Representable posets and their order components
aa r X i v : . [ m a t h . L O ] M a y REPRESENTABLE POSETS AND THEIR ORDERCOMPONENTS
M. E. ADAMS AND D. VAN DER ZYPEN
R´esum´e.
Un ensemble partiellement ordonn´e P est repr´esentable si ilexiste un (0 , P . Dans cet article, nous voulons d´emontrerque si tous les composants d’ordre de P sont repr´esentables, P estrepr´esentable aussi. En plus, nous montrons que, bien que la topologied’ intervalle de chaque composant soit compacte, il existe un ensemblepartiellement ordonn´e qui est repr´esentable et qui poss`ede un composantd’ordre non-repr´esentable. Introduction
A poset is said to be representable if it is isomorphic to the poset of primeideals of a bounded distributive lattice (that is a distributive lattice with alargest element 1 and a smallest element 0). The question of which posetsare representable essentially dates back to Balbes [1] (see also, Balbes andDwinger [2]) and has been considered by a number of authors since (see, forexample, the expository article Priestley [6].)In [5], Priestley proved that the category D of bounded distributive lat-tices with (0 , P ofcompact totally order-disconnected spaces (henceforth referred to as Priest-ley spaces ) with order-preserving continuous maps are dually equivalent.(A compact totally order-disconnected space ( X ; τ, ≤ ) is a poset ( X ; ≤ ) en-dowed with a compact topology τ such that, for x , y ∈ X , whenever x y ,then there exists a clopen decreasing set U such that x ∈ U and y U .)The functor D : D → P assigns to each object L of D a Priestley space( D ( L ); τ ( L ) , ⊆ ), where D ( L ) is the set of all prime ideals of L and τ ( L )is a suitably defined topology (the details of which will not be requiredhere). The functor E : P → D assigns to each Priestley space X the lattice( E ( X ); ∪ , ∩ , ∅ , X ), where E ( X ) is the set of all clopen decreasing sets of X .In particular, a poset ( X ; ≤ ) is seen to be representable iff there exists atopology τ such that ( X ; τ, ≤ ) is a Priestley space.Let ( X ; ≤ ) be a poset. Then we define a relation R on X by setting( x, y ) ∈ R whenever x ≤ y or y ≤ x . Let R ′ be the transitive clo-sure of R . Then R ′ is an equivalence relation. An order component of X is an equivalence class [ x ] R ′ of the relation R ′ for some x ∈ X . Fur-ther, for any Y ⊆ X , let ( Y ] = { x ∈ X | x ≤ y for some y ∈ Y } and[ Y ) = { x ∈ X | x ≥ y for some y ∈ Y } . Should Y = { y } for some y ∈ X , then, for simplicity, we will denote ( Y ] and [ Y ) by ( y ] and [ y ),respectively. Finally, let [ x, y ] = [ x ) ∩ ( y ], S − = { X \ ( x ] | x ∈ X } , and S + = { X \ [ x ) | x ∈ X } . Then S = S − ∪ S + is an open subbase for the socalled interval topology τ i on X (sometimes, in the interest of clarity, τ i willbe denoted τ i ( X ) when we wish to emphasize the poset concerned). It iswell known that if ( X ; τ, ≤ ) is a Priestley space, then τ contains the intervaltopology τ i .Our principal result is the following: THEOREM 1.1.
If the order components of a poset ( X ; ≤ ) are repre-sentable, then so is X . However, even though each order component of arepresentable poset is compact under its interval topology, there exists a rep-resentable poset with an order component which is not representable. The proof of 1.1 will be given in §
2, where we begin in 2.1 by showingthat a poset is compact under its interval topology iff each order componentis compact under its respective interval topology. As observed in 2.2, it fol-lows readily from this that each order component of a representable posetis compact with respect to its interval topology. We then establish in 2.3that if every order component of a poset is representable, then so too is theposet. Finally, we define a countably infinite poset which we show to beorder-isomorphic to an order component of a representable poset in 2.5, butwhich, as we show in 2.6, is not itself representable.For any undefined terms or additional background, we refer the readerto the texts Gr¨atzer [3] and Kelley [4], with each of which our notation isconsistent. 2.
Proof of . Let ( X k ; ≤ k ) k ∈ K be a family of pairwise disjoint nonemptyposets. Then for ( X ; ≤ ) where X = S k ∈ K X k and ≤ = S k ∈ K ≤ k , the fol-lowing are equivalent :( i ) for each k ∈ K , the space ( X k ; τ i ( X k )) is compact ;( ii ) ( X ; τ i ( X )) is compact. EPRESENTABLE POSETS AND THEIR ORDER COMPONENTS 3
Proof.
Assume that (i) holds and let U be an open cover of X = S k ∈ K X k .By Alexander’s subbase lemma, we may assume that U = { X \ ( a ] | a ∈ A } ∪ { X \ [ b ) | b ∈ B } for some subsets A, B ⊆ X . We distinguish two cases:First, there is some k ∈ K such that A ∪ B ⊆ X k . In which case, consider U X k = { X k \ ( a ] | a ∈ A } ∪ { X k \ [ b ) | b ∈ B } . Since ( X k ; τ i ( X k )) is compactby assumption, U X k has a finite subcover { X k \ ( a ] , ..., X k \ ( a r ] } ∪ { X k \ [ b ) , ..., X k \ [ b s ) } , so { X \ ( a ] , ..., X \ ( a r ] } ∪ { X \ [ b ) , ..., X \ [ b s ) } is a finite subcover of U . Sec-ond, there is no k ∈ K such that A ∪ B ⊆ X k . In which case there are w , w ∈ A ∪ B such that w ∈ X k and w ∈ X k ′ for some k = k ′ ∈ K .If w , w ∈ A , then { X \ ( w ] , X \ ( w ] } is a finite subcover of U . If w ∈ A, w ∈ B , then { X \ ( w ] , X \ [ w ) } is a finite subcover of U (similarly for w ∈ B, w ∈ A ). Finally if w , w ∈ B , then { X \ [ w ) , X \ [ w ) } is a finitesubcover of U .Thus, in any case, ( X ; τ i ( X )) is compact.Assume that (ii) holds and let k ∈ K . Assume that U is an open cover of X k . By Alexander’s subbase lemma we may assume that U = { X k \ ( a ] | a ∈ A } ∪ { X k \ [ b ) | b ∈ B } for some subsets A, B ⊆ X k . Consider the following open cover of X = S l ∈ K X l U ∗ = { X \ ( a ] | a ∈ A } ∪ { X \ [ b ) | b ∈ B } . Then U ∗ has a finite subcover { X \ ( a ] , ..., X \ ( a r ] } ∪ { X \ [ b ) , ..., X \ [ b s ) } since X is compact with its interval topology. Thus { X k \ ( a ] , ..., X k \ ( a r ] } ∪{ X k \ [ b ) , ..., X k \ [ b s ) } is a finite subcover of X k . (cid:3) If ( X ; ≤ ) is representable, then, for some topology τ , ( X ; τ, ≤ ) is a Priest-ley space. In particular, ( X ; τ ) is a compact space and, as τ i ⊆ τ , so too is( X ; τ i ). Thus, the following is an immediate consequence of 2.1. LEMMA 2.2.
Each order component of a representable poset is compactwith respect to its interval topology.
We now go on to show that if the order components of a poset are repre-sentable, then so is the poset.
LEMMA 2.3.
Let ( X k , ≤ k ) k ∈ K be a family of pairwise disjoint nonemptyrepresentable posets. Then ( X ; ≤ ) is representable, where X = S k ∈ K X k and ≤ = S k ∈ K ≤ k . M. E. ADAMS AND D. VAN DER ZYPEN
Proof. If K is empty or a singleton, the statement is trivial. So we mayassume that K has more than one element. For any k ∈ K , let τ k be atopology making ( X k ; τ k , ≤ k ) a Priestley space. Fix k ∈ K and x ∈ X k . Wenow build a subbase for a topology on X in three steps. We set: S = S l ∈ K \{ k } τ l ; S = { U ∈ τ k | x / ∈ U } ; S = { U ⊆ X | x ∈ U and U ∩ X k ∈ τ k and, for some k ′ ∈ K \{ k } , U =[ U ∩ X k ] ∪ [ S l ∈ K \{ k,k ′ } X l ] } .Then let τ be the topology having S = S ∪ S ∪ S as a subbase. UsingAlexander’s subbase lemma we check easily that ( X ; τ ) is compact using thefact that any subbase member containing x is, in some sense, large by virtueof the definition of S ⊆ S . Moreover, an easy distinction by cases tells usthat ( X ; τ, ≤ ) is totally order-disconnected. (cid:3) It remains to provide an example of a poset ( P ; ≤ ) which is order isomor-phic to an order component of a representable poset, but is not representableitself.On the set P = { p } ∪ { p i ,...,i n | ≤ n < ω and 0 ≤ i j < ω for 0 ≤ j ≤ n } , inductively define an order relation ≤ as follows.For 0 ≤ j < i < ω , p < p i < p j . For 0 ≤ i < ω , 0 ≤ k ≤ i , and 0 ≤ i < j < ω , p i ,i < p i ,j < p k . For 0 ≤ i , i < ω , 0 ≤ k ≤ i , and 0 ≤ j < i < ω , p i ,k < p i ,i ,i < p i ,i ,j . In general, let 0 < r < ω .For 0 ≤ i , i , . . . , i r < ω , 0 ≤ k ≤ i r , and 0 ≤ i < j < ω , p i ,i ,...,i r − ,i r ,i < p i ,i ,...,i r − ,i r ,j < p i ,i ,...,i r − ,k . For 0 ≤ i , i , . . . , i r +1 < ω , 0 ≤ k ≤ i r +1 , and 0 ≤ j < i < ω , p i ,i ,...,i r ,k < p i ,i ,...,i r ,i r +1 ,i < p i ,i ,...,i r ,i r +1 ,j . EPRESENTABLE POSETS AND THEIR ORDER COMPONENTS 5
To see that ( P ; ≤ ) is a poset, for 0 ≤ n < ω , let P ( n ) = { p } ∪ { p i ,...,i m | ≤ m ≤ n and, for 0 ≤ j ≤ m , 0 ≤ i j < ω } . Thus, P (0) and, for each 0 ≤ n < ω , P ( n + 1) \ P ( n ) are clearly antisym-metric and transitive. Further, x ∈ P ( n ) is comparable with y ∈ P \ P ( n )only if x ∈ P ( n ) \ P ( n −
1) and y ∈ P ( n + 1) \ P ( n ), where it is the casethat x > y and x < y depending on whether n is even or odd, respectively.In particular, ≤ is antisymmetric. Moreover, if n is even, say n = 2 r , then x = p i ,...,i r − ,k and y = p i ,...,i r ,i providing 0 ≤ k ≤ i r and 0 ≤ i < ω ,and if n is odd, say n = 2 r + 1, then x = p i ,...,i r ,k and y = p i ,...,i r +1 ,i providing 0 ≤ k ≤ i r +1 and 0 ≤ i < ω . In particular, ≤ is transitive and, asclaimed, ( P ; ≤ ) is seen to be a countable connected poset. We also note inpassing that, for 0 ≤ i , . . . , i n < ω , [ p i ,...,i n ) and ( p i ,...,i n ] are finite chainsdepending on whether n is even or odd, respectively, a fact that we will referback to later.In order to show that ( P ; ≤ ) is order-isomorphic to an order componentof a representable poset, we will define a suitable order (cid:22) on a compacttotally disconnected space ( C ; τ ) which itself is homeomorphic to the Stonespace of a countable atomless Boolean algebra. To do so, we will need anexplicit description of ( C ; τ ), which we now give.Let Q = ( Q ; ≤ ) denote the rational interval (0 , A, B ) is a
Dedekind cut of Q providing that A and B are disjoint non-empty sets suchthat Q = A ∪ B and, for a ∈ A and b ∈ B , a < b . For a Dedekind cut( A, B ) of Q , A is a gap providing A does not have a greatest element and B does not have a smallest element and, otherwise, it is a jump . Let ( C ; ≤ )denote the set of all decreasing subsets of the rational interval (0 ,
1) orderedby inclusion. Thus, for I ∈ C , if I = ∅ or Q , then I is a jump precisely when I = (0 , r ) or (0 , r ] for some r ∈ Q . Intuitively, ( C ; ≤ ) may be thought ofas the real interval [0 ,
1] where every rational element 0 < r < τ i , denoted henceforth simply by τ , on ( C ; ≤ ) has as a base the open intervals C , [ ∅ , I ) = { J ∈ C : J ⊂ I } ,( I, Q ] = { J ∈ C : I ⊂ J } , and ( I, J ) = { K ∈ C : I ⊂ K ⊂ J } . Itis well-known that ( C ; τ ) is a compact totally disconnected space, whoseclopen subsets are precisely the sets ∅ , C , and finite unions of sets of theform [ I, J ] = { K ∈ C : I ⊆ K ⊆ J } where I = (0 , r ] and J = (0 , s ) for r , s ∈ Q with r < s .Setting Q = ( s i : 0 ≤ i < ω ) to be some enumeration of Q , we nowinductively define a new partial order on C as follows:In C , choose gaps x and, for 0 ≤ i < ω , x i such that x < x i < x j for 0 ≤ j < i < ω, M. E. ADAMS AND D. VAN DER ZYPEN where x is a member of the closure of { x i | ≤ i < ω } , denoted cl ( { x i | ≤ i < ω } ), and set x ≺ x i ≺ x j . Choose clopen intervals ( X i : 0 ≤ i < ω ) such that x i ∈ X i , X i ∩ X j = ∅ whenever i = j , the length of X i , denoted ln ( X i ), is ≤ in the pseudomet-ric obtained from the metric imposed on C by the real metric on (0 , , s ), (0 , s ] X i for any 0 ≤ i < ω .For 0 ≤ i < ω , 0 ≤ k ≤ i , and 0 ≤ i < ω , choose gaps x i ,i ∈ X i suchthat x i ,i < x i ,j < x k for 0 ≤ i < j < ω, where x i ∈ cl ( { x i ,i | ≤ i < ω } ), and set x i ,i ≺ x i ,j ≺ x k . Choose clopen intervals ( X i ,i : 0 ≤ i < ω ) such that x i ,i ∈ X i ,i , X i ,i ∩ X i ,j = ∅ for i = j , X i ,i ⊆ X i , ln ( X i ,i ) ≤ , and (0 , s ),(0 , s ] X i ,i for 0 ≤ i < ω .For 0 ≤ i , i < ω , 0 ≤ k ≤ i , and 0 ≤ i < ω , choose gaps x i ,i ,i ∈ X i ,i such that x i ,k < x i ,i ,i < x i ,i ,j for 0 ≤ j < i < ω, where x i ,i ∈ cl ( { x i ,i ,i | ≤ i < ω } ), and set x i ,k ≺ x i ,i ,i ≺ x i ,i ,j . Choose clopen intervals ( X i ,i ,i : 0 ≤ i < ω ) such that x i ,i ,i ∈ X i ,i ,i , X i ,i ,i ∩ X i ,i ,j = ∅ for i = j , X i ,i ,i ⊆ X i ,i , ln ( X i ,i ,i ) ≤ , and (0 , s ),(0 , s ] X i ,i ,i for 0 ≤ i < ω .In general, let 0 < r < ω .For 0 ≤ i , i , . . . , i r < ω , 0 ≤ k ≤ i r , and 0 ≤ i < ω , choose gaps x i ,i ,...,i r ,i ∈ X i ,i ,...,i r such that x i ,i ,...i r − ,i r ,i < x i ,i ,...i r − ,i r ,j < x i ,i ,...,i r − ,k for 0 ≤ j < i < ω, where x i ,...,i r ∈ cl ( { x i ,...,i r ,i | ≤ i < ω } ), and set x i ,i ,...i r − ,i r ,i ≺ x i ,i ,...i r − ,i r ,j ≺ x i ,i ,...,i r − ,k . Choose clopen intervals ( X i ,i ,...,i r ,i : 0 ≤ i < ω ) such that x i ,i ,...,i r ,i ∈ X i ,i ,...,i r ,i , X i ,i ,...,i r ,i ∩ X i ,i ,...,i r ,j = ∅ for i = j , X i ,i ,...,i r ,i ⊆ X i ,i ,...,i r , ln ( X i ,i ,...,i r ,i ) ≤ r +1 , and (0 , s r +1 ), (0 , s r +1 ] X i ,i ,...,i r ,i for 0 ≤ i <ω . For 0 ≤ i , i , . . . , i r +1 < ω , 0 ≤ k ≤ i r +1 , and 0 ≤ i < ω , choose gaps x i ,i ,...,i r +1 ,i ∈ X i ,i ,...,i r +1 such that x i ,i ,...,i r ,k < x i ,i ,...i r ,i r +1 ,i < x i ,i ,...i r ,i r +1 ,j for 0 ≤ i < j < ω, EPRESENTABLE POSETS AND THEIR ORDER COMPONENTS 7 where x i ,...,i r +1 ∈ cl ( { x i ,...,i r +1 ,i | ≤ i < ω } ), and set x i ,i ,...,i r ,k ≺ x i ,i ,...i r ,i r +1 ,i ≺ x i ,i ,...i r ,i r +1 ,j . Choose clopen intervals ( X i ,i ,...,i r +1 ,i : 0 ≤ i < ω ) such that x i ,i ,...,i r +1 ,i ∈ X i ,i ,...,i r +1 ,i , X i ,i ,...,i r +1 ,i ∩ X i ,i ,...,i r +1 ,j = ∅ for i = j , X i ,i ,...,i r +1 ,i ⊆ X i ,i ,...,i r +1 , ln ( X i ,i ,...,i r +1 ,i ) ≤ r +1) , and (0 , s r +2 ), (0 , s r +2 ] X i ,i ,...,i r +1 ,i for 0 ≤ i < ω .Elsewhere on C , let (cid:22) be trivial. Thus, since ( X ; (cid:22) ) is order-isomorphicto ( P ; ≤ ), ( C ; (cid:22) ) is a poset whose order components consist precisely of X = { x } ∪ { x i ,...,i n | ≤ n < ω and 0 ≤ i j < ω for 0 ≤ j ≤ n } and 2 ω singletons. LEMMA 2.4. ( C ; τ, (cid:22) ) is a Priestley space.Proof. As ( C ; (cid:22) ) is a poset and ( C ; τ ) is a compact totally disconnectedspace, it remains to show that, for u, v ∈ C , whenever u v there exists aclopen decreasing set U such that u ∈ U and v U .Since ( X ; (cid:22) ) is order-isomorphic to ( P ; ≤ ), we set, for 0 ≤ n < ω , X ( n ) = { x } ∪ { x i ,...,i m | ≤ m ≤ n and, for 0 ≤ j ≤ n, ≤ i j < ω } and observe that, as [ x i ,...,i n ) or ( x i ,...,i n ] is a finite chain depending onwhether n is even or odd, respectively, it follows from the choice of elementsin X \ X ( n ) that, for 0 ≤ i , . . . , i n < ω , S ( X i ,...,i n − ,k : 0 ≤ k ≤ i n ) isclopen increasing or decreasing, accordingly.Consider u, v ∈ C with u v . In each case we will exhibit a clopendecreasing set U such that u ∈ U and v U .If u < v , then u ≤ (0 , s ] < v for some s ∈ Q . Since (cid:22) is compatible with ≤ , set U = ( ∅ , (0 , s ]]. Henceforth, we assume that u > v and, in particular, u and v are incomparable under (cid:22) .Suppose there is an infinite sequence ( i k : 0 ≤ k < ω ) such that u ∈ X i ,...,i k for any 0 ≤ k < ω . Then, by choice, u is a gap and, since ln ( X i ,...,i n ) ≤ n , v X i ,...,i n for some 0 ≤ n < ω . Without loss of gener-ality, we may assume that n is even. Set U = S ( X i ,...,i n ,l : 0 ≤ l ≤ i n +1 ).By the above observation, U is clopen decreasing, u ∈ U , and, since U ⊆ X i ,...,i n , v U .Likewise, if there is an infinite sequence ( j k : 0 ≤ k < ω ) such that v ∈ X j ,...,j k for any 0 ≤ k < ω , then v is a gap and, since ln ( X j o ,...,j m ) ≤ m , u X j ,...,j m for some 0 ≤ m < ω . We may assume, again with no loss ingenerality, that m is odd. Set U = C \ S ( X j ,...,j m ,l : 0 ≤ l ≤ j m +1 ). Then, M. E. ADAMS AND D. VAN DER ZYPEN U is clopen decreasing, v U , and, since U ⊆ C \ X j ,...,j m , u ∈ U .Suppose, for some finite sequence ( i k : 0 ≤ k ≤ n ), u ∈ X i ,...,i n , but u X i ,...,i n ,l for any 0 ≤ l < ω . Then, providing u = x i ,...,i n , it is nothard to see that there exists a clopen set U such that u ∈ U , v U , andeach element of U is incomparable under (cid:22) to any other element of ( C ; (cid:22) ),whereby U is decreasing. Were it the case that u X l for any 0 ≤ l < ω ,then a similar set may be defined unless u = x .Likewise, suppose it is the case that, for some finite sequence ( j k : 0 ≤ k ≤ m ), v ∈ X j ,...,j m , but that v X j ,...,j m ,l for any 0 ≤ l < ω . Then,providing v = x j ,...,j m , there exists a clopen set V such that v ∈ V , u V ,and each element of V is incomparable under (cid:22) to any other element of( C ; (cid:22) ). In this case, set U = C \ V . Likewise, unless v = x , a similar setmay be defined whenever v X l for any 0 ≤ l < ω .Thus, it now remains to consider the eventuality that u = x or x i ,...,i n for some ( i k : 0 ≤ k ≤ n ) and v = x or x j ,...,j m for some ( j k : 0 ≤ k ≤ m ).Observe that, by hypothesis, since v < u , u = x is impossible and, hence,we need only consider u = x i ,...,i n for some ( i k : 0 ≤ k ≤ n ). Further, if v = x , then, by hypothesis, u = x i ,...,i n for some n >
0. Since v X i and u = x i , u ∈ U = S ( X i ,k : 0 ≤ k ≤ i ) ⊆ X i , which, as observed above, isclopen decreasing. Thus, in addition, we may assume that v = x j ,...,j m forsome ( j k : 0 ≤ k ≤ m ).A number of possibilities still remain to be considered.Suppose first that n ≤ m .Consider i k = j k for all 0 ≤ k ≤ n . Then, by hypothesis, m ≥ n + 2 and,since u > v , n is even. Thus, V = S ( X i ,...,i n ,j n +1 ,l : 0 ≤ l ≤ j n +2 ) is clopenincreasing v ∈ V , and u V . Set U = C \ V .Suppose i k = j k for all 0 ≤ k < n , but i n = j n . Then, by hypothesis, m ≥ n +1. Suppose n is even. Were it the case that i n > j n , then it would fol-low that u < v , contrary to hypothesis. Thus, we may assume that i n < j n .But then it follows that m ≥ n + 2. Thus, v ∈ V = S ( X i ,...,i n − ,j n ,j n +1 ,l :0 ≤ l ≤ j n +2 ) which is clopen increasing and, since V ⊆ X i ,...,i n − ,j n , u V .Suppose n is odd. Thus, v ∈ V = S ( X i ,...,i n − ,j n ,l : 0 ≤ l ≤ j n +1 ), which isclopen increasing, and again, since V ⊆ X i ,...,i n − ,j n , u V . In either case,set U = C \ V .Consider, for some 0 ≤ k ≤ n − i l = j l for all 0 ≤ l < k , but i k = j k .If k is even, then u ∈ U = S ( X i ,...,i k − ,i k ,l : 0 ≤ l ≤ i k +1 ) which is clopendecreasing and, since U ⊆ X i ,...,i k − ,i k and v ∈ X i ,...,i k − ,j k , v U . If k EPRESENTABLE POSETS AND THEIR ORDER COMPONENTS 9 is odd, then v ∈ V = S ( X i ,...,i k − ,j k ,l : 0 ≤ l ≤ j k +1 ) which is clopenincreasing and, since V ⊆ X i ,...,i k − ,j k and u X i ,...,i k − ,j k , u V . In thiscase, set U = C \ V .It remains to consider n > m .Suppose i k = j k for all 0 ≤ k ≤ m . Then, by hypothesis, n ≥ m + 2 and,since u > v , m is odd. Hence, u ∈ U = S ( X j ,...,j m ,i m +1 ,l : 0 ≤ l ≤ i m +2 )which is clopen decreasing, whilst v U .Consider i k = j k for all 0 ≤ k < m , but i m = j m . By hypothesis, n ≥ m + 1. Suppose m is even. Then, u ∈ U = S ( X j ,...,j m − ,i m ,l : 0 ≤ l ≤ i m +1 )which is clopen decreasing, and, since U ⊆ X j ,...,j m − ,i m , v U . Suppose m is odd. Were i m < j m , then it would follow that u < v , contrary tohypothesis. Thus, we may assume that i m > j m and, so, n ≥ m + 2. Hence, u ∈ U = S ( X j ,...,j m − ,i m ,i m +1 ,l : 0 ≤ l ≤ i m +2 ) which is clopen decreasing,and, since it is also the case that U ⊆ X j ,...,j m − ,i m , v U .Finally, it remains to consider the case that, for some 0 ≤ k ≤ m − i l = j l for all 0 ≤ l < k , but i k = j k . However, the same argument holds,word for word, as given in the analogous case when n ≤ m . (cid:3) Since the order components of ( C ; τ, (cid:22) ) consist of precisely X = { x } ∪{ x i ,...,i n | ≤ n < ω and 0 ≤ i j < ω for 0 ≤ j ≤ n } and 2 ω singletons and,by choice, ( X ; (cid:22) ) is order-isomorphic to ( P ; ≤ ), the following is an immedi-ate consequence of 2.4. LEMMA 2.5. ( P ; ≤ ) is order-isomorphic to an order component of a rep-resentable poset. The proof of 1.1 will be complete once we have established the following.
LEMMA 2.6. ( P ; ≤ ) is not representable.Proof. Suppose, contrary to hypothesis, that ( P ; ≤ ) is representable and let( P ; τ, ≤ ) be a Priestley space for some topology τ .We claim that, for x ∈ P , there is a sequence ( x i : 0 ≤ i < ω ) such thateither, for 0 ≤ j < i < ω , x i < x j and x is the greatest lower bound of { x i | ≤ i < ω } or, for 0 ≤ i < j < ω , x i < x j and x is the least upperbound of { x i | ≤ i < ω } .To justify the claim, we consider the various possibilities. If x = p , thensetting x i = p i yields, for 0 ≤ j < i < ω , p < p i < p j . Moreover, for y ∈ P \ P (0), [ y ) ∩ P (0) is finite. In particular, p is the greatest lower bound of { p i | ≤ i < ω } . Similarly, for x = p i ,...,i n , let x i = p i ,...,i n ,i for0 ≤ i < ω . If n is even, then, for 0 ≤ i < j < ω , p i ,...,i n ,i < p i ,...,i n ,j < p i ,...,i n . Since p i ,...,i n is the greatest lower bound of [ p i ,...,i n ,i ) and, for y ∈ P \ P ( n +1), ( y ] ∩ P ( n + 1) is finite, it follows that p i ,...,i n is the least upper boundof { p i ,...,i n ,i | ≤ i < ω } . Likewise, if n is odd, then, for 0 ≤ j < i < ω , p i ,...,i n < p i ,...,i n ,i < p i ,...,i n ,j . Since p i ,...,i n is the least upper bound of ( p i ,...,i n ] and, for every y ∈ P \ P ( n + 1), [ y ) ∩ P ( n + 1) is finite, it follows that p i ,...,i n is the greatestlower bound of { p i ,...,i n ,i | ≤ i < ω } .Using the above claim, we now show that every x ∈ P is an accumulationpoint.To see this, say x is the greatest lower bound of { x i | ≤ i < ω } where,for 0 ≤ j < i < ω , x i < x j . For 0 ≤ i < ω , there exists a clopen increasingset V i such that x i ∈ V i and x i +1 V i . Clearly, { V i | ≤ i < ω } is an opencover of S = { x i | ≤ i < ω } with no finite subcover. In particular, S is notclosed. Choose y ∈ cl ( S ) \ S . If y x , then there is a clopen decreasing set U with y ∈ U and x U , from which it follows that U ∩ S = ∅ , contradicting y ∈ cl ( S ). If y > x , then y is not a lower bound of S , as x is the greatest.In particular, for some 0 ≤ n < ω , x n y . It follows that there is a clopendecreasing set U with x n ∈ U and y U . Thus, S ⊆ { x , . . . , x n } ∪ U , whichis a closed set. On the other hand, y ∈ P \ ( { x , . . . , x n } ∪ U ), contradictingthe fact that y ∈ cl ( S ). We conclude that y = x and, in particular, that, asclaimed, x is an accumulation point. As similar argument holds in the casethat x is the least upper bound of { x i | ≤ i < ω } where, for 0 ≤ i < j < ω , x i < x j .Suppose then that L is a bounded distributive lattice such that ( D ( L ); τ ( L ) , ⊆ ) (recall the notation introduced in §
1) is homeomorphic and order-isomorphicto ( P ; τ, ≤ ). For a , b ∈ L , there correspond clopen decreasing sets A , B ,respectively. Suppose a < b . Then A ⊂ B and it is possible to choose x ∈ B \ A . Since x is an accumulation point, there exists a distinct element y ∈ B \ A . Say, without loss of generality, x y . Then there exists aclopen decreasing set U with x ∈ U and y U . Set C = A ∪ ( B ∩ U ).Then C is a clopen decreasing set such that A ⊂ C ⊂ B . In particular, C corresponds to an element c ∈ L such that a < c < b . We concludethat ( Q ; ≤ ) the rational interval (0 ,
1) is embeddable in L , that is, ( Q + ; ≤ )the rational interval [0 ,
1] is a (0 , L . If one such embeddingis denoted by f + : Q + −→ L , then f corresponds to continuous order-preserving map D ( f ) : D ( L ) −→ D ( Q + ) which is also onto. That is, thereis a mapping from P onto D ( Q + ). Since D ( Q + ) is uncountable and P is EPRESENTABLE POSETS AND THEIR ORDER COMPONENTS 11 countable, this is impossible and, as required, we conclude that ( P ; ≤ ) isnot representable. (cid:3) References [1] R.Balbes,
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