Domain Representations Induced by Dyadic Subbases

We study domain representations induced by dyadic subbases and show that a proper dyadic subbase S of a second-countable regular space X induces an embedding of X in the set of minimal limit elements of a subdomain D of {0,1,â¥}Ï. In particular, if X is compact, then X is a retract of the set of limit elements of D.


Introduction
From a computational point of view, it is natural to consider a subbase of a second-countable T 0 space X as a collection of primitive properties of X through which one can identify each point of X.In this way, by fixing a numbering of the subbase, one can represent each point of X as a subset of N and construct a domain representation of X in the domain P ω of subsets of N [3,17].Note that P ω is isomorphic to the domain of infinite sequences of the Sierpinski space t1, Ku.
On the other hand, each regular open set A (i.e., an open set which is equal to the interior of its closure) of a topological space X divides X into three parts: A, the exterior of A, and their common boundary.Therefore, one can consider a pair of regular open subsets which are exteriors of each other as a pair of primitive properties and use a subbase which is composed of such pairs of open sets in representing the space.Such a subbase is called a dyadic subbase and a dyadic subbase of a space X induces a domain representation of X in the domain T ω of infinite sequences of T " t0, 1, Ku.In [20] and [12], the authors introduced to a dyadic subbase the properness property which expresses a kind of orthogonality between the components and studied domain representations of Hausdorff spaces induced by proper dyadic subbases.In this representation, the domain is fixed to T ω and an embedding ϕ S of a Hausdorff space X in T ω is derived from a proper dyadic subbase S of X.
In this paper, we derive from a dyadic subbase S a domain (i.e., an ω-algebraic pointed dcpo) D S and a bounded complete domain p D S which are subdomains of T ω containing ϕ S pXq as subspaces.The domain D S has the following properties.(1) If X is a strongly nonadhesive Hausdorff space (Definition 5.4), then the set LpD S q of limit (i.e., non-compact) elements of D S has minimal elements.(2) If X is regular, then ϕ S is an embedding of X in the set of minimal elements of LpD S q. (3) If X is compact, then there is a retraction ρ S from LpD S q to X.That is, every infinite strictly increasing sequence in KpD S q represents a point of X through ρ S and pD S , LpD S q, ρ S q is the kind of domain representations studied in [19].The domain p D S also has the properties (1) to (3) and, in addition, it is bounded complete.
We study properties of representations for second-countable Hausdorff spaces and investigate which property holds under each of the above-mentioned conditions.Therefore, a space in this paper means a second-countable Hausdorff space unless otherwise noted.We are mainly interested in the case where X is a regular space because the corresponding domain representations have good properties as we mentioned above.In addition, it is proved in [11] that every second-countable regular space has a proper dyadic subbase and in [12] that every dense-in-itself second-countable regular space has an independent subbase, which is a proper dyadic subbase with an additional property.
We review proper dyadic subbases and their properties in the next section, and we study TTE-representations and domain representations in T ω derived from (proper) dyadic subbases in Section 3. We introduce the domains D S and p D S in Section 4, and present the strongly nonadhesiveness condition in Section 5. Then we study domain representations in these domains for the case X is regular in Section 6.Finally, in Section 7, we study the small inductive dimension of the T 0 -spaces LpD S q and Lp p D S q based on a result in [19].

Preliminaries and Notations:
Bottomed Sequences: Let N be the set of non-negative integers and ¾ be the set t0, 1u.Let T be the set t0, 1, Ku where K is called the bottom character which means undefinedness.The set of infinite sequences of a set Σ is denoted by Σ ω .Each element of T ω is called a bottomed sequence and each copy of 0 and 1 which appears in a bottomed sequence p is called a digit of p.A finite bottomed sequence is a bottomed sequence with a finite number of digits, and the set of all finite bottomed sequences is denoted by T ˚.We sometimes omit K ω at the end of a finite bottomed sequence and identify a finite bottomed sequence with a finite sequence of T. The set of finite sequences of ¾ is denoted by ¾ ˚.
We define the partial order relation Ď on T by K Ď 0 and K Ď 1, and its product order on T ω is denoted by the same symbol Ď, i.e., for every p, q P T ω , p Ď q if ppnq Ď qpnq for each n P N. Then ¾ ω is the set of maximal elements of T ω .We consider the T 0 -topology tH, t0u, t1u, t0, 1u, Tu on T, and its product topology on T ω .We write domppq " tk : ppkq ‰ Ku for p P T ω .For a finite bottomed sequence e P T ˚, the length |e| of e is the maximal number n such that epn ´1q ‰ K.We denote by p| n the finite bottomed sequence with dompp| n q " domppq X t0, 1, . . ., n ´1u such that p| n Ď p.That is, p| n pkq " ppkq if k ă n and p| n pkq " K if k ě n.Note that the notation p| n is used with a different meaning in [20].
The letters a and b will be used for elements of ¾, c for elements of T, i, j, k, l, m, n for elements of N, p and q for bottomed sequences, and d and e for finite bottomed sequences.We write c ω " pc, c, ¨¨¨q P T ω for c P T. We denote by prn :" as the bottomed sequence q such that qpnq " a and qpiq " ppiq for i ‰ n.
Topology: Throughout this paper, X denotes a second-countable Hausdorff space unless otherwise noted.Therefore, if X is regular, then X is separable metrizable by Urysohn's metrization theorem.Recall that a subset U of X is regular open if U is the interior of its closure.
A filter F on the space X is a family of subsets of X with the following properties.
(2) If A P F and A Ď A 1 Ď X, then A 1 P F.
(3) If A, B P F, then A X B P F. Let Vpxq denote the family of neighbourhoods of x P X.For a filter F on X and a point x P X, if we have Vpxq Ď F then we say that F converges to x.
A family B of subsets of X is called a filter base if it satisfies H R B, B ‰ H, and that for all A, B P B there exists C P B such that C Ď A X B. A filter generated by a filter base B is defined as the minimum filter containing B. We say that a filter base converges to x P X if it generates a filter which converges to x.
We denote by cl Y A, bd Y A, and ext Y A the closure, boundary, and exterior of a set A in a space Y , respectively, and we omit the subscript if the space is obvious.Domain Theory: Let pP, Ďq be a partially ordered set (poset).We say that two elements p and p 1 of a poset P are compatible if p Ď q and p 1 Ď q for some q P P , and write p Ò p 1 if p and p 1 are compatible.For p P P and A Ď P , we define Òp " tq : q Ě pu, Óp " tq : q Ď pu, ÒA " YtÒq : q P Au, and ÓA " YtÓq : q P Au.Therefore, we have ÓÒp " tq : q Ò pu.We say that A is downwards-closed if A " ÓA, and upwards-closed if A " ÒA.
A subset A of a poset P is called directed if it is nonempty and each pair of elements of A has an upper bound in A. A directed complete partial order (dcpo) is a partially ordered set in which every directed subset A has a least upper bound (lub) \A.A dcpo is pointed if it has a least element.
Let pD, Ďq be a dcpo.A compact element of D is an element d P D such that for every directed subset A, if d Ď \A then d P ÓA.An element of D is called a limit element if it is not compact.We write KpDq for the set of compact elements of D, and LpDq for the set of limit elements of D.
For x P D, we define K x " KpDq X Óx.A dcpo D is algebraic if K x is directed and \K x " x for each x P D, and it is ω-algebraic if D is algebraic and KpDq is countable.In this paper, a domain means an ω-algebraic pointed dcpo.The Scott topology of a domain D is the topology generated by tÒd : d P KpDqu.In this paper, we consider a domain D as a topological space with the Scott topology.A poset is bounded complete if every subset which has an upper bound also has a least upper bound.T ω is a bounded complete domain such that KpT ω q " T ˚.
An ideal of a poset P is a directed downwards-closed subset.The set of ideals of P ordered by set inclusion is denoted by IdlpP q.The poset IdlpP q becomes a domain called the ideal completion of P if P is countable.We have an order isomorphism KpIdlpP qq -P for each countable poset P with a least element.On the other hand, for a domain D, we have IdlpKpDqq -D.Therefore, KpDq, the set of compact elements of D, determines the structure of D. We say that an ideal of KpDq is principal if its least upper bound is in KpDq.An infinite strictly increasing sequence d 0 Ĺ d 1 Ĺ d 2 Ĺ ¨¨¨in KpDq determines a non-principal ideal te P KpDq : e Ď d i for some iu of KpDq and thus determines a point of LpDq.
A poset P is a conditional upper semilattice with least element (cusl) if it has a least element and every pair of compatible elements has a least upper bound.If P is a cusl, then IdlpP q is a bounded complete domain.For background material on domains, see [8,1,15].
Let X be a T 0 -space and B " tB n : n P Nu be a subbase of X indexed by N. Consider the representation δ B :Ď N ω Ñ X such that δ B ppq " x if and only if tppkq : k P Nu " tn P N : x P B n u. δ B is called a standard representation of X with respect to B. Any representation which is continuously equivalent to a standard representation is admissible [23,14,22].

Domain representation:
Let D be a domain, D R a subspace of D, and µ a quotient map from D R onto X.The triple pD, D R , µq is called a domain representation of X.Note that we do not require D to be bounded-complete or each element of D R to be total (i.e., condense) in this paper.See [2,3] for the notion of totality.A domain representation is called a retract domain representation if µ is a retraction, and a homeomorphic domain representation if µ is a homeomorphism.
A domain representation pD, D R , µq of X is upwards-closed if D R is upwards-closed and µpdq " µpeq for every d Ě e P D R .A domain representation pD, D R , µq is called admissible if for every pair pE, E R q of a domain E and a dense subset E R Ď E and for every continuous function ν : E R Ñ X, there is a continuous function φ : E Ñ D such that νpxq " µpφpxqq holds for all x P E R .A domain representation E " pE, E R , νq reduces continuously to a domain representation D " pD, D R , µq if there is a continuous function φ : E Ñ D such that φpE R q Ď D R and νpxq " µpφpxqq for all x P E R .For more about (admissible) domain representations, see [3,9,16,17].

Proper dyadic subbases
Recall that a space means a Hausdorff space unless otherwise noted.Definition 2.1.A dyadic subbase S of a space X is a family tS n,a : n P N, a P ¾u of regular open sets indexed by N ˆ¾ such that (1) S n,1 is the exterior of S n,0 for each n P N and (2) it forms a subbase of X.
Note that we allow duplications in S n,a and therefore, for example, a one point set X " txu has a dyadic subbase S n,0 " X, S n,1 " H pn " 0, 1, . ..q.Note also that this definition is applicable also to non-Hausdorff spaces, though we only consider the case X is Hausdorff in this paper.We denote by S n,K the set XzpS n,0 Y S n,1 q.Since S n,0 is regular open, we get bd S n,0 " bd S n,1 " S n,K .Note that S n,K is defined differently in [20].
A topological space is called semiregular if the family of regular open sets forms a base of X.It is immediate that a regular space is semiregular.From the definition, a space with a dyadic subbase is a second-countable semiregular space.On the other hand, it is shown in [20] that every second-countable semiregular space has a dyadic subbase.From a dyadic subbase S, we obtain a topological embedding ϕ S : X Ñ T ω as follows.
We denote by x the sequence ϕ S pxq P T ω and denote by r X the set ϕ S pXq Ď T ω if there is no ambiguity of S.
In the sequence x, if xpnq " a for n P N and a P ¾, then this fact holds for some neighbourhood A of x because S n,a is open.On the other hand, if xpnq " K, then every neighbourhood A of x contains points y 0 and y 1 with ỹ0 pnq " 0 and ỹ1 pnq " 1.Therefore, if xpnq " K, then every neighbourhood A of x does not exclude both of the possibilities xpnq " 0 and xpnq " 1.
Example 2.2 (Gray subbase).Let I " r0, 1s be the unit interval and let X 0 " r0, 1{2q and X 1 " p1{2, 1s be subsets of I.The tent function is the function t : I Ñ I defined as tpxq " " 2x px P cl X 0 q, 2p1 ´xq px P cl X 1 q.
We define the dyadic subbase G as G n,a " tx : t n pxq P X a u for n P N and a P ¾.The map ϕ G is an embedding of the unit interval in T ω [7,18].If x is a dyadic rational number other than 0 or 1, then ϕ G pxq has the form eK10 ω for e P ¾ ˚, and it is in ¾ ω otherwise.Figure 1 shows the Gray subbase, with the gray lines representing G n,0 and the black lines representing G n,1 .
For a dyadic subbase S and p P T ω , let Sppq " denote the corresponding subsets of X.Note that, for x P X and p P T ω , x P Sppq ô xpkq Ď ppkq for k P domppq ô p Ò x. (2.4) For e P T ˚" KpT ω q, Speq is an element of the base generated by the subbase S. We denote by B S the base tSpeq : e P T ˚uztHu.On the other hand, LpT ω q is the space in which X is represented as the following proposition shows.
Proposition 2.3.Suppose that S is a dyadic subbase of a space X.
(1) Let x, y be distinct elements in X.Since X is T 1 , there exists e P T ˚such that x P Speq and y R Speq.From (2.3), we have e Ď x and e Ď ỹ.So we get x Ď ỹ, therefore, y R Spxq.
(2) Suppose that dompxq is finite.Then Spxq is an open set and thus txu is a clopen set which contradicts the fact that x is on the boundary of S n,a for n R dompxq.
Definition 2.4.We say that a dyadic subbase S is proper if cl Speq " Speq for every e P T ˚.
If S is a proper dyadic subbase, then Speq is the closure of the base element Speq.Therefore, by (2.4), the sequence x codes not only base elements to which x belongs but also base elements to whose closure x belongs.Proposition 2.5.Suppose that S is a proper dyadic subbase of a space X.
(1) If x P X and p Ě x, then the family tSpeq : e P K p u is a filter base which converges to x P X.
(2) If x ‰ y P X, then x P S n,a and y P S n,1´a for some n P N and a P ¾.That is, x and y are separated by some subbase element.(3) If x P X and p Ě x, then Sppq " txu.
(4) If p P ¾ ω , then Sppq is either a one-point set txu for some x P X or the empty set. Proof.
(1) Since we have x Ò e for every e P K p , we obtain cl Speq " Speq ‰ H. Therefore, we get H R tSpeq : e P K p u. (2) Since X is Hausdorff, there is e P T ˚such that x P Speq and y R cl Speq " Speq.That is, e Ď x and e / Ò ỹ by (2.3) and (2.4).Therefore, we get x / Ò ỹ. (3) From (2), we have Spxq " txu.We get Sppq Ď Spxq " txu from p Ě x, and Sppq Q x from p Ò x.
(4) Let p P ¾ ω .If p Ě x for some x P X, then Sppq is a one-point set txu by (3).If p Ğ x for all x P X, then p / Ò x, because p is maximal.Therefore, Sppq is empty.
[20] contains an example of a non-proper dyadic subbase for which Proposition 2.5 (1) to (4) do not hold.
Finally, we define a property of a dyadic subbase which is stronger than properness.
Definition 2.6.An independent subbase is a dyadic subbase such that Speq is not empty for every e P T ˚.
The Gray subbase in Example 2.2 is an independent subbase and we show many independent subbases as examples of proper dyadic subbases.When S is an independent subbase, we have Spdq Ě Speq if and only if d Ď e.In particular, Spdq ‰ Speq if d ‰ e.Therefore, for an independent subbase S, the poset pB S , Ěq ordered by reverse inclusion is isomorphic to T ˚.3.Representations and domain representations derived from dyadic subbases We study some representations and domain representations of a space X derived from a (proper) dyadic subbase of X.
We introduce two representations.The first one is immediately derived from a dyadic subbase.If S is a dyadic subbase of X, then the inverse ϕ ´1 S of the embedding ϕ S is a representation of X with the alphabet Γ " t0, 1, Ku where K is considered as an ordinary character of Γ.Each point is represented uniquely with this representation and it is easy to show that ϕ ´1 S :Ď Γ ω Ñ X is an admissible representation if and only if S n,K " H for every n.
The second one is derived from a proper dyadic subbase.If S is a proper dyadic subbase of X, from Proposition 2.5 (3), we have a map ρ S from Ò r X Ă T ω to X such that ρ S ppq is the unique element in Sppq.In particular, from Proposition 2.5 (4), ρ S restricted to the maximal elements ¾ ω is a partial surjective map from ¾ ω to X, that is, it is a representation of X which we denote by ρ 1 S .Example 3.1.For the Gray subbase G of I, ρ 1 G is a total function from ¾ ω to I which is called the Gray expansion of I [18].ρ 1 G is equivalent to the binary expansion through simple conversion functions.
As this example suggests, we consider that ρ 1 S is a generalization of the binary expansion representation to a proper dyadic subbase S. We study its continuity in Proposition 3.4.It is not admissible in general as the following proposition shows.Proposition 3.2.Suppose that S is a proper dyadic subbase of a space X. ρ 1 S is admissible if and only if S n,K " H for every n.S is continuous by Proposition 3.4 below.Since S n,K is empty, x P S n,0 or x P S n,1 holds for every x P X.Therefore, one can construct a reduction from the standard representation of X with respect to an enumeration of the subbase S to ρ 1 S .Next, we study domain representations.We start with a general construction of a domain representation from a base of a space.Suppose that B is a base of a space X such that H R B, X P B, and B is closed under finite non-empty intersection.For the domain D B obtained as the ideal completion of the poset pB, Ěq with the reverse inclusion and for the map ιpxq " tU P B : x P U u from X to D B , ι is a homeomorphic embedding of X in D B .Therefore, pD B , ιpXq, ι ´1 q is a homeomorphic domain representation which is known to be admissible [3,9,17].
We introduce two domain representations derived from (proper) dyadic subbases.The first one is pT ω , r X, ϕ ´1 S q, which is defined for a space X with a dyadic subbase S. Since ϕ S is an embedding, it is a homeomorphic domain representation.In particular, if S is an independent subbase, then the poset B S is isomorphic to the poset T ˚.Therefore, the domain D B S is isomorphic to T ω and thus the domain representations pT ω , r X, ϕ ´1 S q and pD B S , ιpXq, ι ´1 q coincide.However, if S is a dyadic subbase which is not independent, then the poset T ˚provides only a "notation" of the base B S , and a set Spdq may be the same as Speq for d ‰ e P T ˚.We show that pT ω , r X, ϕ ´1 S q is an admissible domain representation even for this case.Proposition 3.3.If S is a dyadic subbase of a space X, then pT ω , r X, ϕ ´1 S q is an admissible domain representation.
Proof.Suppose that E R is a subset of a domain E and µ is a continuous map from E R to X. Define a function φ : KpEq Ñ T ω as φpeqpnq " a if and only if µpÒe X E R q Ď S n,a .Since φ is monotonic, it has a continuous extension to E, which is a continuous function from E to T ω .It is also denoted by φ.We show that the function φ satisfies ϕ ´1 S pφppqq " µppq for p P E R .We have φppqpnq " \ ePKp φpeqpnq.Therefore, for a P ¾, φppqpnq " a if and only if pDe P K p qpφpeqpnq " aq, if and only if pDe P K p qpµpÒe X E R q Ď S n,a q, if and only if µppq P S n,a .Therefore, φppq " ϕ S pµppqq.
The other domain representation is pT ω , Ò r X, ρ S q, which is defined for a regular space X with a proper dyadic subbase S. Suppose that S is a proper dyadic subbase of a space X.From Proposition 2.5, ρ S is a map from Ò r X to X.We have ϕ S pρ S ppqq Ď p and ρ S pϕ S pxqq " x.Therefore, pT ω , Ò r X, ρ S q is an upwards-closed retract domain representation if and only if ρ S is a quotient map.Blanck showed in Theorem 5.10 of [3] that if a topological space has an upwards-closed retract Scott domain representation, then it is a regular Hausdorff space.Therefore, pT ω , Ò r X, ρ S q is a domain representation only if X is regular.We show this fact as a corollary to the following equivalence.Proposition 3.4.Let S be a proper dyadic subbase of a space X.The followings are equivalent.
Proof.p1 ñ 2q: Let p P Ò r X and x " ρ S ppq.Since tSpx| n q : n P Nu is a neighbourhood base of x in X and tÒp| m X Ò r X : m P Nu is a neighbourhood base of p in Ò r X, it suffices to show that for every n, there is m such that ρ S pÒp| m X Ò r Xq Ď Spx| n q.Since X is regular, there is m ą n such that x P Spx| m q Ď cl Spx| m q Ď Spx| n q.Then, for all q P Ò r X such that q Ě p| m , we have ϕ S pρ S pqqq Ò x| m because ϕ S pρ S pqqq Ď q Ě p| m Ě x| m .Thus, ρ S pqq P Spx| m q.Therefore, ρ S pqq P Spx| m q " cl Spx| m q Ď Spx| n q. p2 ñ 3q: Immediate.p3 ñ 1q: Suppose that x P X and n P N. For each p P Òx X ¾ ω , since ρ 1 S is continuous on p, there exists e p P K p such that ρ 1 S pÒe p X ¾ ω q Ď Spx| n q.It means that Spe p q Ď Spx| n q.
Here, we can assume that x| |ep| Ď e p by replacing e p with e p \ x| |ep| .Note that Òe p X ¾ ω for p P Òx X ¾ ω is an open cover of Òx X ¾ ω and Òx X ¾ ω is compact because it is homeomorphic to ¾ j for some j ď ω.Therefore, for some finite subset tp 0 , . . . ,p h´1 u of Òx X ¾ ω , we have Y iăh Òe p i Ě Òx X ¾ ω .Let m be the maximal length of e p i for i ă h and let l " m ´| dompx| m q|.Let d 0 , . . ., d 2 l ´1 be sequences of length m obtained by filling the first l bottoms of x| m with 0 and 1.We have pÒd 0 Y . . .Y Òd 2 l ´1q X ¾ ω " Òx| m X ¾ ω .Therefore, Y iă2 l ´1 Spd i q " Spx| m q.On the other hand, for each i ă 2 l , there is j ă h such that d i Ě e p j .Therefore, Spd i q Ď Spe p j q Ď Spx| n q.Thus, we have Spx| m q Ď Spx| n q.Since Spx| m q " cl Spx| m q, it means that X is a regular space.
Corollary 3.5.Suppose that S is a proper dyadic subbase of a space X.The triple pT ω , Ò r X, ρ S q is a domain representation if and only if X is regular.In this case, it is an admissible retract domain representation.
Proof.Suppose that X is regular.From Proposition 3.4, ρ S is a retraction with right inverse ϕ S .Therefore, ρ S is a quotient map.Since r X Ď Ò r X and ϕ ´1 S is a restriction of ρ S to r X, the identity map on T ω is a reduction map from the admissible domain representation pT ω , r X, ϕ ´1 S q to pT ω , Ò r X, ρ S q.
4. Domains D S and p D S In the previous section, we studied domain representations in the domain T ω .In the following sections, we study domain representations in subdomains D S and p D S of T ω .Before that, we consider the domain E S which is defined as the closure of r X in T ω .It is easy to show that the triple pE S , r X, ϕ ´1 S q is a dense domain representation of X and, if in addition S is proper and X is regular, then pE S , Ò r X, ρ S q is a dense admissible retract domain representation of X.In these domain representations, we have Speq ‰ H for every e P KpE S q and the family tSpeq : e P K p u forms a filter base for every p P LpE S q.In this sense, one can say that E S does not contain superfluous elements.However, E S is identical to T ω if S is an independent subbase and the domain E S does not have information about X in this case.We consider further restrictions of T ω and define the domains D S and p D S as follows.(2) We define the poset p K S Ď T ˚as p K S " tp| m : p P Ò r X, m P Nu and define p D S " Idlp p K S q.
For the Gray-subbase G of I, we have K1 P K G because ϕ G p1{2q " K10 ω , but K0 R K G and KK1 R K G . Figure 2 shows the structure of D G " p D G .We have K S Ď p K S Ď T ˚and D S Ď p D S Ď T ω for a dyadic subbase S. We also have X Ď D S and Ò X Ď p D S .
(1) If S is a dyadic subbase of a space X, then r X is dense in D S .(2) If S is a proper dyadic subbase of a space X, then r X is dense in p D S . Proof.
(1) r X is dense in D S because Speq is not empty for every e P K S .(2) By Proposition 2.5(1), Speq is not empty for every e P p K S .
The domains D S and p D S are not equal in general as the following example shows.
Example 4.3.Let W be the space obtained by glueing four copies of I at one of the endpoints.That is, W " ¾ ˆ¾ ˆI{" for " the equivalence relation identifying pa, b, 0q for a, b P ¾.Let z be the identified point.That is, z " rpa, b, 0qs for a, b P ¾.Let R be the dyadic subbase defined as R 0,c " tcu ˆ¾ ˆp0, 1s{ ", R 1,c " ¾ ˆtcu ˆp0, 1s{ ", R n`2,c " ¾ ˆ¾ ˆGn,c { ", for n P N and c P ¾.We have z " KK0 ω P LpD R q and ab0 ω P LpD R q for a, b P ¾.However, aK0 ω R LpD R q for a P ¾ and Kb0 ω R LpD R q for b P ¾.On the other hand, we have aK0 ω P Lp p D R q for a P ¾ and Kb0 ω P Lp p D R q for b P ¾.
is a dyadic subbase.Note also that cl A pA X P q " A X cl X P .Therefore, for d P T ˚, we have Therefore, T is proper.
Proposition 5.10.Suppose that X is a strongly nonadhesive space and S is a proper dyadic subbase of X.
(2) The poset p K S is finite-branching. Proof.
(1) Let e P T ˚.By applying Proposition 5.8 to the proper dyadic subbase T on Speq in Lemma 5.9, succpKq is finite in the poset K T .Since K T is identical to Òe in K S , succpeq is finite in K S .(2) In this proof, succpdq for d P K S means succpdq in K S .Let e P p K S and let k be the maximal length of elements in Y dPÓeXK S succpdq, which exists by (1).Suppose that, for some n ě k and a P ¾, ern :" as P p K S .Then, for some x P X and p Ě x, p| n`1 " ern :" as.Therefore, x| n Ď e.For d 0 " x| n , let m ą n be the least integer such that x| m Ľ d 0 .The set succpd 0 q contains x| m and we have contradiction.
Theorem 5.11.Suppose that X is a strongly nonadhesive space and S is a proper dyadic subbase of X.
(1) LpD S q has enough minimal elements and minpLpD S qq is compact.
(2) Lp p D S q has enough minimal elements and minpLp p D S qq is compact.
Note that, as Proposition 5.5 shows, Theorem 5.11 is applicable to all the Urysohn spaces, in particular, to regular spaces.Note also that the premise of Theorem 5.11 is not a necessary condition for LpD S q to have enough minimal elements.For example, for the space A and the dyadic subbase S A in Example 5.6, the domain D S A has enough minimal elements and ϕ S A pAq Ď minpLpD S A qq.
It is shown in [21] that there is a Hausdorff space X and an independent subbase S of X such that D S is equal to T ω and therefore LpD S q does not have enough minimal elements.

Domain representations in minimal-limit sets
Now, we show that X is embedded in minpLpD S qq and minpLp p D S qq for the case S is a proper dyadic subbase of a regular space X.We start with new notations and a small lemma.Definition 6.1.For a dyadic subbase S of a space X, p P T ω , and n P N, we define S n ex ppq Ď X and Sn ex ppq Ď X as follows.
S n ex ppq " x such that e " q| n .Theorem 6.3.Suppose that S is a proper dyadic subbase of a regular space X and D P tD S , p D S u.If p P LpDq and p is compatible with x in T ω , then p Ě x.In particular, r X Ď minpLpDqq.
Proof.Suppose that p P LpT ω q satisfies p Ò x and p Ě x.There is an index n P N such that xpnq ‰ K and ppnq " K.We assume that xpnq " 0. That is, x P S n,0 .Since X is regular and S is proper, x P Speq Ď cl Speq " Speq Ď S n,0 for some e P KpD S q.We can assume that e " x| m for some m ą n such that ppm ´1q ‰ K.
We have Speq " č kPdompeq pS k,epkq Y S k,K q, Sm ex ppq " Therefore, since p Ò e, we have S n,0 Ě Speq Ě Sm ex ppq Ě S m ex ppq.On the other hand, since ppnq " K, we have S n,0 X Sm ex ppq " H. Therefore, we can conclude that both Sm ex ppq " Sm ex pp| m q and S m ex ppq " S m ex pp| m q are empty.Thus, by Lemma 6.2, we have p| m R K S and p| m R p K S .Then, from Lemma 4.7, we have p R D S and p R p D S .
Theorem 6.4.Suppose that S is a proper dyadic subbase of a compact Hausdorff space X and D P tD S , p D S u.We have r X " minpLpDqq and X is a retract of LpDq.

Figure 1 :
Figure 1: Gray subbase of the unit interval I.

Proof. 1 S
Only if part: suppose that ρ 1 S is admissible and x P S n,K .Theorem 12 of[4] says that every admissible representation has a continuously equivalent open restriction.Suppose that δ :Ď ¾ ω Ñ X is such an open restriction of ρ and x " δppq.Let a " ppnq.Sinceδ is an open map, δptq P ¾ ω : qpnq " auq is an open neighbourhood of x, and since δ is a restriction of ρ 1 S , δptq P ¾ ω : qpnq " auq Ď ρ 1 S ptq P ¾ ω : qpnq " auq " S n,a Y S n,K .Therefore, S n,a Y S n,K is a neighbourhood of x, which contradicts with x P S n,K .If part: since the base B S is composed of closed and open sets, X is regular and therefore ρ 1
,K , Lemma 6.2.Let e P T ˚and n " |e|.(1) S n ex peq ‰ H if and only if e P K S .(2) Sn ex peq ‰ H if and only if e P p K S .Proof.(1) x P S n ex peq if and only if e " x| n .(2) x P Sn ex peq if and only if x| n Ď e if and only if there exists q Ě