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Applied General Topology

c© Universidad Politécnica de Valencia

Volume 9, No. 2, 2008

pp. 281-292

Scott-representability of some spaces of

Tall and Mǐskin

Harold Bennett and David Lutzer
∗

Abstract. In this paper we show that a variation of a technique

of Mǐskin and Tall yields a cocompact completely regular Moore space

that is Scott-domain-representable and has a closed Gδ-subspace that

is not Scott-domain-representable. This clarifies the general topology

of Scott-domain-representable spaces and raises additional questions

about Scott-domain representability in Moore spaces.

2000 AMS Classification: Primary 54E30; Secondary 54D70,06B35, 06F30,
54H12, 54D20

Keywords: domain, Scott-domain, Scott-domain-representable space, Moore
space, complete Moore space, cocompact, Čech-complete, subcompact, Cho-
quet complete.

1. Introduction

A domain is a continuous poset (P, ⊑) in which each non-empty directed
subset has a supremum. A Scott domain is a domain in which each nonempty
bounded set has a supremum. (For more details, see Section 2.) Representing
mathematical objects as the set of maximal elements of a domain or of a Scott
domain is an idea that originated in theoretical computer science.

Every domain carries a natural topology, called the Scott topology, and a
topological space is said to be domain representable (respectively, Scott-domain-
representable) if it is homeomorphic to the set of maximal elements of a do-
main (respectively, a Scott domain) with the relative Scott topology. In recent
years, topologists have come to see domain representability and Scott-domain
representability as strong completeness properties associated with the Baire
category theorem. For example, every subcompact regular space is domain-
representable [4] and every domain-representable space is Choquet complete
[8], and therefore a Baire space. (See Section 2 for definitions.)

∗Corresponding author.



282 H. Bennett and D. Lutzer

The basic general topology of domain-representable spaces is fairly well un-
derstood. For example, while domain-representability is an open-hereditary
property, it is not closed-hereditary (because if X is any completely regular
space that is not domain-representable, then the space obtained from βX by
isolating all points of βX − X is domain-representable [3] and contains X as
a closed subspace). Similarly, Scott-domain-representability is open-hereditary
and not closed-hereditary (as can be seen by applying the same βX construc-
tion described above). Further, any Gδ-subspace of a domain-representable
space is domain-representable, as shown in [3], so it is natural to ask whether
Gδ-subspaces of Scott-domain-representable spaces inherit Scott-domain rep-
resentability. Among metrizable spaces, the answer is “Yes,” because if X is a
Scott-domain representable metric space, then X is completely metrizable. Let
Y be a Gδ-subset of X. Then Y is also completely metrizable so that a recent
result of Kopperman, Kunzi, and Waszkiewicz [7] shows that Y is Scott-domain
representable. The first goal of this paper is to show that, without metrizabil-
ity, Scott-domain-representability is not inherited by (closed) Gδ-subspaces.
Furthermore, our example is a Moore space, a particularly nicely-behaved type
of generalized metric space.

It was already known that the equivalence among metric spaces of essen-
tially all strong completeness properties (complete metrizability, Scott-domain-
representability, Čech-completeness, cocompactness, subcompactness, and
domain-representability) breaks down outside of the metric space category.
But there is still a rich theory of completeness in the wider class of Moore
spaces, and results due to K. Martin, Tall, Rudin, Bennett, Lutzer, and Reed
show that

• among Moore spaces, domain-representability is equivalent to subcom-
pactness [4] and is equivalent to Rudin-completeness [2] which is strictly
weaker than Moore-completeness [6];

• for completely regular Moore spaces, Moore completeness is equivalent
to Čech-completeness [2];

• there is a completely regular Moore space that is Čech-complete but
not cocompact [11] and not Scott-domain-representable [5];

• if a Moore space is Scott-domain-representable, then it is completely
regular and Moore-complete, Čech complete [8], and cocompact [7];

Additional equivalents of domain-representability among Moore spaces that
involve the strong Choquet game are given in [5]. The second goal of this
paper is to explore the role of Scott-domain representability in the class of
completely regular Moore spaces and we show that a certain Moore space X0
(due to Mǐskin [10]) is Scott-domain-representable and contains a closed Gδ-
subspace Z (due to Tall [11]) that is not Scott-domain representable. This
example raises a natural question about completeness and representability in
Moore spaces, namely:

Question 1.1. Is Scott-domain-representability equivalent to cocompactness
among completely regular Moore spaces?



Scott-representability of some spaces of Tall and Miškin 283

Kopperman, Kunzi, and Waszkiewicz [7] have characterized Scott-domain-
representability in any completely regular space as being a combination of
cocompactness and a bi-topological condition (“pairwise complete regularity”),
but is not yet clear how to apply their characterization in the Moore space con-
text. A natural place to look for counterexamples to Question 1.1 is in Mǐskin’s
construction of a cocompact Moore space, mentioned above. In Section 3 we
show that some of Mǐskin’s spaces are Scott-domain-representable, but we do
not know the answer to the following:

Question 1.2. Is it true that each of Mǐskin’s spaces in [10] is Scott-domain-
representable?

In this paper we show that a certain Čech-complete Moore space constructed
by Tall embeds as a closed subspace of a Scott-domain representable Moore
space. To what extent is this a general phenomenon? More precisely, we have:

Question 1.3. Does each completely regular, Čech-complete Moore space X
embed in a Moore space Y (X) that is Scott-domain-representable? What if X
is required to be a dense subspace of Y (X)? What if X is required to be a closed
subspace?

Basic definitions appear in Section 2. Section 3 gives the basic constructions
due to Tall and Mǐskin, and shows that, with some additional restrictions, one
of Mǐskin’s spaces is Scott-domain-representable and has a closed Gδ-subspace
that is not. Throughout the paper, we reserve the symbols R, Q, and P for the
usual sets of real, rational, and irrational numbers.

2. Basic definitions

A space X is cocompact if it is T1 and has a collection C of closed subsets
with the following two properties:

a) if D is a centered 1 subcollection of C, then
⋂
D 6= ∅;

b) if U is an open subset of X and x ∈ U , then some C ∈ C has x ∈
Int(C) ⊆ C ⊆ U .

Note that the members of C might not be the closures of their interiors, even
when the interiors are non-void. If one insists that members of C are the
closures of their interiors, i.e., are regularly-closed sets, then one obtains a
different notion called regular cocompactness. The Sorgenfrey line, for example,
is cocompact but not regularly cocompact [2].

Cocompactness was introduced by de Groot and his colleagues [1]. Another
strong completeness first studied by the Amsterdam school is subcompactness,
where we say that a space X is subcompact if X has a base B with the property
that

⋂
F 6= ∅ whenever F ⊆ B has the property that if B1, B2 ∈ F, then some

B3 ∈ F has cl(B3) ⊆ B1 ∩ B2.

1A collection D is centered if
⋂
{Di : i ≤ n} 6= ∅ whenever {Di : i ≤ n} is a finite

subcollection of D.



284 H. Bennett and D. Lutzer

To define domain-representability and Scott-domain-representability, we be-
gin with a poset (S, ⊑). A subset E ⊆ S is directed if for each e1, e2 ∈ E some
e3 ∈ E has e1, e2 ⊑ e3. If sup(E) ∈ S whenever E is a nonempty directed sub-
set of S, then S is a dcpo (“directed-complete partial order”). Given a, b ∈ S,
we write a ≪ b to mean that whenever E ⊆ S is a directed set with b ⊑ sup(E),
then some e ∈ E has a ⊑ e. The set ⇓(b) is defined to be {a ∈ S : a ≪ b}. In
case ⇓(b) is directed and has b as its supremum for each b ∈ S, we say that S is
continuous. If S is a continuous dcpo, then we say that S is a domain. If the
domain S has the additional property that every nonempty bounded subset of
S has a supremum in S, then we say that S is a Scott domain. Among domains,
Scott domains are easily characterized:

Lemma 2.1. A domain (S, ⊑) is a Scott domain if and only if sup({a, b})
exists whenever a, b ∈ S and a, b ⊑ c for some c ∈ S.

Proof. To prove the nontrivial half of the lemma, suppose E is a nonempty
bounded subset of S. Let f ∈ S be an upper bound for E. If e1, e2, e3 ∈ E,
then sup({e1, e2}) ∈ S and f is an upper bound for {sup({e1, e2}), e3} in S
so that sup(sup(e1, e2), e3) ∈ S. It is easy to show that sup(sup(e1, e2), e3) =
sup(sup(ei, ej ), ek) for each permutation i, j, k of 1, 2, 3, so that the supremum
of each three-element subset of the bounded set E is well-defined. Similarly,
sup(F ) is a well-defined point of S for each non-empty finite subset F ⊆ E.
Now let D := {sup(F ) : ∅ 6= F ⊆ E and |F | < ω}. Then D is a directed
subset of S so that, S being a domain, sup(D) ∈ S. Clearly sup(D) = sup(E)
as required. �

Every poset (S, ⊑) can be endowed with a special topology called the Scott
topology in which a set U is open if and only if it satisfies both (i) if x ⊑ y
and x ∈ U , then y ∈ U , and (ii) if E ⊆ S is a nonempty directed set with
sup(E) ∈ U , then E ∩ U 6= ∅. In a domain S, the collection of all sets
⇑(a) := {b ∈ S : a ≪ b} is a base for the Scott topology on S. The set
of maximal elements of a domain S is denoted by max(S). If a topological
space X is homeomorphic to the subspace max(S) of some domain S with the
relative Scott topology, then we say that X is domain-representable. If S is
a Scott-domain and X is homeomorphic to max(S), then we say that X is
Scott-domain-representable.

Kopperman, Kunzi, and Waszkiewicz [7] have characterized Scott-domain-
representable spaces as being the cocompact spaces that also satisfy a certain
bi-topological condition. A short, direct proof of cocompactness of any Scott-
domain-representable space is possible and we give it here. A central tool is
the following Interpolation Lemma [9].

Lemma 2.2. Suppose a ≪ c in a domain S. Then some b ∈ S has a ≪ b ≪ c.

Lemma 2.3. Let S be a Scott domain. For each p ∈ S, let ↑(p) = {q ∈ S :
p ⊑ q}. Then each set ↑(p) ∩ max(S) is a relatively closed subset of max(S).

Proof. Suppose that x ∈ max(S) is a limit point of ↑(p) ∩ max(S). Then for
each q ≪ x, ⇑(q) ∩ ↑(p) 6= ∅. Consequently p and q have a common extension,



Scott-representability of some spaces of Tall and Miškin 285

so that r(p, q) := sup{p, q} is in S. Let E := {r(p, q) : q ≪ x}. We claim
that E is a directed set. For suppose that r(p, q1), r(p, q2) ∈ E. Because
⇓(x) is directed, some q3 ∈ ⇓(x) has q1, q2 ⊑ q3. Then r(p, q3) ∈ E and
r(p, qi) ⊑ r(p, q3) for i = 1, 2. Because S is a dcpo, sup(E) ∈ S so that some
z ∈ max(S) has sup(E) ⊑ z. Recall that as a subspace of S, max(S) is a T1-
space. Therefore, if z 6= x, then some q4 ≪ x has z 6∈ ⇑(q4). Because q4 ≪ x,
Lemma 2.2 gives q5 ∈ S with q4 ≪ q5 ≪ x. But q5 ≪ x forces r(p, q5) ∈ E so
that q4 ≪ q5 ⊑ r(p, q5) ⊑ sup(E) ⊑ z. Therefore z ∈ ⇑(q5) ⊆ ⇑(q4), contrary
to our choice of q4. Therefore, p ⊑ sup(E) = z = x showing that x ∈ ↑(p) as
required. �

Our next result appears in [7]. We present an easy direct proof.

Corollary 2.4. Suppose S is a Scott domain. Then the subspace of maximal
elements of S is cocompact.

Proof. First, the subspace max(S) of S is T1. Second, let C = {max(S) ∩ ↑(p) :
p ∈ S}. In the light of Lemma 2.3, each member of C is a closed subset
of max(S). To verify the first part of the cocompactness definition, suppose
that D ⊆ C is a centered collection. Write D = {max(S) ∩ ↑(a) : a ∈ A}.
Then, given any finite set F := {a1, · · · , ak} ⊆ A we know that ↑(a1) ∩ · · · ∩
↑(ak) 6= ∅ because D is centered, so that sup(F ) ∈ S by Lemma 2.1. Let

 := {sup(F ) : ∅ 6= F ⊆ A and |F | < ω}. Then  is directed, so sup(Â) ∈ S,

say sup(Â) = b ∈ S. Then ∅ 6= ↑(b) ∩ max(S) ⊆
⋂
D, as required.

To verify the second part of the definition of cocompactness, it is enough to
consider a point x in a basic open set max(S)∩⇑(q). The Interpolation Lemma
2.2 provides a point p ∈ S with q ≪ p ≪ x. Then ⇑(p) is a neighborhood of x
with ⇑(p) ⊆ ↑(p) ⊆ ⇑(q) so that x is in the relative interior of ↑(p) ∩ max(S)
which is contained in the closed set ↑(p)∩max(S) ⊆ max(S)∩⇑(q), as required
to show that max(S) is cocompact. �

3. A variation of spaces of Tall and Mǐskin

Tall and Mǐskin began their constructions with a countable subset of the
plane that had uncountably many limit points on the x-axis. We need more
control and so we replace that countable set by a binary tree T with ω-many
levels and use its branch space Y in place of the limit points on the x-axis. This
tree may be embedded in the upper half plane in such a way that its branch
space corresponds in a natural way to an uncountable set (the Cantor set) on
the x-axis. Therefore, the space we will construct is one of the spaces due to
Mǐskin.
Description of the space X and the subspace X0:

a) The tree T : Let T be a binary tree with ω-many levels. Denote the
unique minimal element of T by 0̄. The level of any d ∈ T in our
tree is denoted by lv(d) and T (n) = {d ∈ T : lv(d) = n} so that
T =

⋃
{T (n) : 0 ≤ n < ω}.



286 H. Bennett and D. Lutzer

b) The branches of T : Let Y be the set of all branches of T , i.e., each
y ∈ Y is a maximal linearly ordered subset of T . We let e(y, n) denote
the unique element of the branch y that lies at level n of the tree T .
Thus, for example, e(y, 0) = 0̄ for each y ∈ Y and if y1, y2 ∈ Y have
e(y1, n) = e(y2, n), then e(y1, k) = e(y2, k) for each 0 ≤ k ≤ n.

c) The space X: Let T ∗ := {(d, S) : d ∈ T, ∅ 6= S ⊆ Y }. The underlying
set of our space is X = T ∗ ∪Y and the set X is topologized by isolating
each point of T ∗ and by using the sets

N (y, n) = {y} ∪ {(d, S) ∈ T ∗ : lv(d) ≥ n, d ∈ y, and y ∈ S}

as basic neighborhoods of y ∈ Y . Equivalently, N (y, n) = {y} ∪
{(e(y, k), S) ∈ T ∗ : n ≤ k < ω and y ∈ S}.

d) The subspace X0 = X − {(0̄, Y )}. This is a closed and open subspace
of X and will be the space used in our example. However, X0 is not
needed until the very end of Section 3.

As in [10], the space X is a cocompact, Čech-complete, completely regular
Moore space and therefore so is its closed and open subspace X0. The subspace
X0 contains a closed (and hence Gδ) -subspace Z := {(d, S) ∈ T

∗ : (d, S) 6=
(0̄, Y ), |S| < ω} ∪ Y that is homeomorphic to one of the spaces constructed
by Tall in [11]. Consequently, Z is not cocompact and therefore is not Scott-
domain-representable (by Corollary 2.4). What remains is to prove that X0 is
Scott-domain representable. Our first step is to define the Scott domain that
comes very close to representing X, and then to work around the “almost” part
of that statement to show that X0 is Scott-domain representable. We begin by
constructing our poset.
The poset (S, ⊑)

e) The sets I(B, k): Let ∅ 6= B ⊆ Y and let k ≥ 0. If |B| = 1, then let
I(B, k) := N (y, k) where y is the unique point of B. If |B| ≥ 2, then
let

I(B, k) := {(d, S) ∈ T ∗ : lv(d) ≥ k, B ⊆ S, and ∀y ∈ B, d ∈ y}.

Note that the condition d ∈ y is equivalent to e(y, lv(d)) = d, a fact
that will be used later.

f) Let S = {{t} : t ∈ X} ∪ {I(B, k) : I(B, k) 6= ∅, ∅ 6= B ⊆ Y, 0 ≤ k <
ω} and let ⊑ denote reverse inclusion. Consequently, if t ∈ X, then
I(B, k) ⊑ {t} means t ∈ I(B, k).

Remark 3.1. If |B| ≥ 2, one can prove that I(B, k) =
⋂
{N (y, k) : y ∈ B},

and that was the way we initially thought of the sets I(B, k). However, that
fact is not really needed in our construction.

The next example illustrates how the sets I(B, k) can behave. It introduces
special notations, and parts c), d), and e) will be very important tools in the
proofs of later lemmas in this section.

Example 3.2. Let d′, d′′ be the two points of T (1) and recall that 0̄ is the
unique point of T (0). Let y′, y′′ ∈ Y have e(y′, 1) = d′ and e(y′′, 1) = d′′ (so



Scott-representability of some spaces of Tall and Miškin 287

that y′, y′′ are two branches of T that disagree at level 1 of the tree). Note that
y′ ∩ y′′ = {(0̄, Y )} = T (0). Then

a) A set of the form I(B, k) can be empty. For example, I({y′, y′′}, 1) =
∅ = I(Y, 1) because if (d, S) ∈ I({y′, y′′}, 1) then lv(d) ≥ 1 and
e(y′, lv(d)) = d = e(y′′, lv(d)). Because T is a tree and lv(d) ≥ 1
we must have d′ = e(y′, 1) = e(y′′, 1) = d′′ so that y′ ∩ y′′ contains
some element of T at or above level 1, which is false.

b) I({y′, y′′}, 0) is the infinite set {(0̄, S) : y′, y′′ ∈ S ⊆ Y } and I(Y, 0) is
the singleton set {(0̄, Y )}.

c) For any I(B, k), if (d, S) ∈ I(B, k) then (d̂, S) ∈ I(B, k) where d̂ is the
unique predecessor of d in level k of the tree T .

d) For any I(B, k), if (d, S) ∈ I(B, k) then (d, B) ∈ I(B, k) and (d, S′) ∈
I(B, k) whenever S ⊆ S′ ⊆ Y . In particular (d, Y ) ∈ I(B, k).

e) From b), c) and d), the only way that |I(B, k)| = 1 is for k = 0 and
B = Y , and then I(Y, 0) = {(0̄, Y )}.

Lemma 3.3. For any B ⊆ Y and any k ≥ 0, |I(B, k)∩Y | ≤ 1. If |B| ≥ 2 then
I(B, k) ⊆ T ∗ and π1[I(B, k)] is finite, where π1 : T

∗ → T is first coordinate
projection.

Proof. The first two assertions follow directly from the definition of the sets
I(B, k), so we prove only the final assertion. Because |B| ≥ 2 we may choose
distinct y1, y2 ∈ B. Then there is some integer L such that e(y1, L) 6= e(y2, L)
so that e(y1, j) 6= e(y2, j) for each j ≥ L. Therefore, if (d, S) ∈ I(B, k), we know
that d ∈ T and lv(d) < L, and there are only finitely many such points. �

Lemma 3.4. The maximal elements of S are the singleton sets {x} where
x ∈ X.

Lemma 3.5. If ∅ 6= B1 ⊆ B2 ⊆ Y with and k1 ≤ k2, then I(B1, k1) ⊑
I(B2, k2). Furthermore if I(B1, k1) ⊑ I(B2, k2) 6= ∅, then B1 ⊆ B2 and
k1 ≤ k2.

Proof. First suppose that B1 ⊆ B2 and k1 ≤ k2. If |B2| = 1 then B1 = B2.
Let y be the unique point of B2. Then k1 ≤ k2 gives I(B2, k2) = N (y, k2) ⊆
N (y, k1) = I(B1, k1) and hence I(B1, k1) ⊑ I(B2, k2). In case B2 has at least
two points, then I(B2, k2) ⊆ T

∗ so that each element of I(B2, k2) has the form
(d, S) where lv(d) ≥ k2 and d ∈ B2. Hence lv(d) ≥ k2 ≥ k1 and d ∈ y for each
y ∈ B2. Because B1 ⊆ B2, we have (d, S) ∈ I(B1, k1), as required.

To prove the second claim, note that I(B1, k1) ⊑ I(B2, k2) gives I(B2, k2) ⊆
I(B1, k1) because ⊑ is reverse inclusion. Now fix any (d, S) ∈ I(B2, k2). Then

lv(d) ≥ k2, B2 ⊆ S, and d ∈ y for all y ∈ B2. Let d̂ be the unique predecessor

of d at level k2 of the tree T . Then (see Example 3.2), (d̂, S) ∈ I(B2, k2) ⊆

I(B1, k1) so that k2 = lv(d̂) ≥ k1. Thus k1 ≤ k2. Next, Example 3.2 shows
that since (d, S) ∈ I(B2, k2), (d, B2) ∈ I(B2, k2) ⊆ I(B1, k1) so that B1 ⊆ B2,
as required. �



288 H. Bennett and D. Lutzer

Lemma 3.6. Let E := {I(Bα, kα) : α ∈ A} be a directed subset of (S, ⊑) that
contains no maximal element of itself. Let C =

⋃
{Bα : α ∈ A}.

a) If |C| = 1 then the set {kα : α ∈ A} is unbounded, and sup(E) = {y}
where y is the unique point of C. (Note that in this case, y ∈ Y .)

b) If |C| ≥ 2, then {kα : α ∈ A} is bounded and sup(E) = I(C, L) where
L = max{kα : α ∈ A}.

Proof. In case (a), it is clear that {y} is an upper bound for E, and that no other
{z} for z ∈ Y can be an upper bound for E. In addition, each Bα = {y}. If
the set {kα : α ∈ A} is bounded, let kβ be its largest member. Then I(Bβ , kβ )
is the maximal member of E, contrary to hypothesis. Therefore {kα : α ∈ A}
is unbounded, and now it is clear that sup E = {y}.

To prove (b), fix distinct y1, y2 ∈ C and choose αi ∈ A with yi ∈ Bαi for
i = 1, 2. Using directedness of E, find β ∈ A with I(Bαi , kαi ) ⊑ I(Bβ , kβ ).
Then I(Bβ , kβ ) 6= ∅ so that by Lemma 3.5 yi ∈ Bαi ⊆ Bβ . According to
Lemma 3.3, the set F := π1[I(Bβ , kβ )] is finite.

Next, we claim that some d ∈ F has d ∈ π1[I(Bα, kα)] for each α ∈ A.
For contradiction, suppose that corresponding to each d ∈ F there is some
γ(d) ∈ A with d 6∈ π1[I(Bγ(d), kγ(d))]. Directedness of E provides some η ∈ A
with I(Bβ , kβ ) ⊑ I(Bη, kη) and such that I(Bγ(d), kγ(d)) ⊑ I(Bη, kη) for each

of the finitely many d ∈ F . Choose any (d̄, S) ∈ I(Bη, kη). Then I(Bβ , kβ ) ⊑
I(Bη, kη) yields I(Bη, kη) ⊆ I(Bβ , kβ) so that d̄ ∈ π1[I(Bβ , kβ )] = F . Be-
cause d̄ ∈ F we know that γ(d̄) is defined and d̄ 6∈ π1[I(Bγ(d̄), kγ(d̄)]. Because

I(Bγ(d̄), kγ(b̄)) ⊑ I(Bη, kη) we have (d̄, S) ∈ I(Bη, kη) ⊆ I(Bγd̄ , kγb̄ ) and that is

impossible because we know that d̄ 6∈ π1[I(Bγ
d̄
, kγ(d̄))].

At this stage of the argument, we know that there is some d0 ∈ F with
d0 ∈ π1[I(Bα, kα)] for each α ∈ A. Then for some Sα ⊆ Y we have (d0, Sα) ∈
I(Bα, kα). Because Bα ⊆ C, part (c) of Example 3.2 shows that (d0, C) ∈
I(Bα, kα). Consequently lv(d0) ≥ kα and we conclude that lv(d0) is an upper
bound for the set {kα : α ∈ A}. Let L be the largest member of the set
{kα : α ∈ A}. Note that lv(d0) ≥ L.

Next we claim that (d0, C) ∈ I(C, L). Consider the membership criteria for
I(C, L). We already know that lv(d0) ≥ L and obviously C ⊆ C, so all we must
show is that d0 ∈ y for each y ∈ C. Fix any y ∈ C. Then there is some α ∈ A
with y ∈ Bα. From above we know that (d0, C) ∈ I(Bα, kα) so that y ∈ Bα
gives d0 ∈ y as required. Now we know that I(C, L) 6= ∅ so that I(C, L) ∈ S.

According to Lemma 3.5, I(C, L) is an upper bound for E. To complete
the proof that I(C, L) = sup(E), we consider any upper bound G ∈ S for E
and we will show that I(C, L) ⊑ G. With I(Bβ , kβ ) as defined in the second
paragraph of this proof, we have I(Bβ , kβ ) ⊑ G so that G ⊆ I(Bβ , kβ ). Hence
G ⊆ I(Bβ , kβ ) ⊆ T

∗ so that either G has the form G = I(H, m) or else
G = {(e, S)} ∈ max S. In the first case, Lemma 3.5 shows that I(Bα, kα) ⊑
I(H, m) implies Bα ⊆ H and kα ≤ m for each α ∈ A, so that C ⊆ H and
L = max{kα : α ∈ A} ≤ m. Hence I(C, L) ⊑ I(H, m) = G, as claimed. In
the second case, where G = {(e, S)}, we will show that (e, S) ∈ I(C, L). Note



Scott-representability of some spaces of Tall and Miškin 289

that I(Bα, kα) ⊑ G = {(e, S)} gives (e, S) ∈ I(Bα, kα) so that lv(e) ≥ kα and
Bα ⊆ S for each α and therefore C ⊆ S and lv(e) ≥ max{kα : α ∈ A} = L.
Furthermore, if y ∈ C then y ∈ Bα for some α ∈ A so that (e, S) ∈ I(Bα, kα)
guarantees that e ∈ y. Therefore, I(C, L) ⊑ G, as required. to show that
I(C, L) = sup(E). �

Lemma 3.7. In S, we have I(B1, k1) ≪ I(B2, k2) if and only if B1 is a finite
set, B1 ⊆ B2, and k1 ≤ k2.

Proof. First suppose I(B1, k1) ≪ I(B2, k2). Then I(B1, k1) ⊑ I(B2, k2) so that
B1 ⊆ B2 and k1 ≤ k2. We let F be the collection of all finite subsets of B2.
Then E := {I(F, k2) : F ∈ F} is a directed subset of S and I(B2, k2) = sup E
so that I(B1, k1) ≪ I(B2, k2) gives I(B1, k1) ⊑ I(F1, k2) for some F1 ∈ F,
showing that B1 ⊆ F1. Since F1 is finite, so is B1.

For the converse, suppose that B1 is a finite set and B1 ⊆ B2 and k1 ≤ k2
(so that I(B1, k1) ⊑ I(B2, k2)), and suppose that E = {I(Bα, kα) : α ∈ A} is a
directed subset of S with I(B2, k2) ⊑ sup(E). If E contains a maximal element
of itself, there is nothing to prove, so assume that E contains no maximal
element.

Let C :=
⋃
{Bα : α ∈ A}. There are several cases to consider. In case

|C| ≥ 2, Lemma 3.6 gives

I(B1, k1) ⊑ I(B2, k2) ⊑ sup E = I(C, L)

where L is the largest member of the bounded set {kα : α ∈ A}, say L = kγ
for some γ ∈ A. Then I(B1, k1) ⊑ I(B2, K2) ⊑ I(C, L) gives B1 ⊆ B2 ⊆ C.
Therefore, each y in the finite set B1 is a point of C =

⋃
{Bα : α ∈ A}, so we

may find α(y) ∈ A with y ∈ Bα(y). Directedness of the collection E allows us to
find β ∈ A with I(Bα(y), kα(y)) ⊑ I(Bβ , kβ) for each y in the finite set B1 and
therefore y ∈ Bα(y) ⊆ Bβ . Therefore B1 ⊆ Bβ. Once again using directedness,
find δ ∈ A with I(Bγ , kγ ), I(Bβ , kβ ) ⊑ I(Bδ, kδ). Then B1 ⊆ Bβ ⊆ Bδ and

k1 ≤ max{kα : α ∈ A} = L = kγ ≤ kδ ≤ L.

Therefore I(B1, k1) ⊑ I(Bδ, kδ) ∈ E as required.
The remaining case is where |C| = 1, say C = {z}. Then Bα = {z} for

each α ∈ A. Because E contains no maximal element of itself, Lemma 3.6
shows that sup E = {z} and that {kα : α ∈ A} is unbounded. Choose µ ∈ A
with kµ > k1. Then I(Bµ, kµ) = N (z, kµ) ⊆ N (z, k1) = I(B1, k1) so that
I(B1, k1) ⊑ I(Bµ, kµ) ∈ E as required. �

Lemma 3.8. Suppose S ∈ S and y ∈ Y . Then S ≪ {y} if and only if
S = I({y}, k) for some k ≥ 0.

Proof. Suppose S = I({y}, k). By Lemma 3.7, I({y}, k) ≪ I({y}, k) ⊑ {y},
so we know that S = I({y}, k) ≪ {y}. For the converse, suppose S ∈ S has
S ≪ {y}. Then S ⊑ {y} so that y ∈ S. By Lemma 3.3, either S = I({y}, k) or
else S = {y}. If S = {y} let E := {I({y}, k) : k ≥ 0}. This is a directed set in
S with sup E = {y} and yet no member I({y}, k) ∈ E has S = {y} ⊑ I({y}, k).
Therefore, S must have the form S = I({y}, k) as claimed. �



290 H. Bennett and D. Lutzer

Lemma 3.9. For t ∈ X − Y, {t} ≪ {t} provided t 6= (0̄, Y ).

Proof. Write t = (d, S) with (d, S) 6= (0̄, Y ). To show that {t} ≪ {t}, suppose
{t} ⊑ sup E where E is a directed subset of S. Maximality of {t} in S (see
Lemma 3.4) shows that sup(E) = {t}.

If E contains a maximal member, there is nothing to prove, so for contra-
diction, suppose E contains no maximal member of itself. Then the collection
E must be of the form E = {I(Bα, kα) : α ∈ A}.

Write C =
⋃
{Bα : α ∈ A}. If |C| = 1, then C = {y} ⊆ Y , so that Lemma

3.6 shows sup E = {y} and hence {y} = {t}. That is impossible because y ∈ Y
and t ∈ X − Y . Therefore |C| ≥ 2.

Because |C| ≥ 2, from Lemma 3.6 we know that the set {kα : α ∈ A} is
bounded and sup E = I(C, L) where L is the maximal element of the bounded
set {kα : α ∈ A}. Then {t} = sup(E) = I(C, L) so that I(C, L) is a singleton.
Part (e) of Example 3.2 shows that the set I(C, L) can be a singleton if and
only if C = Y and L = 0, and then I(C, L) = {(0̄, Y )}, forcing us to conclude
that t = (0̄, Y ), which is false. This contradiction completes the proof of the
lemma. �

Corollary 3.10. The poset (S, ⊑) is continuous.

Proof. Consider any element S ∈ S. If S ≪ S, then S ∈ ⇓(S), so that ⇓(S) is
directed with sup(⇓(S)) = S. So suppose S ≪ S is false. Then Lemmas 3.8
and 3.9 show that one of the following three statements must be true:

(i) S = I(B, k) where B is infinite, or
(ii) S = {y} for some y ∈ Y , or
(iii) S = {(0̄, Y )}.

If S = I(B, k) where B is infinite, let F be the collection of all finite subsets
of B. Then, by Lemma 3.7, ⇓(I(B, k)) = {I(F, j) : j ≤ k, F ∈ F}, which is
directed and has I(B, k) as its supremum, as required. In case S = {y} for
some y ∈ Y , then ⇓(S) = {I({y}, k) : k ≥ 1} which is also directed and has
supremum S = {y}, as required. The case where S = {(0̄, Y )} is actually a
special case of item (i) because {(0̄, Y )} = I(Y, 0) as noted in Example 3.2,
above. �

Lemma 3.11. (S, ⊑) is a Scott domain.

Proof. Suppose U1, U2 ∈ S have a common extension. We may assume that
neither Ui is maximal in S (so that Ui = I(Bi, ki) for i = 1, 2) and that neither
of U1, U2 is contained in the other. Then there is some (d, S) ∈ I(B1, k1) ∩
I(B2, k2). Let C = B1 ∪ B2. Because neither of U1, U2 is contained in the
other, |C| ≥ 2 and (d, S) ∈ I(C, max(k1, k2)) yields I(C, max(k1, k2)) 6= ∅ so
that I(C, max(k1, k2)) ∈ S. Clearly I(C, max(k1, k2)) is an upper bound for
U1 and U2.

To show that I(C, max(k1, k2)) is the least upper bound of U1 = I(B1, k1)
and U2 = I(B2, k2), consider any upper bound U3 ∈ S for U1 and U2. From
Ui ⊑ U3 we obtain U3 ⊆ U1∩U2. Because |C| ≥ 2 we know that U3 ⊆ U1∩U2 ⊆



Scott-representability of some spaces of Tall and Miškin 291

X − Y , so that U3 cannot have the form {y} for some y ∈ Y . Therefore either

U3 = I(D, j) for some D and some j, or else U3 = {(d̂, Ŝ)} ∈ max(S). In the
first case Bi ⊆ D and j ≥ ki for i = 1, 2 so that C ⊆ D and max(k1, k2) ≤ j and
therefore (see Lemma 3.5) I(C, max(k1, k2)) ⊑ U3. In the second case, where

U3 = {(d̂, Ŝ)} ∈ max(S), for i = 1, 2 we know that (d̂, Ŝ) ∈ I(Bi, ki) so that

lv(d̂) ≥ ki, Bi ⊆ Ŝ, and that for each y ∈ Bi, y ∈ d̂. Hence I(C, max(k1, k2)) ⊑
U3. Therefore I(C, max(k1, k2)) = sup(U1, U2) as required. �

There is a natural-looking function that sends each x ∈ X to the element
{x} ∈ S. This mapping is 1-1, onto, and continuous from X to max(S), and
it is tempting to think that the function is an a homeomorphism from X onto
max(S). Unfortunately, it is not. The point (0̄, Y ) ∈ X is isolated in X, but the
point {(0̄, Y )} is not an isolated point of max(S). We are lucky that (0̄, Y ) is
the only “bad” point for the natural mapping. Recall that X0 = X − {(0̄, Y )}.
Then we have:

Lemma 3.12. The function h : X0 → max(S)−{{(0̄, Y )}} given by h(t) = {t}
is a homeomorphism from X0 onto the open subspace max(S) − {{(0̄, Y )}} of
max(S) with the relative Scott topology.

Proof. Clearly the function h is 1-1 and h[X0] = max(S)−{{(0̄, Y )}}. To prove
that h is continuous, it is enough to consider what happens at non-isolated
points of X0, i.e., at points y ∈ Y . Suppose h(y) ∈ ⇑(p) ∩ max(S) where
p ∈ S. Then Lemma 3.8 guarantees that p = I({y}, k) = N (y, k) for some k.
We claim that that h[N (y, k + 1)] ⊆ ⇑(p). Apply Lemma 3.9 to show that if
(d, S) ∈ N (y, k+1) then (d, S) 6= (0̄, Y ) so that h((d, S)) = {(d, S)} ≪ {(d, S)}.
Then note that p ⊑ {(d, S)} ≪ {(d, S)} so that h(d, S) ∈ ⇑(p) as required.

To prove that h is an open mapping onto max(S) −{{(0̄, Y )}}, the first step
is to recall Lemma 3.9 which shows that if t ∈ X − Y with t 6= (0̄, Y ), i.e.,
if t is an isolated point of X0, then in S, {t} ≪ {t} so that h(t) = {t} is an
isolated point of max(S). Second, consider any non-isolated point y ∈ X0 and
note that for k ≥ 1, h[N (y, k)] = max(S) ∩ ⇑(I({y}, k). Therefore h is an open
mapping onto max(S) − {{(0̄, Y )}} as required. �

Our next lemma shows that X0 is Scott-domain-representable.

Lemma 3.13. The subspace X0 = X −{(0̄, Y )} is Scott-domain-representable.

Proof. Because S is a Scott domain, we know that its subspace max(S) is
Scott-domain-representable. It is easy to check that for any domain D, the
subspace max(D) is T1. Therefore we see that for our Scott domain S, the
set max(S)−{{(0̄, Y )}} is an open subspace of the Scott-domain-representable
space max(S). Now recall that any non-empty, relatively open subset of a
Scott-domain representable space is also Scott-domain representable, and that
completes the proof. �



292 H. Bennett and D. Lutzer

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Received July 2007

Accepted October 2007

Harold Bennett (bennett@math.ttu.edu)
Texas Tech University, Lubbock, TX 79409, USA.

David Lutzer (lutzer@math.wm.edu)
College of William & Mary, Williamsburg, VA 23187, USA.