Long colimits of topological groups I: Continuous maps and homeomorphisms
aa r X i v : . [ m a t h . GN ] N ov Long colimits of topological groups I: Continuous mapsand homeomorphisms *Rafael Dahmen and G´abor Luk´acsNovember 28, 2019
Abstract
Given a directed family of topological groups, the finest topology on their union makingeach injection continuous need not be a group topology, because the multiplication may failto be jointly continuous. This begs the question of when the union is a topological group withrespect to this topology. If the family is countable, the answer is well known in most cases. Westudy this question in the context of so-called long families, which are as far as possible fromcountable ones. As a first step, we present answers to the question for families of group-valuedcontinuous maps and homeomorphism groups, and provide additional examples.
1. Introduction
Given a directed family { G α } α ∈ I of topological groups with closed embeddings as bonding maps,their union G = S α ∈ I G α can be equipped with two topologies: the colimit space topology definedas the finest topology T making each map G α → G continuous, and the colimit group topology ,defined as the finest group topology A making each map G α → G continuous. The former is alwaysfiner than the latter, which begs the question of when the two topologies coincide.We say that { G α } α ∈ I satisfies the algebraic colimit property (briefly, ACP ) if T = A , that is,if the colimit of { G α } α ∈ I in the category Top of topological spaces and continuous maps coincideswith the colimit in the category
Grp ( Top ) of topological groups and their continuous homomor-phisms.The family { G α } α ∈ I satisfies ACP if and only if ( G, T ) is a topological group. In general, theinversion ( G, T ) → ( G, T ) is continuous, and the multiplication m : ( G, T ) × ( G, T ) → ( G, T ) is separately continuous, but need not be jointly continuous. Thus, { G α } α ∈ I satisfies ACP if andonly if m is continuous.It is well known that ACP holds if: (1) the bonding maps are all open; or (2) I is countableand each G α is locally compact Hausdorff ([10, Theorem 4.1] and [8, Propositions 4.7 and 5.4]).Yamasaki showed that for countable families of metrizable groups, these two are the only caseswhere ACP holds. * R. Dahmen and G. Luk´acs / Long colimits of topological groups I
Theorem 1.1. ([15, Theorem 4]) Let { G n } n ∈ N be a countable family of metrizable topologicalgroupswithclosed embeddingsas bondingmaps.If(a) thereis n suchthat G n is notlocallycompact,and(b) forevery n thereis m > n such that G n is notopenin G m ,thenthecolimitspacetopology T isnotagrouptopologyon G = S G n ,thatis, { G n } n ∈ N doesnotsatisfyACP.While ACP is well understood for countable families of metrizable groups, Yamasaki’s the-orem does not hold in the uncountable case. For example, if I = ω and the bonding maps areisometric embeddings, then ACP holds (Corollary 2.11). Similarly, if I = ω and G α = Q β<α M β ,where each M α is metrizable, then ACP holds, and the resulting group is the Σ -product of the M β (Corollary 2.6).Motivated by these examples, the aim of the present paper is to examine directed families thatare as far as possible from countable ones. A directed set ( I , ≤ ) is long if every countable subsetof I has an upper bound in I . It may be tempting to conjecture that all long families of topologicalgroups satisfy ACP; this, however, is not the case (see Example 2.15). Nevertheless, long familiesprovide a number of interesting examples where ACP holds.We say that a Hausdorff space X is a long space if the set K ( X ) of compact subsets of X ordered by inclusion is long, that is, every σ -compact subset of X has a compact closure. Theproduct L ≥ := ω × [0 , equipped with order topology generated by the lexicographic order iscalled the Closed Long Ray . The
Long Line L is obtained by gluing together two copies of theClosed Long Ray, one with the reverse order and one with the usual order, at the boundary points (0 , . The connected spaces L ≥ and L share many of the properties of the zero-dimensionalspace ω : they are locally compact Hausdorff, first countable, and normal, but neither metrizablenor paracompact. Both L ≥ and L are long spaces, and any cardinal of uncountable cofinality withthe order topology is a long space.For a topological space X and a topological group M , we put C cpt ( X, M ) for the set ofcontinuous compactly supported M -valued functions on X , where the support of f : X → M is cl X { x ∈ X | f ( x ) = e M } . We equip C cpt ( X, M ) with pointwise operations and the uniform topol-ogy. The next theorem shows that for a long space X , the uniform topology is the “natural” one on C cpt ( X, M ) . Theorem A.
Let X be a long space and M be a metrizable group. For K ∈ K ( X ) , let C K ( X, M ) denote the subgroup of C cpt ( X, M ) consisting of functions with support in K . Then the family { C K ( X, M ) } K ∈ K ( X ) satisfies ACP and colim K ∈ K ( X ) C K ( X, M ) = C cpt ( X, M ) . (1)Recall that a space X is pseudocompact if every continuous real-valued map on X is bounded.One says that X is ω -bounded if every countable subset of X is contained in a compact subset.Long spaces have previously been studied under the names strongly ω -bounded and σC -bounded ([13] and [12]). The relationship between a Hausdorff space X being a long space and othercompactness-like properties is as follows:compact = ⇒ long = ⇒ ω -bounded = ⇒ pseudocompact , (2) . Dahmen and G. Luk´acs / Long colimits of topological groups I and for metrizable spaces they are all equivalent. Nyikos showed that an ω -bounded space need notbe long ([13]). The next theorem shows that the requirement of X being long cannot be omittedfrom Theorem A. Theorem B.
Let X be a locally compact Hausdorff space and V a non-trivial metrizable topolog-ical vector space. If X is not pseudocompact, then the following statements are equivalent: (i) the family { C K ( X, V ) } K ∈ K ( X ) satisfies ACP; (ii) X ∼ = N (discrete) and V is finite dimensional. The proofs of Theorems A and B are presented in Section 3, and are based on basic propertiesof long colimits established in Section 2.Another interesting way to obtain a family of topological groups indexed by the compact sub-sets of a space X is considering Homeo K ( X ) , the homeomorphisms that are the identity outside agiven compact set K ∈ K ( X ) , equipped with the compact open topology. The notion of support,which plays a central role in Theorem A, can also be defined for homeomorphisms: the support of h : X → X is cl X { x ∈ X | h ( x ) = x } . It is not hard to show that { Homeo K ( R ) } K ∈ K ( R ) does notsatisfy ACP. In fact, a more general negative result, in the flavour of Theorem B, holds. Theorem C.
Let X be a locally compact metrizable space that is not compact, and supposethat Homeo U ( X ) is not locally compact for every non-empty open set U with U compact. Then { Homeo K ( X ) } K ∈ K ( X ) does not satisfy ACP. The proof of Theorem C is presented in Section 4, where we establish a partial analogue ofTheorem B for homeomorphism groups, and show that a stronger, more general version of Theo-rem C holds. Notably, the spaces satisfying the conditions of Theorem C are not long. For certainlong spaces, the following analogue to Theorem A holds.For a Tychonoff space X , we put Homeo cpt ( X ) for the set of compactly supported homeo-morphisms of X . Every compactly supported homeomorphism on X can be extended to a home-omorphism of the Stone- ˇCech compactification βX of X that is the identity on the remainder βX \ X . Thus, the group Homeo cpt ( X ) of compactly supported homeomorphisms of X is a sub-group of Homeo( βX ) , and we may equip Homeo cpt ( X ) with the compact-open topology inducedby βX . Then Homeo K ( X ) = Homeo K ( βX ) is a topological subgroup of Homeo cpt ( X ) for every K ∈ K ( X ) . Definition 1.2.
A Tychonoff space X has the Compactly Supported Homeomorphism Property ( CSHP ) if colim K ∈ K ( X ) Homeo K ( X ) = Homeo cpt ( X ) , (3)where the left-hand side is equipped with the colimit space topology.Clearly, if X has CSHP, then { Homeo K ( X ) } K ∈ K ( X ) satisfies ACP; however, the converseis false. For example, if X is an infinite discrete set, then each Homeo K ( X ) is finite, and so { Homeo K ( X ) } K ∈ K ( X ) satisfies ACP, but X does not have CSHP (see Corollary 5.6). Theorem D.
Suppose that
R. Dahmen and G. Luk´acs / Long colimits of topological groups I (a) X = L ≥ ; or (b) X = L ; or (c) X = κ n × λ ×· · ·× λ j , where κ is an uncountable regular cardinal, λ , . . . , λ j are successorordinals smaller than κ , and n, j ∈ N .Then X has CSHP. The proof of Theorem D is presented in Section 5, where we study CSHP.Finally, we turn to posing some open problems.The directed limit of T -spaces or normal spaces is again T or normal, respectively; however,the directed limit of Hausdorff or Tychonoff spaces need not be Hausdorff or Tychonoff ([9]). Problem I.
Let { G α } α ∈ I be a long directed family of topological groups with closed embeddingsas bonding maps. (a) Is the colimit group topology A Hausdorff? (b)
Is the colimit space topology T Hausdorff?
There is a small gap between the negative statement established in Theorem B and the positiveone in Theorem A. There are pseudocompact spaces that are not long ([13]).
Problem II.
Let X be a pseudocompact Tychonoff space that is not long. Does { C K ( X, R } K ∈K ( X ) satisfy ACP? There is a big gap between the negative statement established in Theorem C (or its general-ization in Section 4) and the positive one in Theorem D. There are many long locally compactTychonoff spaces that are not compact, such as finite powers of the long line or the long ray, but itis not known whether they satisfy CSHP.
Problem III.
Characterize the long, Tychonoff, non-compact spaces X that satisfy CSHP.
2. Long families and tightness
A directed family of topological spaces { X α } α ∈ I is called strict if every bonding map X α → X β isan embedding. A topology S on the set-theoretic union X = S α ∈ I X α is admissible if it makes eachinclusion X α → ( X, S ) an embedding. If there is an admissible topology on X , then the family isstrict. Furthermore, it can easily be seen that if the bonding maps are closed (or open ) embeddings,the colimit space topology T := { U ⊆ X | U ∩ X α is open in X α for all α ∈ I } (4)of a family of spaces is admissible.In this section, we provide a sufficient condition for the colimit space topology to be the onlyadmissible one (Proposition 2.3), and use it to show that ACP holds for many long families oftopological groups.Recall that the tightness t ( X, S ) of a topological space ( X, S ) is the smallest cardinal κ suchthat every point p in the closure of a subset A ⊆ X is in the closure of a subset C ⊆ A with | C | ≤ κ .One says that a space is κ -tight [ countably tight ] if t ( X, S ) ≤ κ [if t ( X, S ) = ℵ ]. For ease ofreference, we start off with a well-known characterization of the tightness. . Dahmen and G. Luk´acs / Long colimits of topological groups I Lemma 2.1.
Let X be a topological space and κ be a cardinal. The following statements areequivalent: (i) t ( X ) ≤ κ ; (ii) X = colim C ≤ X, | C |≤ κ C in Top , that is, for every topological space Y and f : X → Y , if f | C is contin-uous for every subset C ⊆ X with | C | ≤ κ , then f is continuous on X . P ROOF . (i) ⇒ (ii): Let f : X → Y be a map such that f | C is continuous for every subset C ⊆ X with | C | ≤ κ . To show that f is continuous, let A ⊆ X and x ∈ cl A . Since t ( X ) ≤ κ , there is C ⊆ A with | C | ≤ κ such that x ∈ cl C . Put D = C ∪{ x } . By our assumption, f | D is continuous, and so f ( x ) ∈ f (cl D C ) ⊆ cl f ( C ) ⊆ cl f ( A ) . (5)Thus, f (cl A ) ⊆ cl f ( A ) , as desired.(ii) ⇒ (i): For A ⊆ X , put cl κ ( A ) := S { cl C | C ⊆ A, | C | ≤ κ } . It is easily shown that cl κ is aKuratowski closure operator, and as such it defines a topology T κ on X ([5, 1.2.7]). The topology T κ is finer than the original topology T of X , and by (ii), the identity map ( X, T ) → ( X, T κ ) iscontinuous. Therefore, T = T κ , and hence cl A = cl κ A for every A ⊆ X . Definition 2.2.
Given a cardinal κ , a directed set ( I , ≤ ) is κ -long if every subset C ⊆ I with | C | ≤ κ has an upper bound in I . One says that I is long if it is ℵ -long. A directed family of spaces orgroups is κ -long [long] if it is indexed by a κ -long [long] set.A totally ordered set is long if and only if its cofinality is not ω ; however, this equivalencefails for ordered sets that are not totally ordered. For example, ω × ω with the product order hascofinality ω , but it is not long, because ω ×{ } has no upper bound. Proposition 2.3.
Let { X α } α ∈ I be a κ -long family of topological spaces. For an admissible topology S on X = S α ∈ I X α , the following are equivalent: (i) t ( X, S ) ≤ κ ; (ii) S = T (the colimit space topology), and t ( X α , S ) ≤ κ for all α ∈ I . P ROOF . (i) ⇒ (ii): One has S ⊆ T , because the embeddings X α → ( X, S ) induce a continuousmap ( X, T ) → ( X, S ) . For the reverse inclusion, we show that ι : ( X, S ) → ( X, T ) is continuous.Let C ⊆ X be such that | C | ≤ κ . Since I is κ -long, there is α ∈ I such that C ⊆ X α . The subspacetopology induced by S on C coincides with the topology induced by X α , because S is admissible.Thus, ι | C : ( C, S | C ) → X α → ( X, T ) (6)is continuous. Therefore, ι is continuous, because ( X, S ) is κ -tight. Hence, S = T . The secondstatement follows from X α being a subspace of ( X, S ) .(ii) ⇒ (i): It follows from Lemma 2.1 that the colimit of κ -tight spaces is κ -tight. Corollary 2.4.
Let { G α } α ∈ I be a κ -long family of topological groups, and endow G = S α ∈ I G α withthe colimit space topology T . If the product G × G is κ -tight, then G is a topological group, and { G α } α ∈ I satisfies ACP. R. Dahmen and G. Luk´acs / Long colimits of topological groups I P ROOF . Put X α = G α × G α for every α ∈ I . Since T × T is admissible on G × G = S α ∈ I X α , byProposition 2.3, G × G coincides with the colimit of the spaces { X α } α ∈ I . Therefore, the multipli-cation G × G → G is continuous.The following construction provides a wealth of examples of countably tight spaces that neednot be first countable. Let { ( Y j , j ) } j ∈ J be a family of spaces with base points. The support of y = ( y j ) ∈ Q j ∈ J Y j is supp( y ) := { j ∈ J | y j = 0 j } . The Σ -product of the family is X j ∈ J ( Y j , j ) := n y ∈ Y j ∈ J Y j | supp( y ) is countable o (7)equipped with the topology induced by the product topology. Theorem 2.5. ([2, 6.16]) Let { ( Y j , j ) } j ∈ J be a family of first countable spaces with base points.Then P j ∈ J ( Y j , j ) is countablytight. Corollary 2.6.
Let { G j } j ∈ J be a family of metrizable topological groups. Put I = [ J ] ≤ ω orderedby inclusion, and for D ∈ I , put G D = Q j ∈ D G j × Q j / ∈ D { e j } with the product topology. Then { G D } D ∈ I satisfies ACP, and colim D ∈ I G D = P j ∈ J ( G j , e j ) . P ROOF . Put G := S D ∈ I G D ⊆ Q j ∈ J G j , and let S denote the product topology on G . By Theorem 2.5, ( G, S ) is countably tight, and it follows from the construction that S is admissible. Thus, byProposition 2.3, S coincides with the colimit space topology T , because the indexing set I islong. Therefore, ( G, T ) is a topological group, being a topological subgroup of Q j ∈ J G j . Hence, T coincides with the colimit group topology. Example 2.7.
Observe that I cannot be replaced with the set of finite subsets of J in Corollary 2.6.Let J = N , and let each G j be any metrizable group that is not locally compact. Then G D is metriz-able but not locally compact for every non-empty finite D ⊆ J and the bonding maps are not open,because the G j are not discrete. Thus, by Theorem 1.1, the colimit space topology is not a grouptopology.Theorem 2.5 has an interesting consequence for permutation groups too. For a set J , let Sym( J ) denote the group of permutations of J equipped with the pointwise topology induced by J J , where J is equipped with the discrete topology. It is well known that Sym( J ) is a topological group.The support of a permutation σ is the set supp( σ ) = { x ∈ J | σ ( x ) = x } . Let Sym ω ( J ) denote thesubgroup of Sym( J ) consisting of permutations with countable support. Corollary 2.8.
Let J be a set, put I = [ J ] ≤ ω ordered by inclusion, and for D ∈ I , put G D = { σ ∈ Sym( J ) | supp( σ ) ⊆ D } , (8) equipped with the subgroup topology. Then { G D } D ∈ I satisfies ACP, and colim D ∈ I G D = Sym ω ( J ) . . Dahmen and G. Luk´acs / Long colimits of topological groups I P ROOF . Consider the family { ( J, x ) } x ∈ J of spaces with base points. The support of a permutation σ ∈ Sym( J ) coincides with the supp( σ ) of σ as an element of J J . Thus, Sym ω ( J ) = Sym( J ) ∩ X x ∈ J ( J, x ) ! . (9)Therefore, by Theorem 2.5, Sym ω ( J ) is countably tight. It follows from the construction that thetopology of Sym ω ( J ) is admissible, and so, by Proposition 2.3, the topology of Sym ω ( J ) coincideswith the colimit space topology T , because the indexing set I is long. Hence, (Sym ω ( J ) , T ) is atopological group, and T coincides with the colimit group topology.We turn now to families of metrizable spaces and groups with well-behaved bonding maps. Theorem 2.9.
Let { ( X α , d α ) } α ∈ I be a strict long family of metrizable spaces. If the bonding maps ( X α , d α ) → ( X β , d β ) are Lipschitz with a fixed constant L > (in particular, if they are isometries),then d ( x, y ) := lim sup α ∈ I d α ( x, y ) = inf α ∈ I sup { d β ( x, y ) | x, y ∈ X β and β ≥ α } (10) is a metric on X = S α ∈ I X α , and it induces the colimit space topology T . In order to prove Theorem 2.9, we need a technical lemma.
Lemma 2.10.
Let ( I , ≤ ) be a long directed set and f : ( I , ≤ ) → ( R , ≤ ) a monotone function. Then f is eventually constant, that is, there is β ∈ I and s ∈ R such that f ( γ ) = s for all γ ≥ β . P ROOF . By replacing f with arctan ◦ f or arctan ◦ ( − f ) , we may assume without loss of gener-ality that f is non-decreasing and bounded. Put s = sup f ( I ) . For every n ∈ N \{ } , there is β n ∈ I such that f ( β n ) > s − n . Since I is a long directed set, there is β ∈ I such that β ≥ β n for every n .By monotonicity, for every γ ≥ β , one has s ≥ f ( γ ) ≥ f ( β n ) > s − n , and thus f ( γ ) = s . P ROOF OF T HEOREM
We show that: (1) d is a metric; and (2) d induces an admissible topol-ogy on X . By Proposition 2.3, it will follow that d induces the colimit space topology on X .For each α ∈ I and x, y ∈ X α , put e d α ( x, y ) = sup β ≥ α d β ( x, y ) . (11)Then d α ≤ e d α ≤ Ld α , and so e d α is a metric on X α that is (strongly) equivalent to d α . For β ≤ γ and x, y ∈ X β , one has e d β ( x, y ) ≥ e d γ ( x, y ) . Thus, by Lemma 2.10, d ( x, y ) = min β ∈ I { e d β ( x, y ) | x, y ∈ X β } , (12)and for every x, y ∈ X , there is β ( x, y ) ∈ I such that d ( x, y ) = e d γ ( x, y ) for every γ ≥ β ( x, y ) . Since I is long, for every countable subset C ⊆ X there is β ( C ) ∈ I such that d ( x, y ) = e d γ ( x, y ) for every x, y ∈ C and γ ≥ β ( C ) . Therefore, d is a metric, and on every countable set C ⊆ X , it induces thesame topology as d γ for γ large enough. Hence, d is admissible, because the maps X α → X γ areembeddings and metric spaces are countably tight. R. Dahmen and G. Luk´acs / Long colimits of topological groups I
Corollary 2.11.
Let { ( G α , d α ) } α ∈ I be a strict long family of metrizable topological groups withbonding maps that are Lipschitz with a fixed constant L > (in particular, isometries). Then { G α } α ∈ I satisfies ACP, and the colimit topology on G = S α ∈ I G α is generated by the metric d ( x, y ) := lim sup α ∈ I d α ( x, y ) = inf α ∈ I sup { d β ( x, y ) | x, y ∈ G β and β ≥ α } . (13) P ROOF . By Theorem 2.9, the colimit space topology T on G is generated by the metric d , andthus T × T is metrizable on G × G ; in particular, it is countably tight. Therefore, by Corollary 2.4,the statement follows. Example 2.12.
The requirement of having a uniform bound for the Lipschitz constants of thebonding cannot be omitted in Theorem 2.9 and Corollary 2.11. Let M be a non-trivial metrizabletopological group, fix a metric d ≤ on M , and put G α := M α = Q γ<α M for α < ω . Since ω iscofinal in [ ω ] ≤ ω , it follows from Corollary 2.6 that colim α<ω G α = P α<ω ( M, e ) , and thus the colimitspace topology is not metrizable. (In fact, it has an uncountable pseudocharacter, and so it cannotbe first countable.)We construct by transfinite induction a metric d α ≤ on each G α such that for every β < α ,the embedding ι βα : G β → G α and the projection π αβ : G α → G β are Lipschitz. Suppose that such { d γ } γ<α have already been constructed. If α = δ +1 is a successor ordinal, we put d α ( x, y ) := max { d δ ( x <δ , y <δ ) , d ( x δ , y δ ) } (14)for x = ( x γ ) γ<α , y = ( y γ ) γ<α ∈ G α . The embedding ι δα is an isometry, and π αδ is Lipschitz withconstant . Thus, by the inductive hypothesis, the composites ι βα = ι βδ ι δα and π αβ = π αδ π δβ areLipschitz for every β < α . If α is a limit ordinal, pick a strictly increasing sequence α i < α suchthat sup α i = α . For every i < ω , pick L i ≥ such that ι α n α i is L i -Lipschitz for every n ≤ i . For x = ( x γ ) γ<α , y = ( y γ ) γ<α ∈ G α , put d α ( x, y ) := X i<ω L i i +1 d α i ( x <α i , y <α i ) . (15)In what follows, for a Lipschitz map f , we put Lip( f ) for the smallest Lipschitz constant for f .Let n < ω . The projection π αα n is Lipschitz, because d α n ( x <α n , y <α n ) ≤ n +1 L n d α ( x, y ) . To seethat the embedding ι α n α is Lipschitz too, suppose that x, y ∈ G α n ⊆ G α . If i < n , then one has d α i ( x <α i , y <α i ) ≤ Lip( π α n α i ) d α n ( x <α n , y <α n ) . On the other hand, if n ≤ i , then ι α n α i ( x <α n ) = x <α i and ι α n α i ( y <α n ) = y <α i , and so d α i ( x <α i , y <α i ) ≤ L i d α n ( x <α n , y <α n ) . Therefore, d α ( x, y ) = X i Let { E α } α ∈ I be a long family of Banach spaces over K (where K ∈ { R , C } ) withisometries as bonding maps. Then: (a) { E α } α ∈ I satisfies ACP; (b) E = S α ∈ I E α with the colimit space topology T is a Banach space; and (c) ( E, T ) is the colimit of { E α } α ∈ I in TVS K and LCTVS K . P ROOF . Statement (a) follows from Corollary 2.11. By Theorem 2.9, the topology T is generatedby the metric d ( x, y ) = min α ∈ I { d α ( x, y ) | x, y ∈ E α } = k x − y k α , where x, y ∈ E α . (19)Thus, the topology T is generated by the norm k x k = k x k α , where x ∈ E α . (20)In particular, ( E, T ) is a locally convex topological vector space, and (c) follows. In order toshow (b), we prove that the normed space ( E, k·k ) is complete. To that end, let { x n } ⊆ E be aCauchy-sequence. Since I is long, there is α ∈ I such that x n ∈ E α for every n . Thus, { x n } is aCauchy-sequence in the Banach space E α , and x n −→ x for x ∈ E α . Hence, x n −→ x in E . Example 2.14. Let J be an uncountable set, put I = [ J ] ≤ ω , and let K ∈ { R , C } . By Corollary 2.13,for ≤ p < ∞ , the space ℓ p ( J, K ) = [ D ∈ I ℓ p ( D, K ) (21)is the colimit of { ℓ p ( D, K ) } D ∈ I in Top , TVS K , and LCTVS K . Example 2.15. Given a strict long family of topological groups, the colimit space topology neednot be a group topology. Let J be an uncountable set, and for every finite subset F ⊆ J , put G F := R F with the Euclidean topology. Put I := [ J ] ≤ ω , and for D ∈ I , put G D := R ( D ) equippedwith the colimit space topology induced by { G F } F ∈ [ D ] <ω . Since each G D is locally compact, G D is a topological group. (In fact, it can be shown that G D carries the box topology.) One has colim D ∈ I G D = colim D ∈ I colim F ∈ [ D ] <ω G F = colim F ∈ [ J ] <ω G F (22)where the colimits are taken in Top . Bisgaard showed that addition is not continuous in the colimitspace topology generated by the family { G F } F ∈ [ J ] <ω ([3, Theorem]). Thus, { G D } D ∈ I does notsatisfy ACP. R. Dahmen and G. Luk´acs / Long colimits of topological groups I 3. Continuous maps with compact support In this section, we prove Theorems A and B concerning groups of continuous functions with com-pact support. We prove Theorem A by establishing the following more elaborate statement. Theorem A ′ . Let X be a long space and M be a metrizable group. For K ∈ K ( X ) , let C K ( X, M ) denote the subgroup of C cpt ( X, M ) consisting of functions with support in K . Then the family { C K ( X, M ) } K ∈ K ( X ) satisfies ACP, and colim K ∈ K ( X ) C K ( X, M ) = C cpt ( X, M ) . (23) Furthermore, if M is a Banach space over K (where K ∈ { R , C } ), then C cpt ( X, M ) is a Banachspace, and it is the colimit of { C K ( X, M ) } K ∈ K ( X ) in TVS K and LCTVS K . P ROOF . Let ρ be a metric on M . For each K ∈ K ( X ) , equip C K ( X, M ) with the sup-metric d K on K , defined by d K ( f , f ) := sup x ∈ K ρ ( f ( x ) , f ( x )) . The bonding maps are isometries, the family { ( C K ( X, M ) , d K ) } K ∈ K ( X ) is strict, and since X is a long space, the family is also long. Thus, byCorollary 2.11, the family satisfies ACP, and the colimit topology on C cpt ( X, M ) = S K ∈ K ( X ) C K ( X, M ) is generated by the metric d ( f , f ) := lim sup K ∈ K ( X ) d K ( f , f ) = lim sup K ∈ K ( X ) (sup x ∈ K ρ ( f ( x ) , f ( x ))) = sup x ∈ X ρ ( f ( x ) , f ( x )) . (24)The last statement follows by Corollary 2.13. Examples 3.1. Let K ∈ { R , C } .(a) For an ordinal α of uncountable cofinality, C cpt ( α, K ) = colim β<α C [0 ,β ] ( α, K ) = colim β<α C ([0 , β ] , K ) .(b) C cpt ( L ≥ , K ) = colim β<ω C [0 ,β ] ( L ≥ , K ) ( ω is identified with ω ×{ } ⊆ L ≥ ).We establish Theorem B by first proving a stronger result using the following observation. Remark 3.2. Let { G α } α ∈ I be a directed family of topological groups that satisfies ACP, and let H be a closed subgroup of colim α ∈ I G α . Then H = colim α ∈ I ( G α ∩ H ) ([5, Proposition 2.4.15]), and thus { G α ∩ H } α ∈ I also satisfies ACP. Theorem B ′ . Let X be a locally compact Hausdorff space and V a non-trivial metrizable topologi-cal vector space. If X contains a regular closed σ -compact non-compact subset, then the followingstatements are equivalent: (i) the family { C K ( X, V ) } K ∈ K ( X ) satisfies ACP; (ii) X ∼ = N (discrete) and V is finite dimensional. P ROOF . (ii) ⇒ (i): Since X is discrete and countable, K ( X ) = [ X ] <ω is countable. Since V isfinite dimensional, the topological vector space C K ( X, V ) ∼ = V | K | is finite dimensional, and hencelocally compact for every K ∈ K ( X ) (cf. [14, Theorem 3.6]). Thus, { C K ( X, V ) } K ∈ K ( X ) , being a . Dahmen and G. Luk´acs / Long colimits of topological groups I countable family of locally compact groups, satisfies ACP ([10, Theorem 4.1] and [8, Propositions4.7 and 5.4]).(i) ⇒ (ii): Since { C K ( X, V ) } K ∈ K ( X ) satisfies ACP, by Remark 3.2, for every closed subgroup H of G := colim K ∈ K ( X ) C K ( X, V ) , one has H = colim K ∈ K ( X ) ( C K ( X, V ) ∩ H ) , (25)and so { C K ( X, V ) ∩ H } K ∈ K ( X ) satisfies ACP. In particular, if A is a closed subspace of X and H is the (closed) subgroup of G consisting of functions whose support is contained in A , then C K ( X, V ) ∩ H = C K ∩ A ( X, V ) , and thus { C K ( X, V ) } K ∈ K ( A ) satisfies ACP.In order to prove that X is discrete, let U ⊆ X be an open subset such that cl X U is compact.We show that U is finite. Let A ⊆ X be a regular closed σ -compact non-compact subspace of X .Then A contains a countable increasing family { K n } ∞ n =1 of compact subsets such that A = ∞ S n =1 K n .Without loss of generality, we may assume that cl X U ⊆ K . By enlarging the K n if necessary, wemay also assume that K n ⊆ int A K n +1 for every n . Then { K n } ∞ n =1 is cofinal in K ( A ) , that is, everycompact subset of A is contained in one of the K n . Thus, by (25), H = colim n C K n ( X, V ) , (26)and in particular, addition is continuous in the colimit space topology on the right-hand side. There-fore, since each C K n ( X, V ) is metrizable, by Theorem 1.1,(a) C K n ( X, V ) is locally compact for every n , or(b) there is n such that C K n ( X, V ) is an open subgroup of C K m ( X, V ) for every m > n .Since A is regular closed and not compact, int X A \ K n is not empty for every n . In particular, thereis a continuous non-zero f : X → R with compact support K such that K ⊆ int X A \ K n . Since { K n } ∞ n =1 is cofinal in K ( A ) , there is m such that K ⊆ K m . Thus, C K n ( X, R ) ( C K m ( X, R ) , and itfollows that C K n ( X, V ) ( C K m ( X, V ) , because V is non-trivial. Consequently, C K n ( X, V ) cannotbe an open subgroup of the (path) connected group C K m ( X, V ) .Therefore, (b) is not possible, and (a) holds: C K n ( X, V ) is locally compact for every n . Since C K n ( X, V ) is a topological vector space, it is locally compact if and only if it is finite dimensional(cf. [14, Theorem 3.6]). Hence, (a) implies that each K n is finite and V is finite dimensional. Inparticular, U ⊆ K is finite, which shows that X is discrete.Since X contains a non-compact subset A , it cannot be finite, and so it remains to show that X is countable. Fix a one-dimensional subspace W of V , and let H denote the (closed) subgroupof G consisting of functions with image in W . Then H ∩ C K ( X, V ) = C K ( X, W ) = R | K | for every K ∈ K ( X ) , and by (25), { R | K | } K ∈ K ( X ) satisfies ACP. Therefore, by Example 2.15, X is count-able.Theorem B follows from Theorem B ′ and the following lemma. Lemma 3.3. If X is a locally compact Hausdorff space and X is not pseudocompact, then X contains a regular closed σ -compact non-pseudocompact subset. R. Dahmen and G. Luk´acs / Long colimits of topological groups I P ROOF . Let f : X → (0 , ∞ ) be a continuous unbounded map. Pick x ∈ X . For every n > , pick x n +1 ∈ X such that f ( x n +1 ) > f ( x n )+2 . Since f is continuous, for every n ∈ N there is a neighbor-hood U n of x n such that U n is compact and f ( U n ) ⊆ ( f ( x n ) − , f ( x n )+1) . To show that { U n } n ∈ N is locally finite, let x ∈ X , and pick a neighborhood V of x such that f ( V ) ⊆ ( f ( x ) − , f ( x )+1) . Itfollows from the choice of the { x n } n ∈ N that V meets at most two members of the family { U n } n ∈ N .Thus, A := [ n ∈ N U n = [ n ∈ N U n (27)is regular closed ([5, 1.1.11]) and σ -compact. The set A is not pseudocompact, because f | A isunbounded.Theorems B and B ′ yield the following corollary. Corollary 3.4. Let X be a non-discrete locally compact Hausdorff space, and let V be a non-trivial metrizable topological vector space. If (a) X is σ -compact and non-compact; or (b) X is not pseudocompact,then the family { C K ( X, V ) } K ∈ K ( X ) does not satisfy ACP. 4. Homeomorphisms with compact support: negative results For a locally compact Hausdorff space, the homeomorphism group Homeo( X ) of X equipped withthe compact-open topology need not be a topological group; however, if X is compact or locallyconnected, then the compact-open topology is a group topology on Homeo( X ) ([1, Theorem 4]).For a compact subset K of X , let Homeo K ( X ) denote the group of homeomorphisms of X thatfix every point outside K (that is, h ( x ) = x for all x ∈ X \ K ), and equip it with the compact-opentopology. Then Homeo K ( X ) is topologically isomorphic to a subgroup of Homeo( K ) , where thelatter is equipped with the compact-open topology, and as such, Homeo K ( X ) is a topologicalgroup.In this section, we consider when the family { Homeo K ( X ) } K ∈ K ( X ) satisfies ACP. We start offby proving a partial analogue of Theorem B. Theorem 4.1. Let X be a locally compact Hausdorff space, and let A ⊆ X be a regular closed, σ -compact, non-compact, metrizable subset. If { Homeo K ( X ) } K ∈ K ( X ) satisfies ACP, then at leastone of the following holds: (i) Homeo M ( X ) is locally compact for every compact metrizable subset M ⊆ X , or (ii) there is a regular closed compact K ⊆ A such that: (1) for all K ∈ K ( A ) with K ⊆ K , the subset Homeo K ( X ) is open in Homeo K ( X ) ; and (2) for all C ∈ K ( A \ K ) , the group Homeo C ( X ) is discrete. . Dahmen and G. Luk´acs / Long colimits of topological groups I P ROOF . We show that if (i) fails, then (ii) holds. To that end, let M be a compact metrizablesubset of X such that Homeo M ( X ) is not locally compact. By replacing M with cl(int M ) ifnecessary, we may assume that M is regular closed. Furthermore, by replacing A with A ∪ M ifnecessary, we may assume that M ⊆ A .Since A is σ -compact, it contains a countable increasing family { K n } ∞ n =1 of compact subsetssuch that A = ∞ S n =1 K n . By enlarging the K n if necessary, we may assume that M ⊆ K n ⊆ int A K n +1 for every n . Then { K n } ∞ n =1 is cofinal in K ( A ) , that is, every compact subset of A is contained inone of the K n .Since { Homeo K ( X ) } K ∈ K ( X ) satisfies ACP, by Remark 3.2, for every closed subgroup H of G := colim K ∈ K ( X ) Homeo K ( X ) , one has H = colim K ∈ K ( X ) (Homeo K ( X ) ∩ H ) , (28)and so { Homeo K ( X ) ∩ H } K ∈ K ( X ) satisfies ACP. In particular, if A is a closed subspace of X and H is the (closed) subgroup of G consisting of homeomorphisms whose support is contained in A ,then Homeo K ( X ) ∩ H = Homeo K ∩ A ( X ) , and thus { Homeo K ( X ) } K ∈ K ( A ) satisfies ACP. By (28), H = colim n Homeo K n ( X ) , (29)and in particular, the group multiplication is continuous in the colimit space topology on the right-hand side. Therefore, since each Homeo K n ( X ) is metrizable, by Theorem 1.1,(a) Homeo K n ( X ) is locally compact for every n , or(b) there is n such that Homeo K n ( X ) is an open subgroup of Homeo K m ( X ) for every m > n .By our assumption, Homeo M ( X ) is not locally compact, and M ⊆ K n for every n . So, (a) is notpossible, and (b) holds. Put K := cl X (int X K n ) . Observe that Homeo K ( X ) = Homeo K n ( X ) .We show that (ii) holds for K .To show (ii)(1), let K ∈ K ( A ) such that K ⊆ K . Since { K n } ∞ n =1 is cofinal in K ( A ) , thereis m > n such that K ⊆ K m . Thus, Homeo K ( X ) is open in Homeo K m ( X ) , and in particular, Homeo K ( X ) is open in Homeo K ( X ) .To show (ii)(2), let C ∈ K ( A \ K ) , and put L := K ∪ C . Since int X K and int X C are disjoint, Homeo K ( X ) commutes with Homeo C ( X ) , and the multiplication map Homeo K ( X ) × Homeo C ( X ) → Homeo L ( X ) (30)is an embedding of topological groups. By (ii)(1), Homeo K ( X ) is open in Homeo L ( X ) . Thus, Homeo K ( X ) is open in Homeo K ( X ) × Homeo C ( X ) , and therefore Homeo C ( X ) is discrete.We establish Theorem C by proving a stronger result. Theorem C ′ . Let X be a locally compact Hausdorff space that is not compact. Each of the follow-ing statements imply the subsequent ones: (i) X is metrizable, locally Euclidean, and has no isolated points; R. Dahmen and G. Luk´acs / Long colimits of topological groups I (ii) X is metrizable and Homeo U ( X ) is not locally compact for every non-empty open set U suchthat U is compact; (iii) X is normal, locally metrizable, and contains (1) a compact subset M such that Homeo M ( X ) is not locally compact; and (2) an infinite discrete closed set T such that for every t ∈ T and every neighborhood V of t with compact closure, Homeo V ( X ) is not discrete; (iv) X contains (1) a compact metrizable subset M such that Homeo M ( X ) is not locally compact; and (2) a regular closed, σ -compact, and metrizable subset A whose interior contains an infinitediscrete closed set T such that for every t ∈ T and every neighborhood V of t with compactclosure, Homeo V ( X ) is not discrete; (v) { Homeo K ( X ) } K ∈ K ( X ) does not satisfy ACP. P ROOF . (i) ⇒ (ii): Let U be a non-empty open set whose closure in X is compact. Since X is locally Euclidean, U contains a non-empty open set V such that there is a homeomorphism h : R n → V for some n ∈ N . Put W := h (( − , n ) . Observe that W = h ([ − , n ) and one has Homeo W ( X ) ∼ = Homeo [ − , n ( R n ) . The group Homeo [ − , n ( R n ) is not locally compact, becauseit contains a copy of Homeo [ − , ( R ) as a closed subgroup, and the latter is known to be non-locally compact (it is not even complete with respect to its left uniformity; see, for example [4]).Since Homeo W ( X ) ⊆ Homeo U ( X ) , it follows that Homeo U ( X ) is not locally compact.(ii) ⇒ (iii): Since X is metrizable but not compact, it contains an infinite closed discrete set T .(iii) ⇒ (iv): Since every compact subset of a locally metrizable space is metrizable, (iii)(1)implies (iv)(1). Without loss of generality, we may assume that T = { t , t , . . . } is countably infinite.In order to show (iv)(2), we construct the set A around T .Let { U i } i ∈ N be a family of open sets in X such that t i ∈ U i and U i is compact for every i ∈ N , and U i ∩ U j = ∅ for i = j . (Such a family exists, because T is discrete and countable.) Put F := X \ S i ∈ N U i .Since F and T are closed disjoint sets in the normal space X , there are disjoint open sets U and V such that T ⊆ U and F ⊆ V . Put W i := U i ∩ U . Then t i ∈ W i and W i is compact, and { W i } i ∈ N islocally finite. Thus, A = [ i ∈ N W i = [ i ∈ N W i (31)is regular closed ([5, 1.1.11]), σ -compact, and its interior contains T . Finally, A is metrizable,because it is locally metrizable and Lindel¨of.(iv) ⇒ (v): Assume that { Homeo K ( X ) } K ∈ K ( X ) satisfies ACP. Then, by Theorem 4.1, either Homeo M ( X ) is locally compact, or A contains a (regular) closed compact subset K such that Homeo C ( X ) is discrete for every C ∈ K ( A \ K ) . By (iv)(1), the first case is not possible, sosuppose that the second case holds. Since T is infinite closed and discrete, T K . Thus, there is t ∈ T such that t / ∈ K . Let V be a neighborhood of t such that V ⊆ A \ K and cl X V is compact,and put C := cl X V . Then, by (iv)(2), Homeo C ( X ) is not discrete. This contradiction shows that { Homeo K ( X ) } K ∈ K ( X ) does not satisfy ACP. . Dahmen and G. Luk´acs / Long colimits of topological groups I 5. The Compactly Supported Homeomorphism Property (CSHP) Our goal in this section is to prove Theorem D. Theorem D. Suppose that (a) X = L ≥ ; or (b) X = L ; or (c) X = κ n × λ ×· · ·× λ j , where κ is an uncountable regular cardinal, λ , . . . , λ j are successorordinals smaller than κ , and n, j ∈ N .Then X has CSHP. We prove parts (a) and (b) of Theorem D by showing that the compact-open topology on Homeo cpt ( L ) is countably tight and coincides with the compact-open topology induced by βX ,and then applying Proposition 2.3. Proposition 5.1. Let X = L ≥ or X = L . Then Homeo cpt ( X ) equipped with the compact-opentopology is countably tight. P ROOF . The statement for X = L ≥ follows from the statement for the case X = L , because Homeo cpt ( L ≥ ) equipped with the compact-open topology is a subspace of Homeo cpt ( L ) equippedwith the compact-open topology.Let X = L , A ⊆ Homeo cpt ( L ) , and suppose that f ∈ cl A . We construct a countable subset C ⊆ A such that f ∈ cl C . Since composition is continuous in the compact-open topology, with-out loss of generality, we may assume that f = id L .For α < ω , put K α = [ − ( α, , ( α, and U α = ( − ( α +1 , , ( α +1 , . The set W α := { h ∈ Homeo cpt ( L ) | h ( K α ) ⊆ U α } (32)is an open neighborhood of id L in the compact open topology, and thus id L ∈ cl( W α ∩ A ) . Since K α is compact and U α is metrizable, the space C ( K α , U α ) is metrizable in the compact-open topology.Consequently, there is a sequence { f ( α ) n } ⊆ W α ∩ A such that f ( α ) n | K α −→ id K α uniformly.We construct now an increasing sequence of ordinals { α m } ⊆ ω such that supp f ( α m ) n ⊆ K α m +1 for every n, m ∈ N . Pick an arbitrary α < ω . Suppose that α m has already been constructed. Since L is a long space, the countable family of compact subsets { supp f ( α m ) n } n ∈ N has an upper bound; inparticular, there is β < ω such that supp f ( α m ) n ⊆ K β for every n ∈ N . Put α m +1 := max( β, α m +1) .Since ( ω , ≤ ) is long, γ := sup m ∈ N α m exists in ω , and supp f ( α m ) n ⊆ K γ for every n, m ∈ N . Put C := { f ( α m ) n | n, m ∈ N } ⊆ A. (33)We show that id L ∈ cl C . Observe that C ⊆ Homeo K γ ( L ) . Pick an order-preserving homeomor-phism ϕ : K γ → [ − , . It suffices to show that id [ − , ∈ cl( ϕCϕ − ) in Homeo([ − , .Let ε > . There is m ∈ N such that a := ϕ ( − ( α m , < − ε and b := ϕ ( α m , > − ε . By ourconstruction, f ( α m ) n | K αm −→ id K αm uniformly, and so ϕf ( α m ) n | K αm ϕ − −→ id [ a,b ] uniformly. Thus, there R. Dahmen and G. Luk´acs / Long colimits of topological groups I is n ∈ N such that | ϕf ( α m ) n | K αm ϕ − ( x ) − x | < ε for every x ∈ [ a, b ] . Since every compactly supportedhomeomorphism of L is order-preserving, for every x ∈ [ b, , one has ≥ ϕf ( α m ) n | K γ ϕ − ( x ) ≥ ϕf ( α m ) n | K γ ϕ − ( b ) > b − ε > − ε, (34)and similarly, for x ∈ [ − , a ] , one has − ≤ f ( α m ) n | K γ ϕ − ( x ) < − ε . Hence, | ϕf ( γ ) n | K αm ϕ − ( x ) − x | < ε for every x ∈ [ − , . P ROOF OF T HEOREM D( A )-( B ). Let X = L ≥ or X = L , and let αX denote its one-point com-pactification. Since X is locally connected, the compact-open topology on Homeo( X ) , and thus on Homeo cpt ( X ) , coincides with the compact-open topology induced by αX ([1, Theorems 1 and 4]).The compact-open topology is admissible with respect to the family { Homeo K ( X ) } K ∈ K ( X ) , andby Proposition 5.1, Homeo cpt ( X ) is countably tight in the compact-open topology. Thus, by Propo-sition 2.3, colim K ∈ K ( X ) Homeo K ( X ) = Homeo cpt ( X ) , (35)where the left-hand side is equipped with the colimit space topology and the right-hand sideis equipped with the compact-open topology. Since Homeo cpt ( X ) is a topological group, thisshows that the family { Homeo K ( X ) } K ∈ K ( X ) satisfies ACP. Therefore, it remains to show thatthe compact-open topology on Homeo cpt ( X ) coincides with the compact-open topology inducedby βX .(a) Since β L ≥ = α L ≥ ([6, 16H6]), the statement follows for X = L ≥ .(b) It is easily seen that β L = L ∪{−∞ , ∞} . (Since L is a pushout of two copies of L ≥ over { } and β preserves pushouts, β L is the pushout of two copies of β L ≥ = α L ≥ over { } .) Since L is open in β L , the compact-open topology on Homeo cpt ( L ) (induced by L ) is coarser than thecompact-open topology induced by βX . In order to prove the converse, let K ⊆ U ⊆ β L , where K is compact and U is open, and consider the subbasic neighborhood of the identity W = { h ∈ Homeo cpt ( L ) | h β ( K ) ⊆ U } , (36)where h β : β L → β L is the extension of h . Since K is compact, it may be covered by finitely manyconnected closed subsets I , . . . , I n of β L such that I j ⊆ U . For j = 1 , . . . , n , put W j := { h ∈ Homeo cpt ( L ) | h β ( I j ) ⊆ U } . (37)One has n T j =1 W j ⊆ W , and so without loss of generality we may assume that K is connected. Wedistinguish among three cases. Case 1. If {−∞ , ∞} ⊆ K , then K = U = β L , and thus W = Homeo cpt ( L ) is open. Case 2. If K ⊆ L , then by replacing U with U ∩ L , one sees that W is open in the compact-open topology. (Compactly supported homeomorphisms of L preserve the order, and thus theirextensions to β L fix ∞ and −∞ .) . Dahmen and G. Luk´acs / Long colimits of topological groups I Case 3. If | K ∩{−∞ , ∞}| = 1 , then without loss of generality we may assume that ∞ ∈ K and −∞ / ∈ K . There are closed connected sets K , K ⊆ K such that K = K ∪ K , K ⊆ L ≥ \{ } , and ∞ / ∈ K . Put U := U ∩ L ≥ \{ } and U = U , and for i = 1 , , put W ′ i := { h ∈ Homeo cpt ( L ) | h β ( K i ) ⊆ U i } . (38)One has W ′ ∩ W ′ ⊆ W . By Case 2, W ′ is open in the compact-open topology. Thus, it remains toshow that W ′ is a neighborhood of the identity in the compact-open topology. Put K ′ := K ∪ ( − K ) and U ′ = U ∪ ( − U ) , where − stands for the involution interchanging the two copies of α L ≥ in β L , and put W ′ := { h ∈ Homeo cpt ( L ) | h β ( K ′ ) ⊆ U ′ } . (39)If h ∈ W ′ , then h β ( K ) ⊆ U ∪ ( − U ) , and the latter union is disjoint, because U ⊆ L ≥ \{ } . Thus,the connected set h β ( K ) is contained in U in its entirety, because h β ( ∞ ) ∈ h β ( K ) ∈ U . There-fore, h ∈ W ′ . This shows that W ′ ⊆ W ′ . For V = β L \ K and C = β L \ U , one has W ′ = { h ∈ Homeo cpt ( L ) | h β ( V ) ⊇ C } = { h − ∈ Homeo cpt ( L ) | h β ( C ) ⊆ V } . (40)Since C, V ⊆ L and the compact-open topology on Homeo cpt ( L ) is a group topology, this provesthat W ′ is open in the compact-open topology.Similarly to the proofs found in Section 2 and the proof of Theorem A ′ , the proofs of parts (a)and (b) of Theorem D rely on the tightness of the groups involved. Unfortunately, this approachcannot be used to prove part (c) of Theorem D, because Homeo K ( X ) need not be countably tightfor a big space X . Example 5.2. Let κ be an ordinal, and λ < κ a limit ordinal of uncountable cofinality (for example, λ = ω and κ = ω ). The set [0 , λ ] = [0 , λ +1) is clopen in X = κ , and thus Homeo [0 ,λ ] ( κ ) = Homeo([0 , λ ]) , (41)where the latter is equipped with the compact-open topology. We show that t (Homeo([0 , λ ])) ≥ cf( λ ) . (42)Consider the subset A := S β<λ Homeo [0 ,β ] ([0 , λ ]) of Homeo([0 , λ ]) . If B ⊆ A such that | B | < cf( λ ) ,then τ := sup S f ∈ B supp( f ) < λ , and thus B ⊆ Homeo [0 ,τ ] ([0 , λ ]) ; in particular, B ⊆ A . On the otherhand, for h : [0 , λ ] → [0 , λ ] defined by h ( α +1) = h ( α +2) and h ( α +2) = α +1 for limit ordinals α and h ( x ) = x otherwise, one can easily see that h / ∈ A but h ∈ A . (For every limit ordinal β < λ , put h β ( x ) = h ( x ) if x ≤ β and h β ( x ) = x if x > β . Clearly, h β ∈ A , and h β → h in the uniform topology,which coincides with the compact-open topology.)The proof of part (c) of Theorem D consists of a number of steps. We first show that CSHP isinherited by clopen subsets, and reduce the proof to the special case of X = κ n . We then provide adescription of the topology of Homeo( K ) for a compact zero-dimensional space K , and use it todescribe open neighborhoods of the identity in Homeo K ( κ n ) . R. Dahmen and G. Luk´acs / Long colimits of topological groups I Lemma 5.3. Let X be a Tychonoff space and A ⊆ X a clopen subset. (a) Homeo cpt ( A ) naturally embeds as a closed topological subgroup into Homeo cpt ( X ) ; and (b) if X has CSHP, then so does A . P ROOF . (a) Since A is clopen, βA embeds as a clopen subset into βX , and every homeomor-phism of βA can be extended to a homeomorphism of βX by defining it as the identity on βX \ βA .Thus, Homeo( βA ) embeds as a topological subgroup into Homeo( βX ) , where the groups areequipped with the respective compact-open topologies. Therefore, Homeo cpt ( A ) embeds as a topo-logical subgroup into Homeo cpt ( X ) , and one can identify Homeo cpt ( A ) = { f ∈ Homeo cpt ( X ) | f | X \ A = id X \ A } . (43)Hence, Homeo cpt ( A ) is closed in Homeo cpt ( X ) .(b) Since X has CSHP, by Remark 3.2, for every closed subgroup H of Homeo cpt ( X ) , one has H = colim K ∈ K ( X ) (Homeo K ( X ) ∩ H ) . (44)In particular, for H = Homeo cpt ( A ) , one obtains Homeo cpt ( A ) = colim K ∈ K ( X ) (Homeo K ( X ) ∩ Homeo cpt ( A )) = colim K ∈ K ( A ) Homeo K ( A ) , (45)and hence A has CSHP.If κ is a cardinal and λ , . . . , λ j < κ are successor ordinals, then κ n × λ ×· · ·× λ j is a clopensubset of κ n + j . Thus, Lemma 5.3 yields the following reduction. Corollary 5.4. Let κ be a cardinal. If the space κ n has CSHP for every n , then κ n × λ ×· · ·× λ j has CSHP for every n and successor ordinals λ , . . . , λ j < κ . For a Tychonoff space X and a finite clopen partition A = { A , . . . , A k } of X (i.e., each A i isclopen), put U ( A ) := { f ∈ Homeo cpt ( X ) | f ( A i ) = A i for all i } , (46)the subgroup of homeomorphisms f such that each A i is f -invariant. Lemma 5.5. Let X be a zero-dimensional space. Then the family { U ( A ) | A is a finite clopen partition of X } (47) is a base at the identity for Homeo cpt ( X ) . P ROOF . Since there is a one-to-one correspondence between clopen partitions of X and of βX ,and Homeo cpt ( X ) is a topological subgroup of Homeo( βX ) (where the latter is equipped with thecompact-open topology), we may assume without loss of generality that X is compact. . Dahmen and G. Luk´acs / Long colimits of topological groups I If A = { A , . . . , A k } is a finite clopen partition of X , then each A i is compact, and U ( A ) = k \ i =1 { f ∈ Homeo cpt ( X ) | f ( A i ) ⊆ A i } (48)is an open neighborhood of the identity in Homeo cpt ( X ) . Conversely, consider a subbasic open set W = { f ∈ Homeo cpt ( X ) | f ( K ) ⊆ U } containing the identity, where K is compact and U is openin X . Since W contains the identity, K ⊆ U , and there is a clopen subset A such that K ⊆ A ⊆ U ,because X is zero-dimensional. Thus, for A := { A , X \ A } , one has U ( A ) ⊆ W . Since the collec-tion of sets of the form U ( A ) is a filter base, this completes the proof. Corollary 5.6. If a Tychonoff space X contains an infinite discrete clopen subset, then X does nothave CSHP. P ROOF . By Lemma 5.3(b), it suffices to show that if X is an infinite discrete space, then it doesnot have CSHP. If X is infinite discrete, then Homeo K ( X ) is finite for every K ∈ K ( X ) , andthus colim K ∈ K ( X ) Homeo K ( X ) is discrete too; however, by Lemma 5.5, Homeo cpt ( X ) is not discrete,because U ( A ) is an infinite group for every finite partition A of X .Our next step is to describe open neighborhoods of the identity in Homeo K ( κ n ) = Homeo( K ) for some special compact open subsets K ⊆ κ n . To that end, we describe finite clopen partitionsof spaces of the form λ ×· · ·× λ n , where λ i are ordinals with cf( λ i ) = ω . (Observe that the space λ ×· · ·× λ n is long if and only if cf( λ i ) = ω for every i .) To that end, we introduce the notion of agrid partition. Definition 5.7. Let λ , . . . , λ n be ordinals, and put X = λ ×· · ·× λ n .(a) For a finite subset F i = { x < · · · < x N } ⊆ λ i , the grid partition of λ i with respect to F i is G λ i F i := { [0 , x ] }∪{ [ x j +1 , x j +1 ] | j = 1 , . . . , N − }∪{ [ x N +1 , · ) } ] , (49)where [ α, β ] := { x ∈ λ i | α ≤ x ≤ β } , and [ α, · ) := { x ∈ λ i | α ≤ x } .(b) For finite subsets F i ⊆ λ i ( i = 1 , . . . , n ), the grid partition of X with respect to { F i } ni =1 is G X { F i } ni =1 := ( n Y i =1 U i | U i ∈ G λ i F i ) . (50) Theorem 5.8. Let λ , . . . , λ n be ordinals with cf( λ i ) = ω , and put X = λ ×· · ·× λ n . For every finiteclopen partition A of X there are finite subsets F i ⊆ λ i such that G X { F i } ni =1 is a refinement of A . In order to prove Theorem 5.8, we first need a lemma. Lemma 5.9. Let λ , . . . , λ n be ordinals, put X = λ ×· · ·× λ n , and let K ⊆ X be a non-empty com-pact subset. Then: (a) K has a maximal element with respect to the lexicographic order on X ; and R. Dahmen and G. Luk´acs / Long colimits of topological groups I (b) if, in addition, K is open, then K is the disjoint union of finitely many compact open sets ofthe form n Q i =1 [ x i , y i ] , where for each i , either x i = 0 or x i is a successor ordinal. P ROOF . Let (cid:22) denote the lexicographic order on X .(a) Let π i : X → λ i denote the canonical projection for i = 1 , . . . , n . We construct K , . . . , K n +1 inductively. Put K = K . Suppose that the compact non-empty sets K , . . . , K i have already beenconstructed. The set π i ( K i ) is compact and non-empty, and so a i := max π i ( K i ) exists. Put K i +1 := π − i ( { a i } ) ∩ K i . (51)Then K n +1 = { ( a , . . . , a n ) } is a singleton, and ( a , . . . , a n ) = max( K, (cid:22) ) .(b) Let a = ( a , . . . , a n ) ∈ K be the maximal element of K with respect to (cid:22) . Since K is open,there is x = ( x , . . . , x n ) ∈ X such that a ∈ n Q i =1 [ x i , a i ] ⊆ K , and either x i = 0 or x i is a successor or-dinal. The set U = n Q i =1 [ x i , a i ] is clopen, and so K ′ := K \ U is again compact open. If K ′ = ∅ , thenwe are done. Otherwise, max( K ′ , (cid:22) ) (cid:22) max( K, (cid:22) ) , and one may repeat the same argument for K ′ . Since each λ i is well-ordered, so is ( X, (cid:22) ) . Therefore, this process will terminate after finitelymany steps.We are now ready to prove Theorem 5.8. P ROOF OF T HEOREM Let A = { A , . . . , A m } be a finite clopen partition of X . Then A ′ := { cl βX A , . . . , cl βX A m } (52)is a partition of βX into compact open subsets.In order to apply Lemma 5.9, we show that βX is a product of ordinals. Each space λ i issequentially compact, because cf( λ i ) = ω . Thus, the product X = n Q i =1 λ i is sequentially compact, andin particular, it is pseudocompact (cf. [5, 3.10.35, 3.10.30]). Therefore, by Glicksberg’s Theorem([7, Theorem 1]), βX = n Y i =1 βλ i . (53)If cf( λ i ) = 1 , then βλ i = λ i , and if cf( λ i ) > ω , then βλ i = λ i +1 ([6, 5N1]). Consequently, βX isalso a product of ordinals.By Lemma 5.9, each cl βX A j is the disjoint union of compact open sets of the form n Q i =1 [ x i , y i ] ,where for each i , either x i = 0 or x i is a successor ordinal. Thus, A ′ has a compact open refinement A ′′ := ( n Y i =1 [ x (1) i , y (1) i ] , . . . , n Y i =1 [ x ( l ) i , y ( l ) i ] ) , (54) . Dahmen and G. Luk´acs / Long colimits of topological groups I where x ( j ) i = 0 or x ( j ) i is a successor ordinal for every i and j . Put F i := { y ( j ) i | j = 1 , . . . , l and y ( j ) i < λ i } . (55)We claim that the grid partition G βX { F i } ni =1 is a refinement of A ′′ . Let M ∈ G βX { F i } ni =1 . Then there are α = ( α , . . . , α n ) and γ = ( γ , . . . , γ n ) in βX such that M = n Q i =1 [ α i , γ i ] . Since A ′′ covers βX , there is k such that α ∈ n Q i =1 [ x ( k ) i , y ( k ) i ] . We show that M ⊆ n Q i =1 [ x ( k ) i , y ( k ) i ] . To that end, it suffices to prove that γ ∈ n Q i =1 [ x ( k ) i , y ( k ) i ] , or equivalently, that γ i ≤ y ( k ) i for every i . Let ≤ i ≤ n . If α i = 0 , then γ i = min F i (or γ i = λ i if F i = ∅ ), and thus γ i ≤ y ( k ) i . Otherwise, α i = y ( r ) i +1 for some r , and γ i is the nextelement of F i (or γ i = λ i ), and thus again γ i ≤ y ( k ) i .This shows that G βX { F i } ni =1 is a refinement of A ′′ . Since F i ⊆ λ i for every i , it follows that G X { F i } ni =1 is a refinement of A .The last step before proving part (c) of Theorem D is a technical proposition that allows one touse grid partitions with a fixed number of points in dealing with open subsets of the colimit spacetopology on colim K ∈ K ( X ) Homeo K ( X ) . Proposition 5.10. Let λ , . . . , λ n be ordinals with cf( λ i ) = ω , and put X = n Q i =1 λ i , and let W be anopen set containing the identity in the colimit space topology on colim K ∈ K ( X ) Homeo K ( X ) . Then thereare m , . . . , m n ∈ N and α (1) ∈ X such that for every α = ( α , . . . , α n ) ≥ α (1) , there are F αi ⊆ α i suchthat | F αi | = m i and U (cid:16) G ↓ α { F αi } ni =1 (cid:17) ⊆ W ∩ Homeo ↓ α ( X ) . (56) P ROOF . Since W is open in the colimit space topology, the intersection W ∩ Homeo ↓ α ( X ) isopen in Homeo ↓ α ( X ) for every α ∈ X . One has Homeo ↓ α ( X ) ∼ = Homeo( ↓ α ) = Homeo cpt ( ↓ α ) , (57)because ↓ α is a compact open subset of the zero-dimensional space X . Thus, by Lemma 5.5, forevery α ∈ X there is a finite clopen partition A α of ↓ α such that U ( A α ) ⊆ W ∩ Homeo ↓ α ( X ) . ByTheorem 5.8 applied to X = ↓ α , each A α has a grid partition refinement G ↓ α { E αi } ni =1 with E i ⊆ α i , andconsequently U ( G ↓ α { E αi } ni =1 ) ⊆ W ∩ Homeo ↓ α ( X ) .We show now that it is possible to choose the grid points E αi in a way that the sizes of E αi stabilize. To that end, let (cid:22) denote the lexicographic order on N n . For every α ∈ X , put ( m α , . . . , m αn ) := min (cid:22) n ( | E α | , . . . , | E αn | ) | E αi ⊆ α i , U (cid:16) G ↓ α { E αi } ni =1 (cid:17) ⊆ W ∩ Homeo ↓ α ( X ) o . (58) R. Dahmen and G. Luk´acs / Long colimits of topological groups I We show that the map f : X → ( N n , (cid:22) ) defined by α ( m α , . . . , m αn ) is monotone. Let α ≤ β ∈ X ,and pick F βi ⊆ β i such that | F βi | = m βi and U ( G ↓ β { F βi } ni =1 ) ⊆ W ∩ Homeo ↓ β ( X ) . For E αi = F βi ∩ α i , onehas U ( G ↓ α { E αi } ni =1 ) ⊆ W ∩ Homeo ↓ α ( X ) and | E αi | ≤ | F βi | = m βi for every i . In particular, ( m α , . . . , m αn ) (cid:22) ( | E α | , . . . , | E αn | ) (cid:22) ( | F α | , . . . , | F αn | ) = ( m β , . . . , m βn ) . (59)Since cf( λ i ) = ω , the directed set ( X, ≤ ) is long, and thus so is its image f ( X ) under the order-preserving map f . Therefore, the countable set f ( X ) has an upper bound ( m , . . . , m n ) ∈ f ( X ) ,and it is the maximal element of f ( X ) . Pick α (1) ∈ X such that f ( α (1) ) = ( m , . . . , m n ) . Then forevery α ≥ α (1) , one has m αi = m i for every i . Hence, one may pick F αi ⊆ α i such that | F αi | = m i and U ( G ↓ β { F αi } ni =1 ) ⊆ W ∩ Homeo ↓ α ( X ) .We turn now to the proof of part (c) of Theorem D. We will show that when the λ i are equal toa regular uncountable cardinal κ , then not only do the cardinalities of the F αi stabilize, but also thesets themselves do. To do so, recall that a subset S of a cardinal κ is called stationary if it intersectsevery closed and unbounded subset of κ . Fodor’s Pressing Down Lemma states that: Theorem 5.11. ([11, Theorem 8.7]) Let κ be a regular uncountable cardinal, and let S ⊆ κ be astationarysubset.If f : S → κ satisfies f ( x ) < x forevery x ∈ S ,thenthereisastationaryset T ⊆ S such that f | T isconstant. P ROOF OF T HEOREM D( C ). By Corollary 5.4, it suffices to prove the statement for X = κ n ,where κ is a regular uncountable cardinal. The colimit space topology on colim K ∈ K ( X ) Homeo K ( X ) isfiner than the topology of Homeo cpt ( X ) . We show the converse.Let W be an open subset of colim K ∈ K ( X ) Homeo K ( X ) and let h ∈ W . Since translation is continuousin the colimit space topology, W := W h − is an open set containing the identity. Thus, by Propo-sition 5.10, there are m , . . . , m n ∈ N and α (1) ∈ X such that for every α ≥ α (1) there are F αi ⊆ α with | F αi | = m i and U (cid:16) G ↓ α { F αi } ni =1 (cid:17) ⊆ W ∩ Homeo ↓ α ( X ) . (60)Let ∆ : κ → κ n denote the diagonal map, and put γ (1) = max ≤ i ≤ n α (1) i , where α (1) = ( α (1)1 , . . . , α (1) n ) . For γ ≥ γ (1) , for every ( i, j ) such that ≤ i ≤ n and ≤ j ≤ m i , let f i,j ( γ ) denote the j -th point in F ∆ γi .Since F ∆ γi ⊆ (∆ γ ) i = γ , one has f i,j ( γ ) < γ . By repeated application of Theorem 5.11, one obtainsa stationary set T ⊆ [ γ (1) , κ ) such that f i,j | T is constant for every ≤ i ≤ n and ≤ j ≤ m i . In otherwords, there are F i ⊆ κ such that F ∆ γi = F i for every γ ∈ T . We show that U (cid:16) G X { F i } ni =1 (cid:17) ⊆ W . (61)Let g ∈ U ( G X { F i } ni =1 ) . Since g has compact support, there is β such that g ∈ Homeo ↓ ∆ β ( X ) . Since T is stationary, in particular, it intersects [ β, κ ) , and so there is γ ∈ T such that β ≤ γ . Therefore, g ∈ U (cid:16) G X { F i } ni =1 (cid:17) ∩ Homeo ↓ ∆ γ ( X ) = U (cid:16) G ↓ ∆ γ { F ∆ γi } ni =1 (cid:17) ( ) ⊆ W . 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