ϕ– in contradiction to the construction of ϕ. Thus (2.2) is false and we have ∀x∈X\A:∃Ux∈U(x)∩τ:∃n∈IN, P1, ..., Pn∈A:Ux∩A⊆ n [ i=1 Pi (2)⇒(3): Note, that (3) is fulfilled with A0:= {Ux|x∈X\A} ∪ A. (3)⇒(1): For a given ultrafilter ϕon Awith ϕ→x∈Xassume ϕ6∈ q−1 τ(A). Then ∀a∈A:∃Ua∈U(a)∩τ:Uc a=X\Ua∈ϕ. With these neighbourhoods define A:= {Ua|a∈A}, which is an open cover of A. By (2) there is an open cover A0⊇Aof Xsuch that ∀O∈A0:∃n∈IN, P1, ..., Pn∈A:Sn i=1 Pi⊇O∩Aholds. Now, ϕ→ximplies ∃O∈A0:O∈ϕ(especially A∩O6=∅follows), and then we have ∃n∈IN, P1, ..., Pn∈A:O∩A⊆Sn i=1 Pi, implying ∃j∈ {1, ..., n}:Pj∈ϕ 286 R. Bartsch and H. Poppe – in contradiction to the construction of A. So, the assumption ϕ6∈ q−1 τ(A) must be false, showing, that every ultrafilter on A, which converges in X, converges in A. Theorem 2.6. Let (X, τ)be a topological space, and let α⊆P(X)consist of weakly relatively complete subsets of X. Then holds for any Awith Cl(X)⊆A⊆P(X): (A0, τα)is compact ⇐⇒ (X, τ)is compact. Proof. According to Lemma 2.1 we need only to show that (A0, τα) is compact, if (X, τ) is compact. So, assuming (X, τ) to be compact, by Proposition 2.4 every weakly relatively complete subset of Xis compact, and we have α⊆K(X). Now we will use Alexander’s Lemma: let Ube a cover of A0, consisting of subbase elements K+A0 i, G−A0 jwith Kicompact and Gjopen. A:= X\(S{G|G−A0∈U}) is closed. By construction, A6∈ G−A0for any G−A0∈U, so for A6=∅there must exist some K+A0 0∈Uwith A∈K+A0 0, yielding that K0⊆S{G|G−A0∈U};K0compact ⇒ ∃G1, ..., Gn∈Uwith K0⊆Sn k=1 Gk, but then {K+A0 0}∪{G−A0 1, ..., G−A0 n}is a cover of A0. If A=∅, then S{Gi|G−A0 i∈U}=X, so from the compactness of Xthe existence of some G−A0 1, ..., G−A0 n∈Uwith X=Sn k=1 Gkfollows. By Lemma 1.1 then Sn k=1 G−A0 k=A0holds.  Many known theorems of compactness w.r.t. the Fell– or the Vietoris–topology follow immediately from the above result. Lemma 2.7. Let (X, τ)be a topological space, A⊆P0(X)with Cl(X)⊆Aand α⊆Cl(X). If R⊆Xis relatively compact in X, then P0(R)∩Ais relatively compact in (A, τα). Proof. Let B:= {O−A i|i∈I, Oi∈τ} ∪ {C+A j|j∈J, Cj∈α}be an open cover of Aby subbase elements of τα. Let O:= Si∈IOi. If O=X, then there exists finitely many i1, ..., in∈Iwith Sn k=1 Oik⊇R, because Ris relatively compact, and thus Sn k=1 O−A ik⊇R−A⊇P0(R)∩A, by Lemma 1.1. If O6=X, then X\Ois nonempty and closed, but not covered by the O−A ifrom B. Thus, there must exist a j0∈Jwith X\O∈C+A j0, implying Cj0⊆O. Now, we have P0(R)∩A= (P0(R)∩C+A j0)∪(P0(R)∩C−A j0), and, of course, P0(R)∩C+A j0is covered just by C+A j0∈B. So, we have to find a finite subcover for (P0(R)∩C−A j0), if this is not empty. Observe, that R∩Cj0is relatively compact in X, because it is a subset of R. Furthermore, {Oi|i∈I}∪{X\Cj0}is an open cover of X. Thus we find again finitely many i1, .., in∈I, s.t. Sn k=1 Oik⊇R∩Cj0(because X\Cj0can be removed from any cover of R∩Cj0without loosing the covering property). Therefore Sn k=1 O−A ik⊇(R∩Cj0)−A, by Lemma 1.1. But P0(R)∩C−A j0⊆(R∩Cj0)−Aholds, because any subset of R, which hits Cj0, automatically hits R∩Cj0. 3. Compact Unions. As an interesting application of a simple set-theoretical property, concerning the +-operator, we want to take a brief look at the naturally arising question, whether a union of compact sets itself is compact. Michael showed in [6] that a union of A short note on hit-and-miss hyperspaces 287 closed sets is compact, if the unifying family is compact w.r.t. the Vietoris-topology. Now, the Vietoris-topology is induced by the upper-Vietoris τ+ V(miss sets: A+Awith Ac∈τ) and τl, but τlis not sufficient to enforce compactness of a union of compact sets, as the following example shows: Let X:= IR, endowed with euclidian topology, M:= {[−m, m]|m∈IN}. Then SM∈MM=IR, is obviously not compact. But every cover of Mwith elements of the defining subbase for τlmust especially cover the element {0}= [0,0] of M, so it must contain a set O−with 0 ∈O. Now, every element of Mcontains the point 0, thus M⊆O−follows. So, Mis compact in τl by Alexander’s subbase Lemma. And unifying compact sets, τlis not necessary, too, as we will see. Proposition 3.1. Let Xbe a set, X⊆P(X)and M⊆X. Then holds [ i∈I C+X i⊇M=⇒[ i∈I Cc i⊇[ M∈M M for every collection Ci, i ∈I. Proof. For every M∈Mthere must exist an iM∈Iwith M∈C+x iM, because of Si∈IC+x i⊇M. Thus M⊆Cc iM⊆Si∈ICc i. In [5] it was shown Lemma 3.2. Let (X, τ)be a topological space and M⊆K(X)compact w.r.t. the upper–Vietoris topology. Then K:= [ M∈M M is compact w.r.t. τ. Applying our simple set-theoretical statement, we get a similar result for unions of relatively compact subsets. Lemma 3.3. Let (X, τ)be a topological space, let Xbe the family of all relatively compact subsets of Xand let M⊆Xbe relatively compact in Xw.r.t. the upper Vietoris topology. Then R:= [ M∈M M is relatively compact in (X, τ). Proof. Let Si∈IOi⊇Xwith Oi∈τ, i ∈Ian open cover of X. Because of the relative compactness of all P∈X, there is a finite subcover Oi1 P, ..., OinP Pfor every P∈X, i.e. OP:= SnP k=1 Oik P⊇M. Of course, OP∈τand so (OP)cis closed w.r.t. τ. Furthermore, P∩Oc P=∅, implying P∈(Oc P)+X. Thus we have X⊆SP∈X(Oc P)+X, where the (Oc P)+Xare open w.r.t. the upper–Vietoris topology. Because of the relative compactness of Xw.r.t. the upper–Vietoris topology, there must exist finitely many P1, ..., Pn∈Xwith M⊆Sn j=1(Oc Pj)+X. Now, from Proposition 3.1 we get R=SM∈MM⊆Sn j=1 OPj, where every OPjis a finite union of members of the original cover {Oi|i∈I}by construction.  288 R. Bartsch and H. Poppe Corollary 3.4. Let (X, τ)be a topological space and let M⊆P0(X)consist of relatively compact subsets of X. If Mis compact w.r.t. the upper–Vietoris topology, then R:= [ M∈M M is relatively compact in (X, τ). Proof. Mis compact and therefore relatively compact in every set, which contains M, especially in the family of all relatively compact subsets of X. So, Lemma 3.3 applies.  References [1] Bartsch, R., Dencker, P., Poppe, H., Ascoli–Arzel`a–Theory based on continuous convergence in an (almost) non–Hausdorff setting, in ”Categorical Topology”; Dordrecht (1996). [2] Beer, G., Tamaki, R.,On hit-and-miss hyperspace topologies, Commentat. Math. Univ. Carol. 34 (1993), No.4, 717-728. [3] Beer, G., Tamaki, R., The infimal value functional and the uniformization of hit–and–miss hyperspace topologies, Proc. Am. Math. Soc. 122, No.2 (1994), 601-612. [4] Comfort, W.W., Negrepontis, S., The Theory of Ultrafilters, Berlin (1974). [5] Klein, E., Thompson, A.C., Theory of correspondences. Including applications to mathematical economics. Canadian Mathematical Society Series of Monographs and Advanced Texts (1984). [6] Michael, E., Topologies on spaces of subsets, Trans. Amer. Math. Soc. 71 (1951), 152–182. [7] Naimpally, S., Hyperspaces and Function Spaces, Q & A in General Topology 9(1991), 33-60. [8] Naimpally, S., All Hypertopologies are Hit-and-Miss, Appl. Gen. Topol. 3, No.1 (2001), 45-53. [9] Poppe, H., Eine Bemerkung ¨uber Trennungsaxiome in R¨aumen von abgeschlossenen Teilmengen topologischer R¨aume, Arch.Math. 16 (1965), 197–199. [10] Poppe, H., Einige Bemerkungen ¨uber den Raum der abgeschlossenen Mengen, Fund.Math. 59 (1966), 159–169. [11] Poppe, H., Compactness in General Function Spaces, Berlin (1974). Received November 2001 Revised September 2002 Ren´ e Bartsch Dept. of Computer Science, Rostock University, Albert-Einstein-Straße 21, 18059 Rostock, Germany E-mail address:[email protected] Harry Poppe Dept. of Mathematics, Rostock University, Universit¨atsplatz 1, 18055 Rostock, Germany E-mail address:[email protected]