1–LABEL:LastPageJul. 14, 2016May 16, 2018

# Uniqueness of directed complete posets based on Scott closed set lattices

Zhao Dongsheng  National Institute of Education, Nanyang Technological
University, Singapore 637616<EMAIL_ADDRESS>and  Xu Luoshan
Department of Mathematics, Yangzhou University, Yangzhou 225002, China
<EMAIL_ADDRESS>
###### Abstract.

In analogy to a result due to Drake and Thron about topological spaces, this
paper studies the dcpos (directed complete posets) which are fully determined,
among all dcpos, by their lattices of all Scott-closed subsets (such dcpos
will be called $C_{\sigma}$-unique). We introduce the notions of down-linear
element and quasicontinuous element in dcpos, and use them to prove that dcpos
of certain classes, including all quasicontinuous dcpos as well as Johnstone’s
and Kou’s examples, are $C_{\sigma}$-unique. As a consequence,
$C_{\sigma}$-unique dcpos with their Scott topologies need not be bounded
sober.

###### Key words and phrases:

dcpo; sober space; Scott topology; $C_{\sigma}$-unique dcpo

###### 2010 Mathematics Subject Classification:

06B30; 06B35; 54C35

##  1\. Introduction

From a result by Drake and Thron in [1], one deduces the following result (see
Fact 3 in the Appendix): a topological space $X$ has the property that $C(X)$
isomorphic to $C(Y)$ implies $X$ is homeomorphic to $Y$ iff $X$ is sober and
$T_{D}$ (every derived set $d(\\{x\\})=cl(\\{x\\})-\\{x\\}$ of point $x\in X$
is closed), where $C(X)$ and $C(Y)$ denote the lattices of closed sets of $X$
and $T_{0}$ space $Y$, respectively (see also [13], line 11-13, page 504).

For any dcpo $P$, let $C_{\sigma}(P)$ denote the lattice of all Scott closed
subsets of $P$ (with the inclusion order). A directed complete poset (or dcpo,
for short) $P$ will be called a $C_{\sigma}$-unique dcpo (or
$C_{\sigma}$-unique, for short) if for any dcpo $Q$, $P$ is isomorphic to $Q$
whenever the lattices $C_{\sigma}(P)$ and $C_{\sigma}(Q)$ are isomorphic. From
a counterexample constructed in [6] recently, we know that not every dcpo is
$C_{\sigma}$-unique. It is therefore natural to ask which dcpos are
$C_{\sigma}$-unique. One of the classic results in domain theory is that a
dcpo $P$ is continuous iff the lattice $C_{\sigma}(P)$ is a completely
distributive lattice (Theorem II-1.14 of [2]). From this it follows that every
continuous dcpo is sober and $C_{\sigma}$-unique. In a similar way, one can
deduce that every quasicontinuous dcpo is sober and $C_{\sigma}$-unique.
Compared with Drake’s and Thron’s result, one naturally asks whether every
$C_{\sigma}$-unique dcpo is sober in their Scott topology.

In [9], Johnstone constructed the first dcpo whose Scott topology is not
sober. Later Isbell [8] constructed a complete lattice whose Scott topology is
not sober. Kou [10] further gave a dcpo whose Scott topology is well-filtered
but not sober. In this paper, we will introduce the concepts of
quasicontinuous element and down-linear element in dcpos. With these concepts
we identify some classes of $C_{\sigma}$-unique dcpos, that include all
quasicontinuous dcpos as well as Johnstone’s and Kou’s examples. The full
characterization of $C_{\sigma}$-unique dcpos is still open.

##  2\. Preliminaries

For any subset $A$ of a poset $P$, let
$\uparrow\\!\\!A=\\{x\in\\!P:y\leq\\!x\mbox{ for some }y\in\\!\\!A\\}$ and
$\downarrow\\!\\!A=\\{x\in\\!P:x\leq\\!y\mbox{ for some }y\in\\!A\\}$. A
subset $A$ is called an upper set if $A=\uparrow\\!\\!A$, and a lower set if
$A=\downarrow\\!\\!A$. A subset $U$ of a poset $P$ is Scott open if (i)
$U=\uparrow\\!U$ and (ii) for any directed subset $D$, $\bigvee D\in U$
implies $D\cap U\not=\emptyset$, whenever $\bigvee D$ exists. All Scott open
sets of a poset $P$ form a topology on $P$, denoted by $\sigma(P)$ and called
the Scott topology on $P$. The complements of Scott open sets are called Scott
closed sets. Clearly, a subset $A$ is Scott closed iff (i)
$A=\downarrow\\!\\!A$ and (ii) for any directed subset $D\subseteq A$,
$\bigvee D\in A$ whenever $\bigvee\\!D$ exists. The set of all Scott closed
sets of $P$ will be denoted by $C_{\sigma}(P)$. The space $(P,\sigma(P))$ is
denoted by $\Sigma\\!P$.

A poset $P$ is directed complete if its every directed subset has a supremum.
A directed complete poset is briefly called a dcpo.

A subset $A$ of a topological space is irreducible if whenever $A\subseteq
F_{1}\cup F_{2}$ with $F_{1}$ and $F_{2}$ closed, then $A\subseteq F_{1}$ or
$A\subseteq F_{2}$ holds. The set of all nonempty irreducible closed subsets
of space $X$ will be denoted by $Irr(X)$.

For any $T_{0}$ topological space $(X,\tau)$, the specialization order
$\leq_{\tau}$ on $X$ is defined by $x\leq_{\tau}y$ iff $x\in cl(\\{y\\})$
where $``cl(\cdot)"$ means taking closure.

######  Remark .

  1. (1)

For any topological space $X$, $(Irr(X),\subseteq)$ is a dcpo. If
$\mathcal{D}$ is a directed subset of $Irr(X)$, the supremum of $\mathcal{D}$
in $(Irr(X),\subseteq)$ equals $cl(\bigcup\mathcal{D})$ (the closure of
$\bigcup\mathcal{D}$), which is the same as the supremum of $\mathcal{D}$ in
the complete lattice of all closed sets of $X$.

  2. (2)

For any $x\in X$, $cl(\\{x\\})\in Irr(X)$. A $T_{0}$ space $X$ is called sober
if $Irr(X)=\\{cl(\\{x\\}):x\in X\\}$, that is, every nonempty irreducible
closed set is the closure of a point.

  3. (3)

Assume that $(X,\tau)$ and $(Y,\eta)$ are topological spaces such that the
open set lattices ${(\tau,\subseteq)}$ and $(\eta,\subseteq)$ of $X$ and $Y$
are isomorphic, then the closed set lattice $(C(X),\subseteq)$ of $X$ and the
closed set lattice $(C(Y),\subseteq)$ of $Y$ are also isomorphic (they are
dual to the corresponding open set lattices). Since irreducibility is a
lattice-intrinsic property of elements, it follows that the posets
$(Irr(X),\subseteq)$ and $(Irr(Y),\subseteq)$ are isomorphic.

For a $T_{0}$ space $X$, a sobrification of $X$ is a sober space $Y$ together
with a continuous mapping $\eta_{X}:X\longrightarrow Y$, such that for any
continuous mapping $f:X\longrightarrow Z$ with $Z$ sober, there is a unique
continuous mapping $\hat{f}:Y\longrightarrow Z$ satisfying
$f=\hat{f}\circ\eta_{X}$. The sobrification of a $T_{0}$ space is unique up to
homeomorphism. Clearly if a space $X$ is sober, then $X$ is homeomorphic to
any sobrification of $X$.

######  Remark .

The following facts about sober spaces and sobrifications are well-known.

  1. (1)

The set $Irr(X)$ of all nonempty closed irreducible sets of a $T_{0}$ space
$X$ equipped with the hull-kernel topology is a sobrification of $X$, where
the mapping $\eta_{X}:X\longrightarrow Irr(X)$ is defined by
$\eta_{X}(x)=cl(\\{x\\})$ for all $x\in X$. The closed sets of the hull-kernel
topology consists of all sets of the form $h(A)=\\{F\in Irr(X):F\subseteq
A\\}$ ($A$ is a closed set of $X$). So the sobrification of a space $X$ is
totally determined by the lattice $C(X)$. (See Exercise V-4.9 of [2] for
details, where the topology was given by means of open sets).

  2. (2)

If $X$ and $Y$ are both sober spaces and the closed set lattices $C(X)$ and
$C(Y)$ are homomorphic, then the sobrification of $X$ and that of $Y$ are
homeomorphic. Hence $X$ and $Y$ are homeomorphic.

  3. (3)

From (1) and (2), we easily deduce that if $Y$ is a sober space, then $Y$ is a
sobrification of a $T_{0}$ space $X$ iff the closed set lattices $C(X)$ and
$C(Y)$ are isomorphic (equivalently, the open set lattice of $Y$ is isomorphic
to that of $X$).

A $T_{0}$ space $X$ will be called Scott sobrifiable if there is a dcpo $P$
such that the Scott space $\Sigma\\!P$ is a sobrification of $X$.

For any $T_{0}$ space $(X,\tau)$, let $\leq_{\tau}$ be the specialization
order on $X$ ($x\leq_{\tau}y$ iff $x\in cl(\\{y\\})$). It is well-known that
the specialization order on the Scott space $\Sigma\\!P$ of a poset $P$
coincides with the original order on $P$. Thus a $T_{0}$ space $(X,\tau)$ is
homeomorphic to $\Sigma\,P$ for some poset $P$ iff $(X,\tau)$ is homeomorphic
to the Scott space $\Sigma(X,\leq_{\tau})$ of the poset $(X,\leq_{\tau})$. The
specialization order on the space $Irr(X)$ (with the hull-kernel topology)
equals the inclusion order of sets. From the above, we can easily deduce the
following fact.

######  Remark .

A $T_{0}$ space $(X,\tau)$ is Scott sobrifiable iff for any Scott closed set
$\mathcal{F}$ of the dcpo $Irr(X)$, there is a closed set $A$ of $X$ such that
$\mathcal{F}=h(A)$, where $h(A)=\\{F\in Irr(X):F\subseteq A\\}$.

A topological space $(X,\tau)$ is called a d-space (or monotone convergence
space) if (i) $X$ is $T_{0}$, (ii) the poset $(X,\leq_{\tau})$ is a dcpo, and
(iii) for any directed subset $D\subseteq(X,\leq_{\tau})$, $D$ converges (as a
net) to $\bigvee\\!D$.

######  Remark .

  1. (1)

Every sober space is a d-space.

  2. (2)

Every Scott space $\Sigma P$ of a dcpo $P$ is a d-space.

  3. (3)

If $(X,\tau)$ is a d-space, then every closed set $F$ of $X$ is a Scott closed
set of the dcpo $(X,\leq_{\tau})$.

######  Lemma 1.

Let $(X,\tau)$ be a d-space. If $\\{x_{i}:i\in I\\}$ is a directed subset of
$(X,\leq_{\tau})$, then the supremum $\sup\\{cl(\\{x_{i}\\}):i\in I\\}$ of
$\\{cl(\\{x_{i}\\}):i\in I\\}$ in $Irr(X)$ equals $cl(\\{x\\})$, where
$x=\bigvee\\{x_{i}:i\in I\\}$.

For more about dcpos, Scott topology and related topics we refer the reader to
[2] and [3].

##  3\. Main results

In this section, we identify some classes of $C_{\sigma}$-unique dcpos, using
irreducible sets, down-linear elements, quasicontinuous elements and the M
property, respectively.

A $T_{0}$ space is called bounded-sober if every nonempty upper bounded (with
respect to the specialization order on $X$) closed irreducible subset of the
space is the closure of a point [14]. Every sober space is bounded-sober, the
converse implication is not true. If $X$ is a $T_{0}$ space such that every
irreducible closed proper  subset is the closure of an element, then $X$ is
bounded-sober. In the following, a dcpo whose Scott topology is sober
(bounded-sober) will be simply called a sober (bounded-sober) dcpo.

######  Lemma 2.

For a bounded-sober dcpo $P$, $\Sigma\\!P$ is Scott sobrifiable if and only if
$P$ is sober.

######  Proof 3.1.

We only need to check that if $\Sigma P$ is not sober, then it is not Scott
sobrifiable. Since $\Sigma\\!P$ is not sober, there is a nonempty irreducible
closed set $F$ such that $F$ is not the closure of any point. From the
assumption that $\Sigma P$ is bounded-sober, one can verify that the set
$\mathcal{F}=\downarrow_{Irr(\Sigma P)}\\{cl(\\{x\\}):x\in F\\}$ consists
precisely of the elements $cl(\\{x\\})$ ($x\in F$), and is a Scott closed set
of $Irr(\Sigma P)$. But any closed set $B$ of $\Sigma P$ containing all
$cl(\\{x\\})(x\in F)$ must contain $F$, thus $h(B)\not=\mathcal{F}$. By Remark
2, $\Sigma P$ is not Scott sobrifiable.

In the following, we shall write $P\cong Q$ if the two posets $P$ and $Q$ are
isomorphic.

######  Theorem 3.

Let $P$ be a sober dcpo. For any bounded-sober dcpo $Q$, if
$C_{\sigma}(P)\cong C_{\sigma}(Q)$ then $P\cong Q$.

######  Proof 3.2.

Let $Q$ be a bounded-sober dcpo such that $C_{\sigma}(P)\cong C_{\sigma}(Q)$.
Then, by Remark 2 (3), $\Sigma\\!P$ is a sobrification of $\Sigma Q$. Thus
$\Sigma Q$ is Scott sobrifiable. By Lemma 2, $\Sigma Q$ is also sober.
Therefore, by Remark 2 (2), $\Sigma P$ and $\Sigma Q$ are homeomorphic, which
then implies $P\cong Q$.

######  Definition 4.

An element $a$ of a poset $P$ is called down-linear if the subposet
$\downarrow\\!\\!a=\\{x\in P:x\leq a\\}$ is a chain (for any
$x_{1},x_{2}\in\downarrow\\!\\!a$, it holds that either $x_{1}\leq x_{2}$ or
$x_{2}\leq x_{1}$).

######  Lemma 5.

Let $(X,\tau)$ be a d-space.

  1. (1)

If $F\in Irr(X)$ is a down-linear element of the poset $Irr(X)$, then there
exists an $x\in X$ such that $F=cl(\\{x\\})$.

  2. (2)

If $F\in Irr(X)$ equals the supremum of a directed set of down-linear elements
of $Irr(X)$, then $F=cl(\\{x\\})$ for some $x\in X$.

######  Proof 3.3.

  1. (1)

First, the set $\\{cl(\\{x\\}):x\in F\\}$ is a subset of $\downarrow\\!\\!F$
in $Irr(X)$, so it is a chain. Thus $\\{x:x\in F\\}$ is a chain of
$(X,\leq_{\tau})$. Since $X$ is a d-space, $\widehat{x}=\sup\\{x:x\in F\\}$
exists. Then, noticing that $F$ is closed, we have $\widehat{x}\in F$ by
Remark 2 (3). Then $F\subseteq cl(\\{\widehat{x}\\})\subseteq F$, implying
$cl(\\{\widehat{x}\\})=F$.

  2. (2)

Let $F$ be the supremum of a directed set of down-linear irreducible closed
sets in $Irr(X)$. Then by (1), $F=\sup\\{cl(\\{x_{i}\\}):i\in I\\}$ in
$Irr(X)$, where $\\{cl(\\{x_{i}\\}):i\in I\\}$ is a directed family. Thus,
$\\{x_{i}:i\in I\\}$ is a directed set of $(X,\leq_{\tau})$. Again, as $X$ is
a d-space, $x=\sup\\{x_{i}:i\in I\\}$ exists. By Lemma 1,
$cl(\\{x\\})=\sup\\{cl(\\{x_{i}\\}):i\in I\\}=F$.

In the following, for a dcpo $P$, we shall use $Irr_{\sigma}(P)$ to denote the
dcpo of all nonempty irreducible Scott closed subsets of $P$. Without
specification, irreducible sets of a poset mean the irreducible sets with
respect to the Scott topology.

######  Theorem 6.

Let $P$ be a dcpo satisfying the following condition

(DL-sup):

for any proper irreducible Scott closed set $F$, $F$ is either a down-linear
element of $Irr_{\sigma}(P)$ or it is the supremum of a directed set of down-
linear elements of $Irr_{\sigma}(P)$.

Then $P$ is $C_{\sigma}$-unique.

######  Proof 3.4.

Let dcpo $P$ satisfy the above condition (DL-sup) and $Q$ be a dcpo such that
$C_{\sigma}(P)\cong~{}C_{\sigma}(Q)$.

  1. (1)

By Lemma 5, if $F\in Irr_{\sigma}(P)$ and $F\not=P$, then $F=cl(\\{x\\})$ for
some point.

  2. (2)

Since $C_{\sigma}(P)\cong C_{\sigma}(Q)$, $Q$ also satisfies condition (DL-
sup). So every nonempty closed irreducible proper subset of $\Sigma\,Q$ is the
closure of a point.

  3. (3)

Let $F$ be a nonempty irreducible closed subset of $P$ with an upper bound
$a$. If $F\not=P$, then $F$ is the closure of some point by (1). Otherwise
$F=P$, thus $a\in P$ is the largest element in $P$, hence
$F=P=\downarrow\\!a=cl(\\{a\\})$. Therefore $\Sigma P$ is bounded-sober.
Similarly $\Sigma Q$ is bounded-sober. If either $\Sigma P$ or $\Sigma Q$ is
sober, then by Theorem 3, $P\cong Q$. Assume now that neither $\Sigma\\!P$ nor
$\Sigma\\!Q$ is sober. Then there is a nonempty irreducible closed set $F$ of
$P$, which is not the closure of a singleton set. But by (1) and (2), $F$
cannot be a proper subset, so $F=P$. Thus $P$ is an irreducible closed set
which does not equal to the closure of any singleton set. Similarly, $Q$ is an
irreducible closed set which is not the closure of any singleton set. Note
that in this case, $P$ and $Q$ are the top elements of $Irr_{\sigma}(P)$ and
$Irr_{\sigma}(Q)$, respectively. Thus $Q\cong\\{cl(\\{y\\}):y\in Q\\}\cong
Irr_{\sigma}(Q)-\\{Q\\}\cong Irr_{\sigma}(P)-\\{P\\}\cong\\{cl(\\{x\\}):x\in
P\\}\cong P$, as desired.

######  Example 7.

In [9], Johnstone constructed the first non-sober dcpo as
$X=\mathbb{N}\times(\mathbb{N}\cup\\{\infty\\})$ with partial order defined by

$(m,n)\leq(m^{\prime},n^{\prime})\Leftrightarrow\mbox{either
}m=m^{\prime}\mbox{ and }n\leq n^{\prime}\\\ \mbox{or }n^{\prime}=\infty\mbox{
and }n\leq m^{\prime}.$

Then

  1. (a)

$(X,\leq)$ is a dcpo, $X$ is irreducible and $X\not=cl(\\{x\\})$ for any $x\in
X$.

  2. (b)

If $F$ is a proper irreducible Scott closed set of $X$, then
$F=\downarrow\\!\\!(m,n)$ for some $(m,n)\in X$.

  3. (c)

If $n\not=\infty$, $\downarrow\\!\\!(m,n)$ is a down-linear element of
$Irr_{\sigma}(X)$. If $n=\infty$, then $\downarrow(m,n)$ is the supremum of
the chain $\\{\downarrow\\!\\!(m,k):k\not=\infty\\}$ whose members are down-
linear.

Hence by Theorem 6, we deduce that dcpo
$X=\mathbb{N}\times(\mathbb{N}\cup\\{\infty\\})$ is $C_{\sigma}$-unique. Thus
an $C_{\sigma}$-unique dcpo need not be sober.

Next, we provide a class of $C_{\sigma}$-unique dcpos via quasicontinuous
elements.

######  Remark 8 (cf. [11]).

Let $A$ be a nonempty Scott closed set of a dcpo $P$. Then

  1. (i)

$A$ is a dcpo.

  2. (ii)

For any subset $B\subseteq A$, $B$ is a Scott closed set of dcpo $A$ iff it is
a Scott closed set of $P$. Thus
$C_{\sigma}(A)=\downarrow_{C_{\sigma}(P)}A=\\{B\in C_{\sigma}(P):B\subseteq
A\\}$.

A finite subset $F$ of a dcpo $P$ is way-below an element $a\in P$, denoted by
$F\ll a$, if for any directed subset $D\subseteq P$, $a\leq\bigvee D$ implies
$D\cap\\!\\!\uparrow\\!\\!F\not=\emptyset$. A dcpo $P$ is quasicontinuous if
for any $x\in P$, the family

$\mathit{fin}(x)=\\{F:F\mbox{ is finite and }F\ll x\\}$

is a directed family (for any $F_{1},F_{2}\in\mathit{fin}(x)$ there is
$F\in\mathit{fin}(x)$ such that $F\subseteq\uparrow F_{1}\cap{\uparrow
F_{2}}$) and for any $x\not\leq y$ there is $F\in\mathit{fin}(x)$ satisfying
$y\not\in\uparrow F$ (see Definition III-3.2 of [2]). Every continuous dcpo is
quasicontinuous.

Every quasicontinuous dcpo is sober (Proposition III-3.7 of [2]). A dcpo $P$
is quasicontinuous iff the Scott open set lattice $\sigma(P)$ of $P$ is
hypercontinuous (Theorem VII-3.9 of [2]). Assume that $P$ is a quasicontinuous
dcpo and $Q$ is a dcpo such that $C_{\sigma}(P)$ is isomorphic to
$C_{\sigma}(Q)$. Then $\sigma(P)$ (it is dual to $C_{\sigma}(P)$) is
isomorphic to $\sigma(Q)$(it is dual to $C_{\sigma}(Q)$, thus $\sigma(Q)$ is
also hypercontinuous, implying that $Q$ is quasicontinuous. Thus both
$\Sigma\\!P$ and $\Sigma\\!Q$ are sober spaces and they have isomorphic closed
set lattices, hence by Theorem 3, we have $P\cong Q$. From this we obtain the
following lemma.

######  Lemma 9.

Every quasicontinuous dcpo is $C_{\sigma}$-unique.

An element $x$ of a dcpo $P$ is called a quasicontinuous element if the sub-
dcpo $\downarrow\\!\\!x$ is a quasicontinuous dcpo.

######  Theorem 10.

Let $P$ be a dcpo. Then $P$ is $C_{\sigma}$-unique if it satisfies the
following two conditions:

  1. (1)

$\Sigma P$ is bounded sober;

  2. (2)

every element of $P$ is the supremum of a directed set of quasicontinuous
elements.

######  Proof 3.5.

Assume that $P$ is a dcpo satisfying the two conditions. Let $Q$ be a dcpo and
$F:C_{\sigma}(P)\longrightarrow C_{\sigma}(Q)$ be an isomorphism. Then $F$
restricts to an isomorphism $F:Irr_{\sigma}(P)\longrightarrow
Irr_{\sigma}(Q)$.

  1. (1)

Let $x\in P$ be a quasicontinuous element. Then $F(\downarrow\\!x)$ is in
$C_{\sigma}(Q)$ and, by Remark 8, $C_{\sigma}(\downarrow\\!x)=\\{B\in
C_{\sigma}(P):B\subseteq\downarrow\\!x\\}={\downarrow_{C_{\sigma}(P)}(\downarrow\\!x)}$
is isomorphic via $F$ to $\downarrow_{C_{\sigma}(Q)}F(\downarrow\\!x)=\\{E\in
C_{\sigma}(Q):E\subseteq
F(\downarrow\\!\\!x)\\}=C_{\sigma}(F(\downarrow\\!x))$ (all Scott closed sets
of $F(\downarrow\\!x)$). Since the dcpo $\downarrow\\!x$ is quasicontinuous,
it is $C_{\sigma}$-unique. Hence the dcpo $\downarrow\\!x$ is isomorphic to
the dcpo $F(\downarrow x)$, implying that there is a largest element in
$F(\downarrow\\!x)$, denoted by $f(x)$. Hence
$F(\downarrow\\!x)=\downarrow\\!f(x)$. It is easily observable that the
mapping $f$ is well defined on the set of quasicontinuous elements of $P$, and
for any two quasicontinuous elements $x_{1},x_{2}\in P$, $f(x_{1})\leq
f(x_{2})$ iff $x_{1}\leq x_{2}$.

  2. (2)

If $x\in P$ is the supremum of a directed set $\\{x_{i}:i\in I\\}$ of
quasicontinuous elements $x_{i}$, then

$\begin{array}[]{lll}F(\downarrow\\!x)&=&F(\sup_{Irr_{\sigma}(P)}\\{\downarrow\\!x_{i}:i\in
I\\})\\\ &=&\sup_{Irr_{\sigma}(Q)}\\{F(\downarrow\\!x_{i}):i\in I\\}\\\
&=&\sup_{Irr_{\sigma}(Q)}\\{\downarrow\\!f(x_{i}):i\in I\\}\\\
&=&\downarrow\\!y_{x},\end{array}$

where $y_{x}=\sup_{Q}\\{f(x_{i}):i\in I\\}$ and $f(x_{i})$ is the element in
$Q$ defined for quasicontinuous elements $x_{i}$ in (1). Let $f(x)=y_{x}$
again.

Thus we have a monotone mapping $f:P\longrightarrow Q$. Following that $F$ is
an isomorphism, we have that $f(x_{1})\geq f(x_{2})$ iff $x_{1}\geq x_{2}$. It
remains to show that $f$ is surjective.

  3. (3)

If $y\in\downarrow\\!\\!f(P)$, then
$\downarrow\\!\\!y\subseteq\downarrow\\!\\!f(x)=F(\downarrow\\!\\!x)$ for some
$x\in P$. Since $F$ restricts to an isomorphism between the dcpos
$Irr_{\sigma}(P)$ and $Irr_{\sigma}(Q)$, there is $H\in Irr_{\sigma}(P)$ such
that $H\subseteq\downarrow\\!x$ and $F(H)=\downarrow\\!y$. But $P$ is bounded-
sober, so $H=\downarrow\\!x^{\prime}$ for some $x^{\prime}\in P$. It follows
that $y=f(x^{\prime})$, implying $y\in f(P)$. Therefore $f(P)$ is a lower set
of $Q$. Also clearly $f(P)$ is closed under sups of directed set, so it is a
Scott closed subset of $Q$.

  4. (4)

Since $F$ is an isomorphism between the lattices $C_{\sigma}(P)$ and
$C_{\sigma}(Q)$, $P$ and $Q$ are the top elements in the respective lattices,
we have that $Q=F(P)=F(\sup_{C_{\sigma}(P)}\\{\downarrow\\!x:x\in
P\\})=\sup_{C_{\sigma}(Q)}\\{F(\downarrow\\!x):x\in
P\\}=\sup_{C_{\sigma}(Q)}\\{\downarrow\\!f(x):x\in P\\}$.

For each $x\in P$, $\downarrow\\!f(x)\subseteq f(P)$ because $f(P)$ is a Scott
closed set of $Q$, it holds then that
$sup_{C_{\sigma}(Q)}\\{\downarrow\\!f(x):x\in P\\}\subseteq f(P)$. Therefore
$Q\subseteq f(P)$, which implies $Q=f(P)$. Hence $f$ is also surjective. The
proof is thus completed.

If $x\in P$ is a down-linear element of a dcpo $P$, then $\downarrow\\!x$ is a
chain, so it is continuous (hence quasicontinuous).

######  Corollary 11.

If $P$ is a dcpo satisfying the following conditions, then $P$ is
$C_{\sigma}$-unique:

  1. (1)

$P$ is bounded-sober.

  2. (2)

every element $a\in P$ is the supremum of a directed set of down-linear
elements.

######  Example 12.

In order to answer the question whether every well-filtered dcpo is sober
posed by Heckmann [5], Kou [10] constructed another non-sober dcpo $P$ as
follows:

Let $X=\\{x\in\mathbb{R}:0<x\leq 1\\}$,
$P_{0}=\\{(k,a,b)\in\mathbb{R}:0<k<1,0<b\leq a\leq 1\\}$ and

$P=X\cup P_{0}.$

Define the partial order $\sqsubseteq$ on $P$ as follows:

  1. (i)

for $x_{1},x_{2}\in X$, $x_{1}\sqsubseteq x_{2}$ iff $x_{1}=x_{2}$;

  2. (ii)

$(k_{1},a_{1},b_{1})\sqsubseteq(k_{2},a_{2},b_{2})$ iff $k_{1}\leq
k_{2},a_{1}=a_{2}$ and $b_{1}=b_{2}$.

  3. (iii)

$(k,a,b)\sqsubseteq x$ iff $a=x$ or $kb\leq x<b$.

If $u=(h,a,b)\in P_{0}$, then $\downarrow\\!\\!u=\\{(k,a,b):k\leq h\\}$ is a
chain. If $u=x\in P_{0}$, then $u=\bigvee\\{(k,x,x):0<k<1\\}$, where each
$(k,x,x)$ is a down-linear element and $\\{(k,x,x):0<k<1\\}$ is a chain. Thus
$P$ satisfies (2) of Corollary 11.

Let $F$ be an irreducible nonempty Scott closed set of $P$ with an upper bound
$v$. If $v=(h,a,b)\in P_{0}$, then
$F\subseteq\downarrow\\!\\!(h,a,b)=\\{(k,a,b):k\leq h\\}$. Take
$m=\bigvee\\{k:(k,a,b)\in F\\}$. Then $F=\downarrow\\!\\!(m,a,b)$, is the
closure of point $(m,a,b)$.

Now assume that $F$ does not have an upper bound in $P_{0}$, then $v=x$ for
some $x\in P_{0}$. If $v\not\in F$, then due to the irreducibility of $F$,
there exist $a,b$ such that $F\subseteq\\{(k,a,b):0<k<1\\}$, which will imply
that $F$ has an upper bound of the form $(m,a,b)$, contradicting the
assumption. Therefore $v\in F$, implying that $F=\downarrow\\!\\!v$ (note that
$F=\downarrow\\!\\!F$ is a lower set) is the closure of point $v$. It thus
follows that $P$ satisfies (1) as well. By Corollary 11, $P$ is
$C_{\sigma}$-unique.

Next, we give another class of $C_{\sigma}$-unique dcpos. In [7], Ho and Zhao
introduced the following notions.

######  Definition 13.

Let $L$ be a poset and $x,y\in L$. The element $x$ is beneath $y$, denoted by
$x\prec y$, if for every nonempty Scott-closed set $S\subseteq L$ with
$\bigvee S$ existing, $y\leq\bigvee S$ implies $x\in S$. An element $x$ of $L$
is called C-compact if $x\prec x$. Let $\kappa(L)$ denote the set of all the
C-compact elements of $L$.

Let $P$ be a poset and $A\subseteq P$ finite. The set $mub(A)$ of minimal
upper bounds of $A$ is said to be complete, if for any upper bound $x$ of $A$,
there exists $y\in mub(A)$ such that $y\leq x$. A poset $P$ is said to satisfy
the property $m$, if for all finite sets $A\subseteq P$, $mub(A)$ is complete.
A poset $P$ is said to satisfy the property $M$, if $P$ satisfies the property
$m$ and for all finite set $A\subseteq P$, $mub(A)$ is finite.

######  Remark 14.

Let $L$ be a complete lattice and $a\in L$ be a C-compact element. If $x,y\in
L$ such that $a\leq x\vee y$, then
$a\leq\bigvee(\downarrow\\!x\cup\\!\downarrow\\!y)$ and
$\downarrow\\!x\cup\\!\downarrow\\!y$ is Scott closed, so
$a\in\downarrow\\!x\cup\\!\downarrow\\!y$, implying $a\leq x$ or $a\leq y$.
Thus $a$ is $\vee$-irreducible.

######  Corollary 15.

For any dcpo $P$, $\kappa(C_{\sigma}(P))\subseteq Irr_{\sigma}(P)$. That is,
all C-compact Scott closed sets are irreducible.

[[4]] Let $P$ be a dcpo. Then

  1. (1)

For all $x\in P$, $\downarrow\\!x\in\kappa(C_{\sigma}(P))$.

  2. (2)

If $P$ satisfies the property $M$, then $A\in\kappa(C_{\sigma}(P))$ iff
$A=\downarrow\\!x$ for some $x\in P$.

######  Theorem 16.

If $P$ is a dcpo satisfying the property $M$ and the condition (2) in
Corollary 11, then $P$ is $C_{\sigma}$-unique.

######  Proof 3.6.

Let $P$ be a dcpo satisfying the condition (2) in Corollary 11 and the
property $M$. Assume that $Q$ is a dcpo and there is an order isomorphism
$H:C_{\sigma}(P)\rightarrow C_{\sigma}(Q)$. Then the restrictions
$H:\kappa(C_{\sigma}(P))\rightarrow\kappa(C_{\sigma}(Q))$ and
$H:Irr_{\sigma}(P)\rightarrow Irr_{\sigma}(Q)$ are all order isomorphisms.

For all $q\in Q$, by Lemma 15(1), $\downarrow\\!q\in\kappa(C_{\sigma}(Q))$,
then $H^{-1}(\downarrow\\!q)={\downarrow\\!x_{q}}$ for a unique $x\in P$ by
Lemma 15(2). Now define a map $h^{\prime}:Q\to P$ such that
$h^{\prime}(q)=x_{q}$ iff $H^{-1}({\downarrow\\!q})={\downarrow\\!x_{q}}$. The
mapping $h^{\prime}$ is monomorphic and order preserving since $H^{-1}$ is.
Note that $\kappa(C_{\sigma}(Q))\cong\kappa(C_{\sigma}(P))\cong P$ is a dcpo.

Now let $x$ be any element of $P$.

  1. (i)

If $x$ is down-linear, then $H(\downarrow\\!\\!x)$ is a linear subset in $Q$
(if $y_{1},y_{2}\in H(\downarrow\\!\\!x)$, then
$h^{\prime}(y_{1}),h^{\prime}(y_{2})\in\downarrow\\!\\!x$), and is Scott
closed. The supremum $\sup_{Q}H(\downarrow\\!x)$ exists and is in
$H(\downarrow\\!x)$. Thus $H(\downarrow\\!x)={\downarrow\\!q_{x}}$ for some
$q_{x}\in Q$.

  2. (ii)

If $x$ is not down-linear, then $x$ is the supremum of a directed set $C$ of
down-linear elements. Since $H$ preserves sups in $\kappa(C_{\sigma}(P))$ and
$\kappa(C_{\sigma}(Q))$, we have that

$\begin{array}[]{lll}H(\downarrow\\!x)&=&H(\downarrow\\!\sup C)\\\
&=&H(\sup_{\kappa(C_{\sigma}(P))}\\{\downarrow\\!c:c\in C\\})\\\
&=&\sup_{\kappa(C_{\sigma}(Q))}\\{H(\downarrow\\!c):c\in C\\}\\\
&=&\downarrow\\!\sup_{Q}\\{q_{c}:\downarrow\\!q_{c}=H(\downarrow\\!c),c\in
C\\}\\\ &=&\downarrow\\!q_{x},\end{array}$

for some $q_{x}\in Q$.

By these facts, we defined a mapping $h:P\to Q$ such that $h(x)=q_{x}$ iff
$F(\downarrow\\!x)=\downarrow\\!q_{x}$. It is then easy to see that $h$ is
monomorphic and order preserving since $H$ is. In addition, it is easy to
verify that $h^{\prime}$ is the inverse of $h$, hence $h$ is an order
isomorphism between $P$ and $Q$, as desired.

Note that Kou’s and Johnstone’s examples of non-sober dcpos do not have the
property $M$.

##  4\. Remarks and some possible further work

We close the paper with some additional remarks and problems for further
exploration.

######  Remark 17.

  1. (1)

If $P$ is a $C_{\sigma}$-unique dcpo and $P^{*}$ is the dcpo obtained by
adding a top element to $P$, then one can show that $P^{*}$ is also
$C_{\sigma}$-unique. Let $X$ be the dcpo of Johnstone. Then $X^{*}$ is
$C_{\sigma}$-unique, but $X^{*}$ is not bounded sober ($X$ is an irreducible
Scott closed set of $X^{*}$ which is not the closure of any point of $X^{*}$).
Thus a $C_{\sigma}$-unique dcpo need not be bounded sober. So, bounded
sobriety is not a necessary condition for a dcpo to be $C_{\sigma}$-unique.

  2. (2)

Recently, Ho, Goubault-Larrecq, Jung and Xi [6] constructed a pair of non-
isomorphic dcpos having isomorphic Scott topologies, showing the existence of
non-$C_{\sigma}$-unique dcpos. Their counterexample also reveals that sobriety
is not a sufficient condition for a dcpo to be $C_{\sigma}$-unique.

In view of the above remarks, to identify larger classes of
$C_{\sigma}$-unique dcpos and formulate a full characterization of
$C_{\sigma}$-unique dcpos will be our future work.

Acknowledgement

  1. (1)

This work was supported by NIE AcRF Project (RI 3/16 ZDS), NSF of China
(11671008) and University Science Research Project of Jiangsu Province
(15KJD110006).

  2. (2)

Two anonymous referees have provided us with many helpful comments and
constructive suggestions for improvement. We are very grateful to them.

## References

  * [1] Drake, D. and Thron, W. J.: On the representation of of an abstract lattice as the family of closed sets of a topological space, Trans. Amer. Math. Soc. 120(1965), 57-71.
  * [2] Gierz, G. et al.: Continuous lattices and Domains, Encyclopedia of Mathematics and Its Applications, Vol.93, Cambridge University Press, 2003.
  * [3] Goubault-Larrecq, J.: Non-Hausdorff topology and Domain Theory: Selected Topics in Point-Set Topology, Vol. 22, Cambridge University Press, 2013.
  * [4] He, Q. Y. and Xu, L. S.: The C-algebraicity of Scott-closed lattices and its applications, Applied Mathematics: A Journal of Chinese Universities (Series A) 369-374. (in Chinese)
  * [5] Heckmann, R.: An upper power domain construction in terms of strongly compact sets, Lecture Notes in Computer Science 598, Springer-Verlag, 1992, 272-293.
  * [6] Ho, W. K., Goubault-Larrecq, J., Jung, A. and Xi, X. Y.: The Ho-Zhao problem, Logical Methods in Computer Science 14(2018), 1:1-19.
  * [7] Ho, W. K and Zhao, D. S.: Lattices of Scott-closed sets, Comment. Math. Univ. Carolin. 50(2009), 297-314.
  * [8] Isbell, J. R.: Completion of a construction of Johnstone, Proc. Amer. Math. Soc. 85(1982), 333-334.
  * [9] Johnstone, P.: Scott is not always sober, Lecture Notes in Math. 871, Springer, Berlin, 1981, 282-283.
  * [10] Kou, H.: Well-filtered dcpos need not be sober, In: Domains and Processes, Semantic Structures in Computation, 1(2001), 41-50.
  * [11] Lawson, J. D. and Xu, L. S.: Posets Having Continuous Intervals, Theoretical Computer Science 316(2004), 89-103.
  * [12] Thron, W. J.: Lattice-equivalence of topological spaces, Duke Math. J. 29(1962), 671-679.
  * [13] Thornton, M. C.: Topological spaces and lattices of lower semicontinuous functions, Trans. Amer. Math. Soc. 181(1973), 495-506.
  * [14] Zhao, D. S. and Fan, T. H.: dcpo-completion of posets, Theoretical Computer Science 411(2010), 2167-2173.

## Appendix A

In this part, for reader’s convenience we present some details of the proof of
a result, essentially due to Drake and Thron, on spaces which are uniquely
determined (among all $T_{0}$ spaces) by means of their closed set lattices.
(This part is requested by one of the referees). In [1], Drake and Thron
proved the following result.

######  Corollary 18.

Every representation family of a $\mathcal{C}$-lattice $(\Gamma,\geq)$ has
exactly one element iff every irreducible element of $\Gamma$ is strongly
irreducible.

Here a lattice is called a $\mathcal{C}$-lattice if it is isomorphic to the
lattice $C(X)$ of all closed sets of a topological space $X$. An element $a$
of a lattice $L$ is called irreducible (strongly irreducible) iff $a$ can not
be expressed as the supremum of a finite (arbitrary) number of elements of
$L$, which are strictly less than $a$.

By the definition of representation families of $\mathcal{C}$-lattices (see
page 58 of [1]) we deduce the following fact, which is equivalent to the above
Corollary 18:

######  Fact 19.

A $T_{0}$ topological space $X$ has the property that (for any $T_{0}$ space
$Y$) $C(X)$ isomorphic to $C(Y)$ implies $X$ is homeomorphic to $Y$ if and
only if every irreducible closed set in the space is strongly irreducible.

A space $X$ is called a $T_{D}$ space iff for any $x\in X$, the derived set
$d(\\{x\\})=cl(\\{x\\})-\\{x\\}$ is a closed set (see Definition 2.1 of [12]).
For example, every $T_{1}$ space is a $T_{D}$ space.

######  Fact 20.

A topological space is both sober and $T_{D}$ iff every irreducible closed set
in the space is strongly irreducible.

######  Proof A.1.

First note that for any $\\{A_{i}:i\in I\\}\subseteq C(X)$, the supremum
$\bigvee_{C(X)}\\{A_{i}:i\in I\\}$ of $A_{i}^{\prime}s$ in the lattice $C(X)$
equals $cl(\bigcup\\{A_{i}:i\in I\\})$.

Assume that the space $X$ is both sober and $T_{D}$. Let $F$ be an irreducible
element of $C(X)$. Then $F=cl(\\{x\\})$ for some $x\in X$ because $X$ is
sober. Let $F=\bigvee_{C(X)}\\{A_{i}:i\in I\\}$ holds, where $A_{i}\in
C(X)(i\in I)$. Then $\bigcup\\{A_{i}:i\in I\\}\subseteq F=cl(\\{x\\})$. Thus
$A_{i}\subseteq cl(\\{x\\})$ for each $i$. If $cl(\\{x\\})\not=A_{i}$ for
every $i$, then $A_{i}\subseteq cl(\\{x\\})-\\{x\\}$, therefore
$\bigcup\\{A_{i}:i\in I\\}\subseteq cl(\\{x\\})-\\{x\\}$. Since
$cl(\\{x\\})-\\{x\\}$ is closed, we have $F=cl(\bigcup\\{A_{i}:i\in
I\\})\subseteq cl(\\{x\\})-\\{x\\}$, which contradicts $F=cl(\\{x\\})$. Hence
$F=cl(\\{x\\})=A_{i}$ for $i$, showing that $F$ is strongly irreducible.

Now assume that every irreducible element of $C(X)$ is strongly irreducible.
Let $F$ be a non empty irreducible member of $C(X)$. Then
$F=\bigvee_{C(X)}\\{cl(\\{x\\}):x\in F\\}$, so $F=cl(\\{x\\})$ for some $x\in
F$ because $F$ is strongly irreducible. It follows that $X$ is sober. Now let
$x\in X$ be any element. Assume that $cl(\\{x\\})-\\{x\\}$ is not closed. Then
$cl(\\{x\\})-\\{x\\}$ is a proper subset of $cl(cl(\\{x\\})-\\{x\\})$. But
trivially $cl(cl(\\{x\\})-\\{x\\})\subseteq cl(\\{x\\})$, thus
$cl(cl(\\{x\\})-\\{x\\})=cl(\\{x\\})$. Thus
$cl(\\{x\\})=\bigvee_{C(X)}\\{cl(\\{y\\}):y\in cl(\\{x\\})-\\{x\\}\\}$. Since
$cl(\\{x\\})$ is irreducible, it is strongly irreducible by the assumption, we
have $cl(\\{x\\})=cl(\\{y\\})$ for some $y\in cl(\\{x\\})-\\{x\\}$, which is
not possible because $X$ is $T_{0}$. Therefore $cl(\\{x\\})-\\{x\\}$ must be
closed. Hence $X$ is $T_{D}$.

From Fact 19 and Fact 20 we derive the following result, first explicitly
stated in [13] (page 504 line 11-13) with no proof (where sober spaces are
called pc spaces).

######  Fact 21.

A space $X$ has the property that (for any $T_{0}$ space $Y$) $C(X)$
isomorphic to $C(Y)$ implies $X$ is homeomorphic to $Y$ iff $X$ is both sober
and $T_{D}$.