aa r X i v : . [ m a t h . GN ] D ec Topological aspects of poset spaces
Carl Mummert ∗ and Frank Stephan † April 3, 2008 ‡ Revised November 13, 2009
Abstract
We study two classes of spaces whose points are filters on partiallyordered sets. Points in MF spaces are maximal filters, while points inUF spaces are unbounded filters. We give a thorough account of thetopological properties of these spaces. We obtain a complete charac-terization of the class of countably based MF spaces: they are preciselythe second-countable T spaces with the strong Choquet property. Weapply this characterization to domain theory to characterize the classof second-countable spaces with a domain representation. Recent work in mathematical logic [10, 11, 12] has led to an interest in certaintopological spaces formed from filters on partially ordered sets. This paperdescribes the general topology of these poset spaces. ∗ [email protected] . C. Mummert was partially supported by a VIGRE gradu-ate traineeship under NSF Grant DMS-9810759 at the Pennsylvania State University. † [email protected]. F. Stephan is supported in part by NUS grant numberR252-000-212-112. ‡ Accepted for publication in the
Michigan Mathematical Journal . An earlier version ofthis paper was released as National University of Singapore School Of Computing technicalreport TRC6/06. G δ subspace of an MF space is again anMF space. We show that UF spaces are closed under taking G δ subspacesbut not closed under binary products. In Section 5, we establish that posetspaces are of the second Baire category and possess the strong Choquet prop-erty. We give a characterization of the class of countably based MF spacesas the class of second-countable T spaces with the strong Choquet prop-erty. In Section 6, we apply the results of Section 5 to domain theory, givinga complete characterization of the second-countable topological spaces thathave a domain representation. Section 7 contains results on the relationshipbetween MF spaces (not necessarily countably based) and semi-topogenousorders. We use semi-topogenous orders to establish a sufficient condition foran arbitrary space to be homeomorphic to an MF space. In Section 8, weshow that every second-countable poset space is either countable or containsa perfect closed set. Our goal in this section is to define the class of poset spaces and show that thisclass includes all complete metric spaces and all locally compact Hausdorffspaces. We first review some basic definitions about partially ordered sets.A poset is a set P with an reflexive, antisymmetric, transitive relation (cid:22) .That is, the following conditions hold for all p , q and r in P .(1) p (cid:22) p .(2) If p (cid:22) q and q (cid:22) p then q = p .(3) If p (cid:22) q and q (cid:22) r then p (cid:22) r .We write p ≺ q if p (cid:22) q and p = q . If there is no r such that r (cid:22) p and r (cid:22) q then we write p ⊥ q .A filter is a subset F of a poset P satisfying the following two conditions.21) For every p, q ∈ F there is an r ∈ F such that r (cid:22) p and r (cid:22) q .(2) For every p ∈ F and q ∈ P if p (cid:22) q then q ∈ F .A filter F is unbounded if there is no r ∈ P such that r ≺ q for every q ∈ F .Furthermore, F is maximal if there is no strictly larger filter containing F .Every maximal filter is unbounded, but in general not every unbounded filteris maximal.For any poset P , we let UF( P ) denote the set of unbounded filters on P and let MF( P ) denote the set of maximal filters on P . We topologize UF( P )with the basis { N p | p ∈ P } , where N p = { F ∈ UF( P ) | p ∈ F } . We give MF( P ) the topology it inherits as a subset of UF( P ); when we workwith spaces of maximal filters we may write N p to denote the set of maximalfilters containing p . To facilitate the exposition, we sometimes identify p ∈ P with the open set N p and identify a subset U of P with the open set S p ∈ U N p .A UF space is a space of the form UF( P ) and an MF space is a spaceof the form MF( P ). UF spaces and MF spaces are collectively referred toas poset spaces . A poset space is countably based if it is formed from acountable poset. It is possible that P is uncountable but MF( P ) or UF( P )is a second-countable space (an example is provided after Theorem 2.3). Wewill show below that every second-countable poset space is homeomorphic toa countably based poset space. This result justifies our terminology. Remark 2.1.
It is sometimes convenient to work with strict partial ordersinstead of the non-strict partial orders defined above. A strict partial orderis a set P with an irreflexive, transitive relation ≺ . Every strict partial order h P, ≺i is canonically associated to non-strict partial order h P, (cid:22)i in which p (cid:22) q if and only if p ≺ q or p = q , and every non-strict partial order arisesin this way. A filter on a strict partial order h P, ≺i is a set F ⊆ P that isupward closed and such that if p, q ∈ F then there is an r ∈ F with r (cid:22) p and r (cid:22) q .It follows immediately from these definitions that if h P, ≺i is a strictpartial order, h P, (cid:22)i is the corresponding non-strict partial order, and F ⊆ P ,then F is a filter in h P, ≺i if and only if F is a filter in h P, (cid:22)i , and vice versa .Moreover, F is a maximal (unbounded) filter in either of these partial ordersif and only if it is maximal (unbounded, respectively) in the other partialorder. 3 topology on the set of maximal (unbounded) filters of a strict partial isdefined in the same way as for a non-strict partial order. Once this definitionis made, it is immediate that for any strict poset h P, ≺i and correspondingnon-strict poset h P, (cid:22)i , the identity map P → P induces a homeomorphismof the topological spaces of maximal (unbounded, respectively) filters of theseposets. For this reason, when it is convenient, we may prove results usingstrict partial orders instead of non-strict partial orders. This technique issound because any example of a poset space obtained from a strict partialorder can be converted to a homeomorphic example obtained from a non-strict poset space, and vice versa .We now present two examples showing that many familiar spaces arehomeomorphic to poset spaces. Theorem 2.2.
Every locally compact Hausdorff space is homeomorphic toan MF space.
Proof . Let X be a locally compact Hausdorff space and let P be the set ofall nonempty precompact open subsets of X . For U , V ∈ P we put U (cid:22) V if U = V or the closure of U is contained in V . If F is a filter and U ∈ F then,because U is precompact, \ F = \ { ¯ V | ¯ V ⊆ U, V ∈ F } is the filtered intersection of non-empty compact sets, and hence is non-empty and compact. Since X is Hausdorff, any two points of X have openneighborhoods whose closures are disjoint. If F is a a maximal filter, thenat most one of these neighborhoods can be in F , which implies that T F isa singleton. Finally, the mapping φ : MF( P ) → X given by F T F hasas its inverse the mapping φ − : x
7→ { p ∈ P | x ∈ N p } . To prove that φ is continuous, fix x ∈ MF( P ) and let U be any openneighborhood of φ ( x ) in X . Because X is locally compact, we may assumewithout loss of generality that U is precompact, because the precompact setsform a basis for the topology. Thus we assume U = N p for some p ∈ P .Now, since φ ( x ) ∈ U , we have p ∈ x , so x ∈ N p . Moreover, for any F ∈ N p in MF( P ), we have φ ( F ) = T F ∈ U . This shows φ is continuous.4o prove that φ − is continuous, let y ∈ X be fixed, and let V be anyopen neighbohood of φ − ( y ) in MF( P ). Without loss of generality we mayassume V = N p for some p ∈ P . Now p itself is some precompact open subset U of X , and for any y ′ ∈ U we have p ∈ φ − ( y ′ ). Thus φ − ( U ) ⊆ V . Thisshows φ − is continuous.As there are non-locally-compact complete separable metric spaces and lo-cally compact Hausdorff nonmetrisable spaces, the next theorem is indepen-dent of Theorem 2.2. A construction similar to that in the next theoremwas used by Lawson [7] to represent complete separable metric spaces in thecontext of domain theory (see Section 6). Theorem 2.3.
For every complete metric space X there is a poset P suchthat X ∼ = UF( P ) and UF( P ) = MF( P ). Moreover, if X is infinite then wemay take the cardinality of P to be that of any dense subset of X . Proof . Let X be a complete metric space; we write B ( x, ǫ ) for the openmetric ball of radius ǫ > x ∈ X . Let A be a dense subsetof X . The poset P is the set of all open balls B ( a, r ) where r is a positiverational number and a ∈ A . For p = B ( a, r ) and p ′ = B ( a ′ , r ′ ) in P we let p ≺ p ′ if and only if d ( a, a ′ ) + r < r ′ . An argument similar to the one in theproof of Theorem 2.2 shows that any unbounded filter on P has a uniquepoint in its intersection. The resulting mapping φ : F T F from UF( P )to X has as its inverse the mapping x
7→ { B ( a, r ) | x ∈ B ( a, r ) , a ∈ A, r ∈ Q + } . Each of these mappings can be shown to be continuous using the samemethod as the proof of Theorem 2.2, using the fact that the open balls in-cluded in P form a basis for X . Finally, since X is a complete metric space,every unbounded filter is maximal (see Theorem 3.1 below for details).If Theorem 2.3 is applied to the real line using the line itself as the densesubset, the resulting poset P will be uncountable, but MF( P ) = UF( P ) willbe homeomorphic to the real line.There are also second-countable nonmetrisable Hausdorff MF spaces. Oneexample is the Gandy–Harrington space from modern descriptive set theory(see [11]). 5 Separation and countability properties
In this section, we determine the separation properties that a poset spacemust satisfy. We then show that every second countable poset space is home-omorphic to a poset space obtained from a countable poset. In Section 8, wewill show that a countably based poset space is either countable or containsa perfect closed set.
Theorem 3.1. (1) Every UF space is T .(2) Every MF space is T .(3) If UF( P ) is T then every unbounded filter on P is maximal and thusUF( P ) = MF( P ). Proof . (1) follows from the fact that distinct filters are distinct as subsetsof P . (2) follows from the fact that no maximal filter can properly containanother maximal filter. To prove (3), suppose UF( P ) is T and let F be anunbounded filter on P . Let G be a filter on P such that F ⊆ G . Clearly G is unbounded. If F = G then there must be a p ∈ P such that F ∈ N p and G N p . This means p ∈ ( F \ G ), which is impossible. Thus F = G ; thisshows that F is maximal. Theorem 3.2.
Suppose that P is a poset such that MF( P ) is second count-able. There is a countable subposet R of P such that the map F R ∩ F is a homeomorphism from MF( P ) to MF( R ). Proof . Suppose that MF( P ) is second countable; thus P contains a count-able subset Q such that { N q : q ∈ Q } is a basis for the topology, becauseevery basis of a second-countable topology contains a countable subclasswhich is also a basis. For n = 0 , , , . . . , we construct a set Q n +1 inductivelyto satisfy the following conditions. • Q n +1 is countable. • Q n ⊆ Q n +1 ⊆ P . • For every F ∈ MF( P ) and every finite subset D ⊆ Q n ∩ F there is a q ∈ Q n +1 such that q (cid:22) d for all d ∈ D .6n order to see that Q n +1 can be taken to be countable, suppose D is a finitesubset of Q n with nonempty intersection. Let E D be the set of all p ∈ P such that p (cid:22) d for every d ∈ D . For every filter F ∈ MF( P ) with D ⊆ F there is an element p ∈ E D ∩ F ; thus { N e : e ∈ E D } is an open cover ofthe intersection of all open sets N d with d ∈ D . Since the given space issecond countable, there is a countable subset F D of E D covering the sameset of maximal filters; if some finite subset D of Q n is not contained in anyfilter then let F D be empty. Now take Q n +1 to be the union of all F D where D ⊆ Q n and D is finite; Q n +1 is also at most countable.Let R = S i Q i . Note that R is countable and { N r | r ∈ R } is a basis forMF( P ). For F ⊆ P we write φ ( F ) for F ∩ R . It is straightforward to verifythat φ ( F ) is a filter for every F ∈ MF( P ), by the construction of R . Because R ⊆ P , every F ∈ MF( R ) extends to some F ′ ∈ MF( P ); then φ ( F ′ ) = F .This shows that φ determines a surjective map Φ from MF( P ) to MF( R ).In order to prove that Φ is injective, it suffices to prove the followingstatement. For maximal filters V, W on P we have V ⊆ W if and only if φ ( V ) ⊆ φ ( W ). Suppose p ∈ V \ W . Then W / ∈ N p and thus W / ∈ N q forall q with N q ⊆ N p . On the other hand, R is a basis and N p is the union ofbasic open sets. Since V ∈ N p there is a r ∈ R with N r ⊆ N p and V ∈ N r .It follows that r ∈ φ ( V ) \ φ ( W ). The other direction of the implication istrivial.This shows that φ is a bijection from MF( P ) to MF( R ). To see that φ iscontinuous, let x ∈ MF( P ) be fixed and let U be an open neighborhood of φ ( x ) = x ∩ R in MF( R ). Without loss of generality we may assume that U is of the form N r for some r ∈ R . Let V = { y ∈ MF( P ) | r ∈ y } be the basicopen set determined by r in MF( P ). Now, because r ∈ φ ( x ) = x ∩ R , we seethat r ∈ x , and thus x ∈ V . Moreover, for any x ′ ∈ V , we have r ∈ x ′ , andso r ∈ x ′ ∩ R , which means φ ( x ′ ) ∈ U . Thus φ is continuous.To see that φ − is continuous, let V be any open subset of MF( P ), andlet φ − ( y ) be in V . Because { N r ⊆ MF( P ) | r ∈ R } is a basis for MF( P ),there is some r ∈ R with φ − ( y ) ∈ N r ⊆ N p . Moreover, any y ′ ∈ MF( R )with r ∈ y ′ will satisfy r ∈ φ − ( y ′ ). Thus, for U = { y ∈ MF( R ) | r ∈ y } , wehave y ∈ U and φ − ( U ) ⊆ V . This shows that φ − is continuous. Corollary 3.3.
An MF space is homeomorphic to a countably based MFspace if and only if it is second countable.
Corollary 3.4.
A UF space is homeomorphic to a countably based UF spaceif and only if it is second countable. 7 roof . Let X = UF( P ) be second countable. Construct a poset R and amap φ in a manner analogous to the proof of Theorem 3.2. We show that φ isa homeomorphism from UF( P ) to UF( R ). It is clear that if F ∈ UF( P ) then φ ( F ) ∈ UF( R ). Every G ∈ UF( R ) extends to some G ′ ∈ UF( P ) and then φ ( G ′ ) = G . Thus φ is well defined and surjective as a map from UF( P ) toUF( R ). To see that φ is injective, suppose that F = G are unbounded filterson P . Without loss of generality we may assume there is some p ∈ G \ F .There is thus some r in R ∩ ( G \ F ), because R is a basis. But r ∈ R ∩ ( G \ F )implies r ∈ φ ( G ) \ φ ( F ), which shows φ ( G ) = φ ( F ). Thus φ is a bijectionfrom UF( P ) to UF( R ). The proof that φ is a homeomorphism is the sameas in the proof of Theorem 3.2. In this section, we show that the class of MF spaces is closed under taking G δ subsets and arbitrary topological products. The class of UF spaces is closedunder taking G δ subspaces, but it not closed under even finite products. Theorem 4.1.
The class of MF spaces is closed under arbitrary topologicalproducts.
Proof . Suppose that we are given a collection hh P i , (cid:22) i i | i ∈ I i of posets.We may assume without loss of generality that each poset has a greatestelement, which we denote by p i . We form a poset P consisting of thosefunctions f from I to S i ∈ I P i such that f ( i ) ∈ P i for all i and f ( i ) = p i forall but finitely many i . For f, g ∈ P we put f (cid:22) g if f ( i ) (cid:22) i g ( i ) for all i .We define a map φ from Q i MF( P i ) to MF( P ) by sending Q i F i to theset of all functions f ∈ P such that f ( i ) ∈ F i for all i . The inverse of φ takes x ∈ MF( P ) and returns and returns Q i x i , where x i = { p ∈ P i | for some q ∈ x, q ( i ) = p } . To see that φ is continuous, let x ∈ Q i MF( P i ) be fixed and let U be abasic open neighborhood of φ ( x ), so U is of the form N p for some p ∈ P .Now p is represented by a function f : I → S i P i that returns the maximalelement of P i for all but finitely many i ∈ I . Thus f determines a basicopen set V in the product topology Q i MF( P i ) such that V is equal, in eachcoordinate i ∈ I , to the open set determined by f ( i ). Then x ∈ V . Suppose8 ′ = Q i x ′ i is any point of Q i MF( P i ) that is in V , meaning that f ( i ) ∈ x ′ i for all i ∈ I . Then φ ( x ′ ) will have the property that p i ∈ x ′ i for each i ∈ I ,which means φ ( x ′ ) ∈ N p . Thus φ is continuous.To see that φ − is continuous, let y ∈ MF( P ) be fixed, and let V beany neighborhood of φ − ( y ) in Q i MF( P i ). By the definition of the producttopology, there is a basic open neighborhood of φ − ( y ) which is obtained asa product Q i V i of open sets V i ⊆ MF( P i ) such that V i = MF( P i ) for allbut finitely many i ∈ I . Moreover, in the finitely many coordinates where V i is a proper subset of MF( P i ), we can find a basic open subset N r ( i ) ⊆ V i such that the projection of φ − ( y ) to coordinate i is in N r ( i ) . For all i where V i = MF( P i ) we let r ( i ) be the greatest element of P i . Now let f be theelement of P such that f ( i ) = r ( i ) for all i ∈ I . Then y ∈ N f (in MF( P )),and any y ′ ∈ N f will satisfy φ − ( y ) ∈ V . Thus φ − is continuous. Corollary 4.2.
Every topological product of countably many countablybased MF spaces is homeomorphic to a countably based MF space.
Proof . Under these hypotheses, the poset constructed in Theorem 4.1 iscountable.
Theorem 4.3.
The class of MF spaces is closed under taking G δ subspaces. Proof . Suppose that h U i | i ∈ N i is a sequence of open subsets of MF( P )and U = T i U i is nonempty. We form a poset Q of pairs h n, p i such that n ∈ N and N p ⊆ T i
There are two posets
P, Q such that MF( P ) = UF( P ),MF( Q ) = UF( Q ), but the topological product MF( P ) × MF( Q ) is not home-omorphic to any UF space. Proof . Let ω denote the least infinite countable ordinal and let ω denotethe least uncountable ordinal. We define P to be the set of functions fromfinite initial segments of ω to { , } and define Q to be the set of functionsfrom countable initial segments of ω to { , } . For both posets the relation (cid:22) is given by extension: p (cid:22) q if, for all α in the domain of q , p ( α ) is definedand takes the value q ( α ).We first show that MF( P ) = UF( P ) and MF( Q ) = UF( Q ). Assume that F is an unbounded filter on P (the argument for Q is parallel). Then allfunctions in F are compatible, that is, they do not contradict each other onany value in the intersection of their domains. There is thus a total limitfunction f , because otherwise there would be a first ordinal α where f isundefined and the function extending f which maps α to 0 would definean element of P which would be a lower bound for the filter F . Since f istotal, all functions mapping the ordinals up to some α in the domain of f tothe corresponding value of f are in the filter. One can see that this filter isalready maximal, because any element outside it but still in P is incompatiblewith this function and adding it would destroy the filter property.Assume now, by way of contradiction, that UF( P ) × UF( Q ) is homeomor-phic to a space UF( R ). We denote by π P , π Q the continuous, open projectionmaps from UF( R ) to its factor spaces. There is a filter F in UF( R ) such that π P ( F ) and π Q ( F ) are the filters generated by the set of all functions in P and Q , respectively, which map all inputs to 0. Now one can select an in-finite sequence r , r , . . . in F such that for each n the projection π P ( N r n )10onsists only of functions which map the first n numbers to 0 and r n +1 (cid:22) r n for all n . The sequence h r i i generates a subfilter G ⊆ F . There is no lowerbound r for G , because otherwise π P ( N r ) would be an open set containingsome basic open set N p such that N p ⊆ π P ( N r n ) for all n ; such a p cannotexist by construction.On the other hand, there is a function f contained in all the open sets π Q ( N r n ) and there are basic open neighbourhoods of f generated by q , q , . . . such that N q n ⊆ π Q ( N r n ) for each n . The basic open sets N q , N q , . . . fix f only on countably many ordinals and thus their intersection is also a basicopen set. So π Q ( G ) is bounded while π Q ( F ) is not and thus G ⊂ F . Itfollows that UF( R ) is not a T space. This contradicts the assumption thatUF( R ) is homeomorphic to MF( P ) × MF( Q ).We note that the previous example is not second countable and that thefailure of second countability was important to the proof. Question 4.5.
Is the class of countably based UF spaces closed under takingfinite (or arbitrary) topological products?We end the section by showing that the class of UF spaces is closed undertaking G δ subspaces. As with the class of MF spaces, this result cannotbe extended to include F σ subspaces. We first prove the result for opensubspaces, which has a much simpler proof. Theorem 4.6.
The class of UF spaces is closed under taking open subspaces.
Proof . Let P be a poset and let U be an open subset of UF( P ). Let R bethe set of all r ∈ P such that N r ⊆ U ; we regard R as a subposet of P . Thenany x ∈ U has a neighborhood N r ⊆ U , where r ∈ R . Thus the restrictionmap φ : x x ∩ R sends each element of U to a filter on R . Note that if thisfilter were not unbounded as a subset of R then it has a lower bound in R and consequently would not be unbounded in P .The inverse map of φ sends each maximal filter on R to it upward closurein P . If φ ( G ) were bounded below by p ∈ P , then in particular p (cid:22) r forsome r ∈ R . Thus N p ⊆ N r ⊆ R , which means p ∈ R and G is not in UF( R ).To set that φ and its inverse are continuous, note that (cid:8) N Pr = { F ∈ MF( P ) | r ∈ F } | r ∈ R (cid:9) is a basis for the restriction of MF( P ) to the subspace U , that (cid:8) N Rr = { F ∈ MF( R ) | r ∈ F } | r ∈ R (cid:9)
11s a basis for MF( R ), and that a point x ∈ U ⊆ MF( P ) is in N Pr if and onlyif φ ( x ) is in N Rr . Theorem 4.7.
The class of UF spaces is closed under taking G δ subspaces. Proof . Let G be the space UF( P ) for some poset P with order ≺ P and let G be a G δ subset of G . Thus there is a descending sequence G , G , . . . ofopen subsets of G such that G ⊇ G ⊇ G ⊇ . . . and G = T n G n . Define R = { p ∈ P | p ∈ F for some F ∈ UF( P ) ∩ G } . For each p ∈ R , let g ( p ) = sup { n ∈ N | N p ⊆ G n } , where g ( p ) = ∞ if N p ⊆ G . Define an order relation ≺ R on R by putting p ≺ R q if p ≺ P q and either g ( q ) < g ( p ) ≤ ∞ or g ( q ) = g ( p ) = ∞ . We will show that theunbounded filters on ( R, (cid:22) R ) are precisely the unbounded filters on P thatare in G and do this by showing the following four claims. Claim 1:
Let F ∈ G ⊆ UF( P ); then F is an unbounded filter in R under (cid:22) R . By definition of R , F ⊆ R . To show that F is a filter on R , fix p, q ∈ F .If g ( p ) or g ( q ) is infinite then p and q have a common extension r under (cid:22) P with g ( r ) = ∞ . Thus r is a common extension of p and q under (cid:22) R .Otherwise, because F ∈ G , there is an r ∈ F with r (cid:22) P p , r (cid:22) P q and N r ⊆ G g ( p )+ g ( q )+1 . Then g ( r ) > g ( p ) + g ( q ), r ≺ R p and r ≺ R q . As (cid:22) R is a restriction of (cid:22) P , F is upward closed under (cid:22) R and F is a filter in R .Furthermore, F must be unbounded in R , because a bound in R would alsobe in a bound P . Claim 2:
Let F ⊆ R be a filter in R ; then either sup { g ( p ) | p ∈ F } = ∞ or F is bounded. Suppose the supremum is n < ∞ instead. There can onlybe one r ∈ F with g ( r ) = n , because F is a filter on R . Because r ∈ R , thereis some F ′ ∈ UF( P ) with r ∈ F ′ and F ′ ∈ G . Thus there is an r ′ ∈ F ′ with g ( r ′ ) > g ( r ) and r ′ ≺ P r ; this means r ′ ≺ R r , which shows that F is boundedin R . Claim 3:
Let F be a bounded filter of P which is also a filter in R ; then F is bounded in R . Let r ∈ P be a lower bound for F . If N r G n for some n then sup { g ( p ) | p ∈ F } < n and F is bounded in R by Claim 2. Otherwise N r ⊆ G , in which case r ∈ R and F is again bounded as a subset of R . Claim 4:
Let F be an unbounded filter in R ; then F is also an unboundedfilter in P . To see this, consider the upward closure F ′ of F in P . F ′ isunbounded in P , by Claim 3. Claim 2 shows that F ′ ∈ G ; thus F ′ ⊆ R . Thedefinition of F ′ shows that F ⊆ F ′ . Fix r ∈ F ′ ; then there must be a p ∈ F p (cid:22) P r . If g ( p ) = ∞ then p (cid:22) R r and so r ∈ F . Otherwise there mustbe a q ∈ F with q (cid:22) P p and g ( q ) > g ( r ). Then it follows from transitivityof (cid:22) P and the definition of ≺ R that q ≺ P r , q ≺ R r and r ∈ F . This shows F ′ = F .Claims 1 and 4 show that the unbounded filters on R are exactly thoseunbounded filters on P which are in G . So the identity map φ : UF( P ) ∩ G → UF( R ) is surjective by Claim 4. As this map is trivially injective, it is thusinvertible. To see that φ and φ − are continuous, let x ∈ UF( P ) ∩ G be fixed.Note that for any r ∈ R , we have r ∈ x if and only if r ∈ φ ( x ), because φ is the identity map on filters. Thus φ ( x ) is in the basic open neighorhood ofUF( R ) determined by r if and only if x is in the basic open neighborhood ofUF( P ) ∩ G determined by r . In this section, we establish that every poset space has the a completenessproperty known as the strong Choquet property. We then characterize theclass of countably based MF spaces as precisely the class of second-countable T spaces with the strong Choquet property. We first establish a weakerproperty. Theorem 5.1.
Every poset space has the property of Baire.
Proof . Let X be MF( P ) or UF( P ). Suppose that h U i | i ∈ N i is a sequenceof dense open sets in X and V is a fixed open set. We construct a sequence h p i | i ∈ N i of elements of P . Let p be such that N p ⊆ V ∩ U . Given p i ,there is an unbounded or maximal filter in N p i ∩ U i +1 . Choose p i +1 such that N p i +1 ⊆ U i +1 ∩ N p i and p i +1 (cid:22) p i . In the end, F = h p i i is a linearly orderedsubset of P . Thus F extends to an element of X . Clearly this element is in V ∩ T i U i .We will now show that every poset space has the strong Choquet property,which is defined using a certain game first introduced by Choquet [1]. Let X be an arbitrary topological space. The strong Choquet game is the Gale–Stewart game (see [4] and [6]) defined as follows. The stages of play arenumbered 0 , , , . . . and both players make a move in each stage. In stage i ,player I plays an open set U i and a point x i such that x i ∈ U i and if i > U i ⊆ V i − . Then player II plays an open set V i such that x i ∈ V i and V i ⊆ U i .13t the end of the game, player I wins if T i U i is empty (or, equivalently, if T i V i is empty). Player II wins if T i U i is nonempty. A position in the gameis a finite (possibly empty) sequence hh U , x i , V , h U , x i , . . . i which is an initial segment of an infinite play of the game following the rulesjust described.A space X has the strong Choquet property if player II has a winningstrategy for the strong Choquet game on X . A winning strategy is a functionthat takes a position after player I has played and tells player II which openset to play, such that if player II follows the winning strategy then player IIwill always win the game regardless of what moves player I makes.The strong Choquet property is strictly stronger than the property ofBaire. Moreover, the class of topological spaces with the strong Choquetproperty is closed under G δ subspaces and arbitrary topological products. Itis known that the class of topological spaces with the property of Baire isnot closed under binary products (an example is provided in [3]). Theorem 5.2.
Every poset space has the strong Choquet property.
Proof . We describe the strategy for player II informally. At the start of thegame, player I plays an open set U and a point x . Player II translates thepoint x into a filter on P , then finds a basic neighbourhood q of x suchthat N q ⊆ U . Player II then plays N q . Now given h x , U i with x ∈ N q ,Player II translates x to a filter on P and then finds a neighbourhood q of x such that q (cid:22) P q and N q ⊆ U . Player II plays N q . Player II continuesthis strategy, always choosing q i +1 (cid:22) P q i . At the end of the game, player IIhas determined { q i | i ∈ N } , a descending sequence of elements of P . Thissequence extends to an element of X which is in T N q i . Player II has thuswon the game.We use the strong Choquet property to obtain the following characterizationof countably based MF spaces. Theorem 5.3.
A topological space is homeomorphic to a countably basedMF space if and only if it is second countable, T and has the strong Choquetproperty. 14e postpone the proof of this theorem temporarily to comment on the hy-potheses involved in the characterization. Clearly, any space X homeomor-phic to a countably based MF space must be T and second countable. Wehave already shown X must also have the strong Choquet property. Thusthe new content of Theorem 5.3 is that the strong Choquet property is suf-ficient for a T second-countable space to be homeomorphic to a countablybased MF space. In the non-second-countable setting, the strong Choquetproperty is not sufficient for a T space to be homeomorphic to an MF space. Example 5.4.
There is a Hausdorff strong Choquet space which is not home-omorphic to any MF space.
Proof . The space X consists of certain functions from ω to { , } . We puta function f in X if and only if there is an ordinal α < ω such that f ( β ) = 0for all β > α . For each f ∈ X and each α < ω , the set { g ∈ X | f ( β ) = g ( β ) for all β < α } is declared to be an open set. The topology on X is the one generated bythese open sets. It is clear that X is a Hausdorff space.It is easy to show that X has the strong Choquet property, as follows.All that player II has to do is to play any basic open subset of the open setplayed by player I which also contains the point given by player I . In theend, the open sets played by player I in the countable number of rounds ofthe game and each round will fix countably many coordinates of a functionin X . In the limit, countably many coordinates are fixed and we can finda point in the intersection of the sets played by I by forcing the remainingcoordinates to map to 0.We now show that X is not homeomorphic to any MF space. Suppose, byway of contradiction, that X ∼ = MF( P ). We construct a transfinite sequence h p α | α < ω i inductively. Let p be any basic open neighbourhood of theconstant 0 function. Given h p α | α < β i , there is a first coordinate γ < ω which is not fixed by any p α ; let f be the function which is 0 except at γ ,and f ( γ ) = 1. Note that any intersection of countably many open sets in X is open. Thus we may choose p β ∈ P such that p β (cid:22) p α for all α < β and f ∈ N p β . Choose any such p β . At the end of this construction, h p α | α < ω i is linearly ordered and thus extends to a maximal filter F . Now the elementof X corresponding to F sends uncountably many ordinals to 1, which isimpossible. 15e now return to the proof of Theorem 5.3, which will occupy the remainderof this section. Let X be a fixed T space with a fixed countable basis anda fixed winning strategy for player II in the strong Choquet game. Our firststep is to define a poset P . The elements of P are called conditions . Acondition is a finite list of the form h A, π , π , . . . , π k i satisfying the following requirements.(1) The set A is a nonempty basic open set from the fixed countable basis.For each condition c we let S ( c ) denote the basic open set A appearingin c .(2) Each π i is a finite (that is, partial) play of the strong Choquet gameon X following the fixed winning strategy s II for player II. We requireeach π i to be of the form h V , x , s II ( V , x ) , V , x , s II ( V , x , V , x ) , . . . ,V r , x r , s II ( V , x , V , x , . . . , V r , x r ) i . Thus each π i ends with an open set, which we will denote by U ( π i ). Itis allowable that π is the empty sequence hi , in which case U ( π ) = X .(3) If a play π is in a condition then so is every initial segment of π thatends with a move by player II.(4) A ⊆ U ( π i ) for each i ≤ k .We define the order ≺ on P as follows. Let c = h A, π , π , . . . , π k i and c ′ = h A ′ , π ′ , π ′ , . . . , π ′ l i be any two conditions. We let c ′ ≺ c if and only if(5) For each finite play π i in c there is a point x n ∈ S ( c ) such that thelonger play π i a h A, x n , s II ( π i a h A, x n i ) i is in c ′ , that is, equals π ′ j for some j ≤ l .(6) A ′ ⊆ A (this is actually a consequence of requirement (5)).Requirement (3) in the definition of a condition allows us to prove that theorder on P is transitive. Because each condition is finite, requirement (5) inthe definition of the order relation ensures c c for all c ∈ P . Thus ≺ is apartial order on P . 16 emma 5.5. For any filter F on P the intersection T c ∈ F S ( c ) is nonempty. Proof . Let h A i | i ∈ N i be an enumeration of all of the basic open setswhich appear as S ( c ) for some c ∈ F ; here we are using the fact that X issecond countable and that each S ( c ) is drawn from a fixed countable basis of X . It is immediate that T c ∈ F S ( c ) equals T i ∈ N A i . We will show the latterintersection is nonempty.We inductively construct a descending sequence of conditions h c i | i ∈ N i and a sequence of finite plays h π i | i ∈ N i so that π i +1 is an immediateextension of π i for each i ∈ N . At stage 0 let c be any condition in F suchthat S ( c ) = A and let π be any finite play in c .At stage i + 1 let c be any condition in F such that S ( c ) = A i . Let c i +1 be a common extension of c and c i in F . It is clear that S ( c i +1 ) ⊆ S ( c ) = A i .Choose π i +1 to be any play in c i +1 which is an immediate extension of π i .Now assume the entire sequence h π i i has been constructed. These partialplays determine an infinite play γ of the strong Choquet game followingthe strategy for player II . Thus the intersection of the open sets played byplayer I in γ is nonempty. By construction, each set A i has a subset playedby player I at some stage of γ . Thus T i A i is nonempty. Lemma 5.6.
Let c and c be two conditions and let x ∈ S ( c ) ∩ S ( c ).There is a condition c such that c ≺ c , c ≺ c and x ∈ S ( c ). Proof . Begin by letting c be empty. For each π in c we put the longer play π a h S ( c ) , x, s II ( π a h S ( c ) , x i ) i into c . For each π in c we put π a h S ( c ) , x, s II ( π a h S ( c ) , x i ) i into c . For each π that has been added to c we add all initial segments of π ending with a move by player II . We then let S ( c ) be a basic open neigh-bourhood of x which is a subset of the open set T π ∈ c U ( π ). This constructionensures that c is a condition satisfying the conclusions of the lemma. Lemma 5.7.
Let F be a maximal filter on P . The intersection T c ∈ F S ( c )contains a single point. 17 roof . By Lemma 5.5 we know that T c ∈ F S ( c ) is nonempty. Suppose that x, y are distinct points in T c ∈ F S ( c ). Let A be a basic open neighbourhoodof x such that y A . We construct a filter G inductively. At stage n weconstruct G n ⊆ P and in the end we let G be the upward closure of S n G n .To begin, let G = F ∪ {h A, hii} . At stage i + 1, we know by induction that x ∈ S ( c ) for every c ∈ G i . Thus we can apply Lemma 5.6 repeatedly so that G i ⊆ G i +1 , every pair of conditions in G i has a common extension in G i +1 and x ∈ S ( c ) for every c ∈ G i +1 .It is immediate from the construction that G = S i G i is a filter whichproperly extends F . This shows that F was not maximal. Proof of Theorem 5.3.
For each F ∈ MF( P ) we denote the single pointin T c ∈ F S ( c ) by φ ( F ). We show that φ is a homeomorphism from MF( P )to X .We first show that φ is an injective map. Suppose that F and F ′ aremaximal filters on P such that x ∈ T c ∈ F S ( c ) and x ∈ T c ∈ F ′ S ( c ). Byfollowing a procedure similar to the proof of Lemma 5.7 we may find a filter G such that F ⊆ G and F ′ ⊆ G . Thus, by maximality, we have F = F ′ = G .We next show that φ is a surjective map. Let x ∈ X be fixed. Let h A i | i ∈ N i be a sequence of basic open sets such that T i A i = { x } . Theexistence of this sequence requires that X be T and first countable. Foreach i ∈ N let c i = h A i , hii . Following a method similar to the proof ofLemma 5.7, we can construct a filter F such that c i ∈ F for each i ∈ N . Let G be an extension of F to a maximal filter. Now S = T c ∈ G S ( c ) is nonemptyby Lemma 5.5 and S ⊆ T i A i = { x } by construction, so φ ( G ) = x .It remains to show that φ is open and continuous. This follows fromLemma 5.6; for each x ∈ X and each condition c , we have c ∈ φ − ( x ) ifand only if x ∈ S ( c ). This shows that X is homeomorphic to MF( P ). ByTheorem 3.2, we may find a countable subposet R of P such that X ishomeomorphic to MF( R ). This completes the proof. In this section, we apply the characterization of countably based MF spacesto characterize those second-countable spaces with a domain representation.Our result gives a complete solution to the so-called model problem forsecond-countable spaces in domain theory.18 domain is a certain type of poset (defined below) and every domainis a topological space with a topology known as the Scott topology. A do-main representation of a topological space X is a domain D such that X ishomeomorphic to the topological space consisting of the maximal elementsof D with the relative Scott topology. The history of such representationsis thoroughly described by Martin [8]. It is known that every complete sep-arable metric space has a domain representation (see Lawson [7]) and thatevery space with a domain representation is T and has the strong Choquetproperty (Martin [8]). We now show that the strong Choquet property issufficient for a T second-countable space to have a domain representation.We summarize the definitions from domain theory that we require; thesedefinitions are explored fully by Gierz et al. [5]. A nonempty subset I of aposet h P, (cid:22)i is directed if every pair of elements in I has an upper bound in I .A poset P is said to be a dcpo (for “directed-complete partial ordering”) ifevery directed subset of P has a least upper bound. Any dcpo D has a secondorder relation ≪ , known as the way below relation, under which q ≪ p ifand only if whenever I ⊆ D is a directed set with p ≪ sup I there is some r ∈ I with q (cid:22) r . For each p ∈ D we put ⇓ p = { q ∈ D | q ≪ p } and ⇑ q = { p ∈ D | q ≪ p } . A dcpo D is continuous if ⇓ p is directed and theequality p = sup ⇓ p holds for every p ∈ D . A domain is a continuous dcpo.A subset B of a domain D is a basis if B ∩ ⇓ p is directed and p = sup( B ∩ ⇓ p )for every p ∈ D . A domain is ω -continuous if it has a countable basis. Anelement p of a dcpo is compact if p ≪ p . A dcpo D is ω -algebraic if there is acountable basis for D consisting of compact elements. The Scott topology ona dcpo D is generated by the basis {⇑ p | p ∈ D } . A domain representation of a space X is a homeomorphism between X and the maximal elements ofa domain with the Scott topology. Theorem 6.1.
A topological space has a domain representation via an ω -algebraic dcpo if and only if the space is second-countable, T and hasthe strong Choquet property. Proof . It can be seen that any space with a domain representation satisfiesthe T separation property and a result of Martin [8] shows that any spacewith a domain representation has the strong Choquet property. Therefore,we only need to prove that a second-countable T strong Choquet space has adomain representation via an ω -algebraic dcpo We use the following lemma,which follows easily from the definitions.19 emma 6.2. Suppose that P is a countable poset. The set of all filters on P ,ordered by inclusion, is an ω -algebraic dcpo D . The maximal filters on P are precisely the maximal elements of D and the compact elements of D areprecisely the principal filters on P . Moreover, the poset topology on MF( P )corresponds exactly to the Scott topology on the maximal elements of D .We showed in Theorem 5.3 that any second-countable T strong Choquetspace is homeomorphic to MF( P ) for a countable poset P . It follows imme-diately from the lemma above that such a space also has a domain represen-tation via an ω -algebraic dcpo.The next corollary follows from the fact that any space with a domain repre-sentation is T and has the strong Choquet property. Although this corollaryis already known, the proof here is new. Corollary 6.3.
If a second-countable space has a domain representationthen it has a representation via an ω -algebraic dcpo.We end this section with several remarks on the relationship between domainrepresentable spaces and MF spaces.A proof of Lemma 6.2 can be modified to show that the collection of allideals on a poset (sometimes called the ideal completion of the poset) formsa domain whose maximal elements in the Scott topology correspond to themaximal ideals of the poset under the Stone topology. All results we haveproved for MF spaces also hold for these spaces of maximal ideals, by duality.The relationship between ideal completions and domain representations hasbeen investigated by Martin [9].A Scott domain is a domain in which every pair of elements with an upperbound has a least upper bound. Lawson [7] has shown that any space witha domain representation via a countably based Scott domain is a completeseparable metric space. It can be seen that posets constructed in Theorem 5.3do not, in general, give Scott domains, even when the posets are constructedfrom formal balls in complete separable metric spaces.The proof of Example 5.4 can be modifed to obtain the following.
Example 6.4.
There is a Hausdorff strong Choquet space that does nothave a domain representation. 20
Semi-Topogenous Orders
In this section, we prove results which give a partial solution to the questionof which arbitrary (not necessarily second countable) topological spaces arehomeomorphic to MF spaces.Suppose that a topological space X is homeomorphic to MF( P ), for someposet P , via a fixed homeomorphism φ . If each element of p ∈ P is replacedby the corresponding open subset φ ( N p ) ⊆ X , the poset order on P willdetermine a corresponding order relation on these subsets of X . Moreover,the collection of all these open subsets forms a basis for the topology on X .It is thus natural to ask whether the existence of a basis with a suitable orderrelation is sufficient for a topological space to be homeomorphic to an MFspace.Cs´asz´ar [2] considered many different types of orders and their connec-tions to topology. The basic concept is that of a semi-topogenous order. Definition 7.1. A semi-topogenous order is a binary relation ⊏ on thepowerset of a topological space X satisfying the following axioms for all u, v, w ⊆ X [2, Chapter 2]: • ∅ ⊏ ∅ and X ⊏ X ; • v ⊏ w ⇒ v ⊆ w ; • u ⊆ v ⊏ w ⇒ u ⊏ w ; • u ⊏ v ⊆ w ⇒ u ⊏ w .Cs´asz´ar considered orders which are only linked to topology, such as the orderwhich says that w is a neighbourhood of v . It might happen that some butnot all open supersets w of a given set v satisfy v ⊏ w . Nevertheless, althoughthis is not made explicit by Cs´asz´ar, it is quite convenient to postulate alsoa connection between the topology and the open spaces.Recall that the open kernel of a set is the union of all its open subsets.We say that the topological space X is generated by the order ⊏ if for each u ⊆ X the set S { o ⊆ X | o ⊏ u } is the open kernel of u . In this case a set w is open if and only if it is the union of all v such that v ⊏ w . It followsthat if v ⊏ w then there is an open o with v ⊆ o ⊆ w ; the converse of thislast implication does not always hold. Every topological space is generatedby some semi-topogenous order, for one can define v ⊏ w to hold if and onlyif there is an open set o with v ⊆ o ⊆ w .21 emark 7.2. There is a close relationship between semi-topogenous ordersand the way below relation ≪ on a continuous dcpo, which was discussedin Section 6. The following true properties of the way below relation areobtained by dualizing the second, third and fourth properties in the definitionof a semi-topogenous order: v ≪ w ⇒ v ≤ wu ≤ v ≪ w ⇒ u ≪ wu ≪ v ≤ w ⇒ u ≪ w The fact that these are dual forms follows from the fact that points in atopological space are minimal as nonempty subsets under ⊆ but are maximalelements of a domain representing the topological space; for this reason, wewrite ≤ for ⊇ and ≪ for ⊐ . The requirement that S { o | o ⊏ u } is the openkernel of u corresponds exactly to the fact that { x | y ≪ x } is the openkernel of an element y of a continuous dcpo with the Scott topology.Thus if a space X has a representation via a continuous dcpo D thenthe dual of the way below relation on D is a semi-topogenous order (exceptthat it is defined only on a subset of the powerset of X ) which generates thetopology on X . Semi-topogenous orders can be viewed as a generalization ofthe way below relation which is applicable to the case when the dcpo is thefull powerset of a topological space. It appears that semi-topogenous ordersare related to auxilliary relations as defined by Gierz et al. [5], although aformal relationship seems difficult to state.A filter in a topological space X is a collection of nonempty subsets that isclosed under finite intersection and under superset. A filter has an open basis if for every w there is an open v in the filter with v ⊆ w . As in general thereneed not be a point contained in the intersections of the sets in a filter, weare interested in a condition on filters that requires their sets to contain acommon point. Our condition that a filter meets a semi-topogenous order willimply that this filter has also an open basis, while a compelteness conditionwill ensure that each filter that meets the order has a nonempty intersection. Definition 7.3.
Let X a space with a semi-topogenous order ⊏ generatingits topology. A filter U on X meets ⊏ if for every w ∈ U there is a v ∈ U with v ⊏ w . X is complete for ⊏ if for every filter U in the space X whichmeets ⊏ there is a point x with x ∈ u for all u ∈ U .22 heorem 7.4. Let X be a T space with a semi-topogenous order ⊏ gener-ating its topology such that X is complete for ⊏ . Then X is homeomorphicto an MF space. Proof . Let P consist of the nonempty open subsets of X and let p ≺ q ifand only if p = q and p ⊏ q . The relation ≺ is obviously transitive andantireflexive, and so it makes P into a poset.For each x ∈ X let U x be the set of all p ∈ P with { x } ⊏ p . If p, q ∈ U x then the open kernel u of p ∩ q contains x and thus there is an open r ⊏ u with x ∈ r . As the open kernel of r again contains x , { x } ⊏ r . So r ∈ U x , r ⊏ p and r ⊏ q . Thus U x is a filter on P .If V is a maximal filter on MF( P ) then V also meets ⊏ . If v generates V then v is open (by definition) and not empty. For every x ∈ v there is a w ⊏ v with x ∈ w ; by maximality w = v . Thus v ⊏ v and every w ⊆ X with v ⊆ w satisfies v ⊏ w and w ∈ V . If there is no single element generating V then there is, for every v, w ∈ V , some u ∈ V with u ≺ v and u ≺ w . Thenit follows that u ⊏ v and u ⊏ w . Furthermore, there is an t ≺ u with t ∈ V ;then it follows that t ⊏ v ∩ w . So V contains all supersets of v ∩ w and thus V is a filter. Furthermore, V meets ⊏ .This means, by assumption, that there is a point x contained in all setsof V . Thus V ⊆ U x and by the maximality of V , V = U x . Therefore, everyfilter U y is contained in a filter U x which is maximal. Due to the T property, y = x ; otherwise U y would contain a p with x / ∈ p in contradiction to thefact that U y ⊆ U x .This shows that the mapping φ : x U x is a bijection from X to themaximal filters on P . To see the φ is open and continuous, first note that if y ∈ X and U is an open set, then y ∈ U if and only if { y } ⊏ U . To see this, fix y ∈ X and any open U such that y ∈ U , which means then { y } ⊆ U . Then,because ⊏ generates the topology and y is trivially in the open kernel of U ,there is some W ⊑ U with y ∈ W . This means { y } ⊆ W ⊏ U , which means { y } ⊏ U by the definition of semi-topogenous orders. The converse directionof the equivalence follows directly from the definition of a semi-topogenousorder.Now, to see that φ is open and continuous, note that for any point x ∈ X and any open set U , we have x ∈ U ⇔ { x } ⊏ U ⇔ U ∈ φ ( x ) ⇔ φ ( x ) ∈ N p , where N p is the basic open subset of MF( P ) corresponding to U .23e do not know whether every MF space has a semi-topogenous order sat-isfying the hypotheses of the previous theorem. We have established thefollowing partial result. Theorem 7.5. If X = MF( P ) and P satisfies ∀ p, q, r [ p ≺ q ∧ N q ⊆ N r ⇒ p ≺ r ] (1)then there is a semi-topogenous order ⊏ generating the topology of X suchthat X is complete for ⊏ . Proof . For any v, w ⊂ X , let v ⊏ w if either v = ∅ , w = X , there is is anopen atom u with v ⊆ u ⊆ w or there are p, q ∈ P with v ⊆ N p , p ≺ q and N q ⊆ w . Note that N p ⊆ N q in the last case.It follows directly from definitions and the present assumptions that ⊏ isa semi-topogenous order. We must show that that ⊏ generates the topologyof X . Let w be an open set and x be a point in w . There is an open set N q with { x } ⊆ N q ⊆ w . In the case that { x } = N q , N q ⊏ w . In the case that { x } 6 = N q there is a further p ≺ q with x ∈ N p : The reason is that givenan y ∈ N q \ { x } , the maximal filter U x belonging to x must contain a p ≺ q which does not contain y by the T axiom. Then { x } ⊆ N p ⊏ w . So w is theunion of all v with v ⊏ w .Now let W be a filter in the topological space X which meets ⊏ . If W contains an r such that N r is atomic, that is, a singleton { x } , then every u ∈ W contains x since otherwise N r ∩ u = ∅ in contradiction to W being afilter.If W does not contain an r such that N r is atomic, then let V be the setof all p ∈ P such that N p ∈ W . Given any p, q ∈ V there is an u such that u ⊏ N p ∩ N q . So there are r, t with u ⊆ N r , N t ⊆ N p ∩ N q and r ≺ t . Itfollows that r ≺ p and r ≺ q . Thus V is the basis of a filter on P ; this filteris contained in a maximal filter on P which is of the form U x for some point x ∈ X . This x is then in N p for all p ∈ V . Let u ∈ W . As W meets ⊏ , thereis a p ∈ V with N p ⊆ u . It follows that x ∈ N p and x ∈ u . So x is a commonpoint of the sets in W .The posets constructed in Theorems 2.2 and 2.3 satisfy condition (1) and thusthey are examples of a poset space that is complete for a semi-topogenousorder generating its topology. 24 xample 7.6. For every complete metric space and every locally compactHausdorff space there exists a semi-topogenous order ⊏ which generates thetopology of X and for which X is complete. Remark 7.7.
Assume that X is a space which is complete for a semi-topogenous order generating its topology. Then one can not only show that X is homeomorphic to an MF space but also that the winning strategy forplayer II is quite easy to obtain. Given any open set u and any point x ∈ u by player I , player II only has to choose an open v with { x } ⊆ v ⊏ u . Itdoes not matter which v with this condition is chosen and the history of thegame can be ignored. The result of the construction will be, at the end of thegame, a basis for a filter which meets ⊏ and thus this filter has a commonpoint.This shows that the “neighbourhood spaces” that we consider here satisfya restricted version of the strong Choquet property. The intuition behind thisrestriction is that one wishes to study non-second-countable spaces by con-sidering “transfinite games.” The role of player I is replaced by consideringfilters instead of descending sequences, and the winning strategy of player II is reduced to a neighbourhood relation ⊏ which could be interpreted as sayingthat if { x } ⊆ v ⊏ u then v is a good move for player II .Indeed, the notion of completeness of spaces with respect to a semi-topogenous order ⊏ is based on this idea: Let the strategy of player II be just to follow ⊏ and let player I build a filter U such that for every w ∈ U there is a v ∈ U which player II might have chosen as a response to w , thatis, a v ⊏ w ; then the intersection of all u ∈ U is not empty. In this section, we establish perfect set theorems for countably based Haus-dorff poset spaces. These theorems show that these spaces are either count-able or have the cardinality 2 ℵ of the continuum. Theorem 8.1.
Any countably based Hausdorff poset space has either count-ably many points or has cardinality 2 ℵ . Corollary 8.2.
Any countably based Hausdorff poset space has either count-ably many points or contains a perfect closed set.25 roof . Any second-countable Hausdorff space of cardinality 2 ℵ contains aperfect closed set. The complement of the perfect closed set is the union ofall the basic open sets from a fixed countable basis which contain fewer than2 ℵ points.To prove Theorem 8.1, we introduce a class of Gale–Stewart games. Thesegames are inspired by the ∗ -games in descriptive set theory (as describedin [6]). For each poset P we define a game which we call the poset stargame on P . There are two players. The play proceeds in stages numbered0 , , , . . . . At stage t , player I plays a pair h p t , p t i ∈ P × P . Player II playsa number n t ∈ { , } . Player I wins the game if the following conditions holdfor all t .(1) p t ⊥ p t .(2) p t +11 (cid:22) p tn t and p t +12 (cid:22) p tn t .Player II wins if player I does not win; there are no ties.A strategy for a player is a function that tells the player what to do at anypossible move of the game. The strategy is a winning strategy if the playerwill win any play of the game in which the player uses the strategy to chooseevery move. It is impossible for both players to have a winning strategy forthe same game. Lemma 8.3.
Let P be a poset. Either player I or player II has a winningstrategy for the poset star game on P . Proof . The set of infinite plays of the poset star game on P that are winningfor player I is closed in the space of all possible plays of the game. (Thisspace is the space of infinite sequences of moves; the set of moves is given thediscrete topology and the space of infinite plays carries the product topol-ogy). The proof follows from a theorem of Gale and Stewart known as closeddeterminacy. Lemma 8.4.
Suppose that X is a Hausdorff poset space based on a countableposet P and player I has a winning strategy for the poset star game on P .Then X has cardinality 2 ℵ . Proof . It suffices to prove the result for MF( P ), which is a subset of UF( P ).Let s I be a winning strategy for player I and let f ∈ { , } N . Consider theplay in which player I follows s I while player II uses f as a guide; that26s, player II plays f ( n ) at stage 2 n . Because s I is a winning strategy forplayer I , this play determines a descending sequence F ( f ) of elements of P .This sequence extends to a maximal filter. For distinct f, g ∈ { , } N thesequences F ( f ) and F ( g ) contain incompatible elements and thus cannotextend to the same filter. Therefore the space MF( P ) has cardinality 2 ℵ . Lemma 8.5.
Let X be a countably based Hausdorff poset space based onthe poset P . If player II has a winning strategy for the poset star game on P then X is countable. Proof . Let s II be a winning strategy for player II . We say that a finite play σ of length 2 k is compatible with s II if s II ( σ [2 i + 1]) = σ (2 i + 2) whenever2 i + 2 ≤ k . We say that a play σ of even length is a good play for a point x if σ is compatible with s II and x is in the open set chosen by player II in thelast move of σ . A good play for x is a maximal play if it cannot be extendedto a longer good play for x ; this means that no matter what pair of disjointopen sets player I plays, s II will direct player II to choose an open set notcontaining x .If player II has a winning strategy then every point x has a maximal play.Note that the empty play is trivially a good play for x . If every good playfor x could be extended to a larger good play for x , then it would be possiblefor player I to win the game by always leaving the game in a position thatis good for x . This play of the game would follow s II , a winning strategy forplayer II , which is a contradiction.If σ is a good play for two points x and y then σ is not a maximal playfor both x and y . For player I could play h U , U i in response to σ , where x ∈ U , y ∈ U and U ∩ U = ∅ . Here we are using the assumption that thetopology of X is Hausdorff.We have now shown that every point in the X has a maximal play andthat no play is maximal for two points. Since the set of maximal plays iscountable, this implies that the set of points in X is countable.We remark that the statement “Every closed subset of a countably basedHausdorff MF space is either countable or has a perfect closed subset” isindependent of ZFC set theory; this result is established in [11].27 cknowledgments We would like to thank the Institute for Mathematical Sciences at the Na-tional University of Singapore for organizing the wonderful ComputationalProspects of Infinity workshop in 2005 which made this work possible. Wealso thank Steffen Lempp and Sasha Rubin for thoughtful comments. Wewould like to thank Jimmie Lawson and Ralph Koppermann for their helpfulcomments on the domain theory results in Section 6.Some of the results presented here appeared in the first author’s PhDthesis [10], supervised by Stephen Simpson at the Pennsylvania State Uni-versity.
References [1] Gustave Choquet.
Lectures on Analysis . W. A. Benjamin, New York,1969.[2] ´Akos Cs´asz´ar.
Foundations of General Topology . Pergamon Press, Ox-ford, 1963.[3] William G. Fleissner and Kenneth Kunen. Barely Baire spaces.
Funda-menta Mathematicae , 101(3):229–240, 1978.[4] David Gale and Frank M. Stewart. Infinite games with perfect infor-mation. In
Contributions to the theory of games, vol. 2 , Annals ofMathematics Studies, no. 28, pages 245–266. Princeton University Press,Princeton, N. J., 1953.[5] Gerhard Gierz, Karl H. Hofmann, Klaus Keimel, Jimmie Lawson,Michael Mislove and Dana Scott.
Continuous lattices and domains , vol-ume 93 of
Encyclopedia of Mathematics and its Applications . CambridgeUniversity Press, 2003.[6] Alexander S. Kechris.
Classical descriptive set theory , volume 156 of
Graduate Texts in Mathematics . Springer-Verlag, 1995.[7] Jimmie Lawson. Spaces of maximal points.
Mathematical Structures inComputer Science , 7(5):543–555, 1997.288] Keye Martin. Topological games in domain theory.
Topology and itsApplications , 129(2):177–186, 2003.[9] Keye Martin. Ideal models of spaces.
Theoretical Computer Science ,305(1–3):277–297, 2003.[10] Carl Mummert.
On the Reverse Mathematics of General Topology . PhDthesis, The Pennsylvania State University, 2005.[11] Carl Mummert. Reverse mathematics of MF spaces.
Journal of Mathe-matical Logic , 305(2):203–232, 2007.[12] Carl Mummert and Stephen G. Simpson. Reverse mathematics and Π comprehension. Bulletin of Symbolic Logic , 11(4):526–533, 2005.[13] Willi Rinow.
Lehrbuch der Topologie . VEB Deutscher Verlag der Wis-senschaften, Berlin (Ost), 1975.
Carl Mummert
Department of MathematicsMarshall UniversityOne John Marshall DriveHuntington, WV 25755USA