# Continuity in Partially Ordered Sets

Authors: Venu G. Menon
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

The notion of a continuous domain is generalized to include posets
which are not dcpos and in which the set of elements way below an
element is not necessarily directed. We show that several of the pleasing algebraic
and topological properties of domains carry over to this setting.


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## Body

## 1. Introduction

Continuous  lattices and their generalizations, continuous domains, have been
studied for more than three decades. Continuous lattices are complete lattices
where each element is the supremum of elements way below it. For a poset to be
a continuous domain, it needs to have sups of only directed sets in addition to
the following two conditions: (i) each element is the sup of elements way below
it, and (ii) for each element, the set of elements way below it is a directed
set. A continuous poset [1] is any poset in which the conditions (i) and (ii)
are satisfied. In a complete lattice, in fact in any sup-semilattice, the
condition (ii) above is automatically satisfied. In [2], the authors have
studied lattices which are not complete but satisfy the conditions (i) and (ii)
above. The purpose of this paper is to study posets which need not be dcpos or
lattices but which satisfy the condition that each element is the sup of
elements way below it. The exact definition will be given in Section 2.Here we recall some basic definitions and terminology
from domain theory; more details can be found in [1]. For x,y∈P, a partially
ordered set, we say that x is way below y (written x≪y), if whenever y≤supD, for a directed
set D, there exists d∈D such that x≤d. A continuous poset  is a partially ordered
set P in which the
following two conditions are satisfied. (i) For each x∈P, x=sup{y:y≪x}; and (ii) for each x∈P, {y:y≪x} is a directed
set. A continuous poset in which every directed set has a least upper bound
(such posets are called dcpos) is called a continuous domain. A continuous
poset which is also a complete lattice is called a continuous lattice. In a
complete lattice, since the second condition above is automatically satisfied,
one needs to verify only the first condition. In any continuous poset P, the way below
relation has the interpolation property, that is, for x,y∈P with x≪y, there exists an
element z∈P, such that x≪z≪y. For more
information about continuous lattices, and continuous domains, the standard
reference is [1]. A subset S of a poset P is called up-complete  if for any directed subset D of S, for which supD exists in P,  supD is contained in S. We use the
following notations: ↓A={x:x≤a for some a∈A};
⇓A={x:x≪a for some a∈A};  and ⇓x=⇓{x}. A subset J of a poset P is called a
lower set if and only if J=↓J. Recall that a
subset U of a poset P is Scott open
if and only if U is an upper set
satisfying the property that if the supremum of any directed set D is in U, then D itself
intersects U. The lower
topology on a poset P has subbasic
closed sets of the form ↑x,x∈P. The join of the
lower topology and the Scott topology on P is the Lawson
topology and is denoted by λ(P).


## 2. C-Posets

The following simple but crucial result is the motivation for our definition of a C-poset.Theorem 2.1.
LetL be a continuous poset. Then for each x,y∈L,
with x≰y,
there exists u∈L
and an up-complete lower set
J⊆P
such that u≰y,
x∉J,
and ↑u∪J=L.Proof.
Suppose thatL is a continuous
poset. Let x,y∈L with x≰y. Then ∃u≪x with u≰y. By the interpolation
property, ∃v∈L such that u≪v≪x. Clearly v≰y. Define J=L∖⇑v. If D is a directed
subset of J, then v is not way
below any element of D. If v≪supD, then there
exists w such that v≪w≪supD. This implies
that ∃d∈D such that v≪w≤d, a
contradiction. This shows that the supremum of D belongs to J, and thus J is up-complete.
It is immediate that u≰y,
x∉J, and ↑u∪J=L.Definition 2.2.
A partially ordered setP is called a C-poset  if for any x,y∈P, with x≰y, ∃u≰y and an
up-complete lower set J such that x∉J, and ↑u∪J=P.Example 2.3.
(i) IfS is an infinite
set and if P is the poset of
all finite and cofinite subsets of S, then P is a C-poset.
Suppose that A, B are in P such that A⊆B. Let x∈A such that x∉B. If we define J={A:A∈P,x∉A}, and u={x}, then this J and u satisfy the
conditions in the definition of a C-poset.
(ii) LetX be a T0 space for which
the lattice of open sets is a continuous lattice. Here X is endowed with
the specialization order. Let L be a continuous domain with a least element
endowed with the Scott topology. Let P=[X→L] be the set of
all Scott continuous functions. It can be shown that P is a C-poset
but not a continuous poset, see [1, page 200].
(iii) LetPi,i∈I be a collection
of C-posets with 0 and 1. Then the product of Pis is a C-poset. If (xi)≰(yi), then there
exists j∈I such that xj≰yj.   Since Pj is a C-poset,
there exists a∈Pj such that a≰yj and an
up-complete lower set K of Pj such that xj∈K and Pj=↑a∪K. Define u=(ui), where ui=0 if i≠j and uj=a. Also define J=↓S, where S={(xi):xi=1 if i≠j and xj∈K}. It is easy to
see that the above u and J satisfy the
conditions in the definition of a C-poset.The next theorem establishes that C-posets do satisfy
the approximation property, that is, each element is the supremum of elements
way below it.Theorem 2.4.
IfP is a C-poset, then for each x∈P, x=sup{z:z≪x}=sup⇓x.Proof.
We need to prove this only forx=0. If x=0, then there exists y∈P such that  x≰y. Therefore, by
the definition of a C-poset, there exists u≰y, and an
up-complete lower set J such that x∈J, and ↑u∪J=P. We will show
that u≪x. Suppose that D is a directed
set in P such that  supD≥x. If D∩↑u=∅, then D⊆J and since J is up-complete, supD∈J. Since J is a lower set,
this forces x∈J, a
contradiction. This shows that u≪x. Thus the set {z:z≪x} is nonempty,
and x is an upper
bound of it. Let w be any upper
bound of that set. We will show that x≰w. Indeed, if x≰w, then ∃u′≰w and an
up-complete lower set J′ such that x∉J′, and ↑u′∪J′=P. But then u′≪x, which
contradicts the assumption that w is an upper
bound of the set {z:z≪x}. Therefore x≰w. This completes
the proof of the theorem.Corollary 2.5.
If a
C-poset is a sup-semilattice, then it is a continuous poset; if a C-poset is a
complete lattice, then it is a continuous lattice.Theorem 2.6.
LetP be a C-poset, and let {xi,k:i∈I,k∈K(i)} be a nonempty
family of elements of P such that {xi,k:k∈K(i)} is a directed
set for all i∈I. Then the following identity holds whenever the specified sups and infs exist:
(2.1)⋀i∈I⋁K(i)xi,k=⋁f∈M⋀i∈Ixi,f(i), where M is the set of
all choice functions f:I→⋃i∈IK(i) with f(i)∈K(i) for all iProof.
Since the right-hand side is always less than or
equal to the left-hand side, we need to prove only the reverse inequality. By
Theorem 2.2, it is enough to show that wheneverx≪⋀i∈I⋁K(i)xi,k, we   have x≪⋁f∈M⋀i∈Ixi,f(i).
Suppose thatx≪⋀i∈I⋁K(i)xi,k. Then, x≪⋁K(i)xi,k, for all i∈I. Therefore, by
the definition of the way below relation, we can choose a g(i)∈K(i) with x≰xi,g(i) for all i∈I. This implies
that x≰⋀i∈Ixi,g(i), and hence x is less than or
equal to the right hand side of the identity. This completes the proof of the
theorem.Letg:P→Q, and let d:Q→P be monotone
functions between posets. The pair (g,d) is called a
Galois connection if for x∈P, and y∈Q,
g(x)≥y if and only if x≥d(y). Here g is called an
upper adjoint, and d is called a
lower adjoint. A monotone function g:P→Q between posets
is called an upper adjoint if there exists a (necessarily unique) monotone
function d:Q→P such that (g,d) is a Galois
connection. Basic properties of Galois connections can be found in [1].Definition 2.7.
A function from a C-poset to another C-poset is
called ahomomorphism  if it is an
upper adjoint which preserves sups of directed sets. A subposet S of a C-poset P is called a subalgebra  if the inclusion map i:S→P is  a homomorphism.Theorem 2.8.
(a) LetP be a C-poset
and let Q be any poset.
If f:P→Q is a surjective
homomorphism, then Q is a C-poset.
(b) IfP is a C-poset and if S is a subalgebra
of P, then S is a C-poset.Proof.
(a) Suppose thatg is the lower
adjoint of f. Let y2≰y1 in Q. Pick x1∈P with f(x1)=y1. Since g is a lower
adjoint of f, y2≰f(x1) implies g(y2)≰x1. Then there
exist u∈P and an
up-complete lower set J such that u≰x1, g(y2)∈J, and P=↑u∪J. Define I:=g−1(J), and v:=f(u). Since the lower
adjoint g preserves sups, I is up-complete,
and y2∈I. If x∈J, then f(x)≰f(x) implies gf(x)≤x, which implies gf(x)∈J. Thus f(x)∈g−1(J)=I. If x∈J, then u≤x which implies f(u)≤f(x). Thus ↑f(u)∪I=Q. This completes
the proof that Q is a C-poset.
(b) Leti:S→P be the
inclusion map and let d be its lower
adjoint. Let x,y∈S such that x≰y. Then there
exists u∈P, and an
up-complete lower set J⊆P, such that u≰y, x∈J, and P=↑u∪J. Since u≰y=i(y), d(u)≰y. Define I:=i−1(J). Since i preserves sups
of directed sets, I is an
up-complete lower set, and x∈I. For t∈S, t∈J, implies i(t)=t∈J, which implies
that u≰t. Thus d(u)≰d(t)=t. Therefore, ↑Sd(u)∪I=S. This completes
the proof that S is a
subalgebra.Notice that the proof of (a) above did not use the
assumption thatf preserves the
sups of directed sets.Definition 2.9.
LetP be a C-poset. A
subposet B of P is called a
basis of P if given any x,y∈P with x≰y, there exist b∈B and an
up-complete lower setJ⊆P such that b≰y, x∉J, and ↑b∪J=P.Theorem 2.10.
IfB is a basis of a
C-poset P,
then for each x∈P,  x=sup{b:b∈B,b≪x}.Proof.
We need to prove this only forx=0. If x=0, then there
exists y∈P such that x≰y. Then, by the
definition of the basis, there exists b∈B such that b≰y, and an
up-complete lower set J such that x∉J, and ↑b∪J=P. It follows
immediately that b≪x. Thus, {b:b∈B,b≪x} is nonempty and x is an upper
bound of it. Suppose that v is any upper
bound of the same set. If x≰v, then there
exist b′∈B and an
up-complete lower set J′ such that x∉J′, and ↑u′∪J′=P. As shown above b′≪x which
contradicts the assumption that v is an upper
bound of {b:b∈B,b≪x}. Thus x≤u. This completes the proof.Proposition 2.11.
(1) LetP,Q be C-posets
with B as a basis of P. If g:P→Q is a surjective
homomorphism, then g(B) is a basis of Q.
(2) IfS is a subalgebra
of a C-poset P, and if B is a basis of P, then d(B) is a basis of S, where d is the lower
adjoint of the inclusion map.Proof.
(1) Letd be the lower
adjoint of g. Suppose that y1,y2∈Q such that y2≰y1. Let x1∈P, such that g(x1)=y1. Since y2≰g(x1),d(y2)≰x1. Since B is a basis of
the C-poset P, there exist b∈B and an
up-complete lower set J⊆P such that b≰x1,d(y2)∈J, and ↑b∪J=P. Let I=d−1(J) and c=f(b). Since the lower
adjoint preserves sups, I is an
up-complete lower set, and y2∉I. If x∈J,
f(x)≤f(x) implies df(x)≰x. This implies
that df(x)∈J, and hence f(x)∈d−1(J)=I. If x∉J, then b≤x, which implies f(b)=c≤f(x). Thus ↑c∪I=Q. This completes
the proof.The proof of (2) is similar and hence omitted.Though algebraic lattices were studied for several decades
before the introduction of continuous lattices, algebraic lattices, and their
generalizations, algebraic domains, have played an important role in domain
theory. Algebraic domains are continuous domains in which every element is the
sup of compact elements (an elementx is compact if x≪x) below it, in
which the set of compact elements below every element is a directed set. We
generalize this notion to posets which are not necessarily dcpos in which the
set of compact elements below any element is not necessarily directed. The
following result is the motivation for our definition.Theorem 2.12.
IfL is an algebraic
poset, then for x,y∈L, with x≰y, there exist u∈L and an
up-complete lower set J⊆L, such that u≰y,x∉J,
↑u∪J=L, and ↑u∩J=∅.Definition 2.13.
A C-poset is called anA-poset  if the condition of Theorem 2.12 is satisfied.Proposition 2.14.
A C-posetP is an A-poset
if and only if the set of all compact elements of P is a basis of P.


## 3. Topology on C-Posets

The wealth of topological structures in continuous
domains and the interplay between the topological and algebraic properties of
continuous domains are well documented [1]. In this section, we look at C-posets
endowed with the Lawson topology.Theorem 3.1.
A C-poset endowed with the Lawson topology is a pospace and hence Hausdorff. An A-poset
endowed with the Lawson topology is totally order-disconnected.Proof.
LetP be a C-poset.
Suppose that x,y∈P such that x≰y. Then, there exist u∈P,  and an
up-complete lower set J such thatu≰y,x∉J,  and ↑u∪J=P. Let U=P∖J, and V=P∖↑u. Then U is a Lawson
open-upper set, and V is a Lawson
open lower set. Clearly x∈U and y∈V. Moreover U∩V=(P∖J)∩(P∖↑u)=P∖(J∪↑u)=∅. This shows that P is a pospace.
IfP is an A-poset,
then for x,y∈P, x≰y implies the
existence of u∈P and an
up-complete lower set J as in the
previous paragraph with the additional condition that ↑u∩J=∅. This means that ↑u is both closed
and open. This completes the proof of the theorem.An inf-semilatticeP is called meet continuous  if for all directed
subsets D of P for which supD exists, and for
all x∈P,
xsupD=supxD.Proposition 3.2.
If a
C-posetP is also an inf-semilattice, then P is meet-continuous.Proof.
LetD be a directed
subset of P such that supD exists, and let x∈P. We need to show
only that x sup D≤supxD since the
reverse inequality is always true. Suppose that x sup D≰supxD. Then there
exist u≰supxD and an
up-complete lower set J such that x sup D∉J and ↑u∪J=P. Since J is a lower set,
this means that x∉J, and supD∉J. Then x∈↑u, and since J is up-complete, ∃d∈D such that d∈↑u. Thus for this d,xd∈↑u which
contradicts u≰supxD. This completes
the proof.Proposition 3.3.
LetP be a C-poset
which is also an inf-semilattice. Then the inf-operation is a continuous
function (L×L,λ(L×L))→(L,λ(L)).Proof.
First, considerV=⋀−1(P∖↑x)={(y,z):x≰y⋀z}. We will show
that V is open. If (y,z)∈V, then x≰y⋀z. Therefore,
there exist a u≰y⋀z and an
up-complete lower set J such that x∉J, and ↑u∪J=P. Let S=(P×P)∖↑(u,u). Then clearly S is open in P×P and (y,z)∈S. We will show
that S⊆V. If (t,r)∈S, then either u≰t or u≰r. Therefore, u≰t⋀r, which implies
that t⋀r∈J. If x≤t⋀r, then, since J is a lower set, x∈J, a
contradiction. Thus x≰t⋀r which means
that S⊆V.
LetO be any Scott-open
subset of P, and consider U=⋀−1(O). We will show
that U={(y,z):y⋀z∈O} is a Scott open
subset of P×P. Suppose that D is a directed
subset of P×P such that supD=(a,b)∈U. Then, a⋀b∈O. Let D1={d1:∃d2 such that (d1,d2)∈D}, and let D2={d2:∃d1 such that (d1,d2)∈D}. Then D1 and D2 are directed
sets such that supD1=a, and supD2=b. Since P is
meet-continuous by Proposition 3.2, a⋀b=supD1⋀supD2=sup(D1D2)∈O. Since D1D2 is a directed
set and since O is Scott-open,
there exist d1∈D1 and d2∈D2 such that d1⋀d2∈O, which implies
that (d1,d2)∈U. This completes
the proof of the proposition.



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