\let\da=\downarrow \def\downarrow{\mathord{\da}}
\def\doubledownarrow{\lower.8pt\hbox{$\da$}\raise.8pt\llap{$\da$}}
\let\ua=\uparrow \def\uparrow{\mathord{\ua}}
\def\doubleuparrow{\lower.8pt\hbox{$\ua$}\raise.8pt\llap{$\ua$}}
\def\Box{\sqcap\llap{$\sqcup$}}

%105
We recall at this point that every topology is a lattice, and indeed a com%-
plete Heyting algebra (remember O-3.22!).  It is therefore meaningful to search
for prime and co-prime elements in $\sigma(L)$ (see 1-3.11, -3.15).  To formulate one
of our conditions it is necessary to speak of the continuity of an operation (the
main topic of  the next section).  We say that the $\sup$ operation is {\it jointly contin%-
uous with respect to the Scott toplogy\/} provided that the mapping
$$(x,y)\mapsto x\vee y : (L,\sigma(L)) \times (L,\sigma(L)) \to (L,\sigma(L))$$
is continuous in the product toplogy.

\medskip
{\bf 1.11. Proposition.} {\it Let $L$ be a complete lattice and $U\subseteq L$.

{\rm (i)}  $U$ is a co-prime in $\sigma(L)$ iff $U\in\sigma(L)$ and $U$ is a filter;

{\rm (ii)}  $U$ is a prime in $\sigma(L)$ and $U\ne L$ if $U = L\setminus\downarrow u$ for some $u\in L$;

{\rm (iii)}  This last condition is also necessary, provided that the $\sup$ operation is
jointly continuous;

{\rm (iv)}  This is the case if $L$ is a continuous lattice.}

\medskip
{\bf Proof.}  (i):  Firstly suppose that $U\in\sigma(L)$ is a filter and that $U$ is not a co-%
prime in $\sigma(L)$.  Thus there are $V,W\in\sigma(L)$ such that $U\subseteq V\cup W$ and elements
$v\in U\setminus V$ and $w\in U\setminus W$.  Since $V$ and $W$ are upper sets we have $vw\not\in V\cup W$.  But
$vw\in U$ since $U$ is a filter.  This is a contradiction.

Secondly, suppose that $U$ is a co-prime in $\sigma(L)$.  To show that $U$ is a
filter, note first that it is an upper set.  Suppose $v,w\in U$.  Then $U \not\subseteq L\setminus\downarrow v$ and
$U \not\subseteq L\setminus\downarrow w$.  By 1.4(ii), the sets $L\setminus\downarrow v$ and $L\setminus\downarrow w$ are Scott open.  Thus, since $U$ is
co-prime,
$$U \not\subseteq ((L\setminus\downarrow v)\cup(L\setminus\downarrow w)) = L\setminus(\downarrow v\cap\downarrow w) = L\setminus\downarrow vw.$$
Thus $U\cap\downarrow vw\ne\emptyset$.  Since $U$ is an upper set, this means $vw\in U$.

(ii): Recall that $L\setminus\downarrow u = L\setminus\{u\}^-$ for all $u$ by 1.4(ii).  But the complements
of singleton closures are prime in any topology.

(iii): Assume that the $\sup$-operation is jointly continuous relative to the
Scott topology, and let $U\in\hbox{PRIME}\;\sigma(L)$, $U\ne L$.  Set $A = L\setminus U$ and $u = \sup
A$.  In order to show that $U = L\setminus\downarrow u$ it suffices to verify $u\in A$.  Since $A$ is
closed under directed $\sup$s by 1.4(i), all we have to verify is that $A$ is directed.

By way of contradiction assume that it is not.  Then there are elements
$a,b\in A$ with $a\vee b\in U$.  By the continuity of the $\sup$-operation we would find
Scott-open neighborhoods $V$ and $W$ of $a$ and $b$, respectively, such that
$V\vee W\subseteq U$.  But since $V$ and $W$ are upper sets, we have $V\vee W = V\cap W$.  Since
%106
$U$ is prime, the relation $V\cap W\subseteq U$ implies $V\subseteq U$ or $W\subseteq U$.  This would entail
$a\in U$ or $b\in U$, which would contradict $a,b\in A = L\setminus U$.

(iv): Finally suppose that $L$ is continuous.  In order to show the continuity
of the $\sup$-operation at $(a,b)$ we pick some $u\ll a\vee b$.  By I-1.6 we have
$$a\vee b = (\sup\doubledownarrow a)\vee(\sup\doubledownarrow b) = \sup(\doubledownarrow a\vee\doubledownarrow b)$$
Since $\doubledownarrow a\vee\doubledownarrow b$ is directed (I-1.2(iii)), we find some $x\ll a$ and $y\ll b$ with $u\ll x\vee y$
(by I-1.19).  But then $\doubleuparrow x$ and $\doubleuparrow y$ are Scott-open neighborhoods of $a$ and $b$,
respectively, such that
$$\doubleuparrow x\vee\doubleuparrow y\subseteq \uparrow(x\vee y)\subseteq \doubleuparrow u.$$
Since the $\doubleuparrow u$ with $u\ll a\vee b$ form a basis of $\sigma(L)$-neighborhoods of $a\vee b$ by Pro%-
position 1.10(i), the desired continuity is established.  $\Box$

\medskip
We can immediately rephrase 1.11 in topological terminology, if we recall
the concept of a {\it sober space\/} (see I-3.36 and the discussion).  Remember that a
non-empty subset $A$ of a topological space $X$ is called {\it irreducible\/} iff it is
closed and not the union of two proper closed subsets (that is, the comple%-
mentary set $X\setminus A \in \hbox{PRIME}\;{\cal O}(X)$; see I-3.11).  A space $X$ is called $\it sober$ iff
every irreducible set $A$ has a unique dense point (that is, $A = \{a\}^-$ with a
unique $a\in A$).  Clearly all singleton closures are irreducible.  (Notice that a sober
space is automatically $T_0$ since $\{x\}^- = \{y\}^-$ always implies $x = y$.)  We now
have the following corollaries of 1.11 with a slight sharpening:

\medskip
{\bf 1.12. Corollary.}  {\it If $L$ is a complete lattice such that the $\sup$-operation is
jointly Scott-continuous, then $(L,\sigma(L))$ is a sober space.}

\medskip
{\bf Proof.}  Immediate from 1.11 and the definitions.  $\Box$

\medskip
{\bf 1.13. Corollary.}  {\it If $L$ is a continuous lattice, then $(L,\sigma(L))$ is a quasicompact
and locally quasicompact sober space.  In particular, $(L,\sigma(L))$ is a Baire space.}

\medskip
{\bf Proof.}  We have to show that a point $x\in L$ has a basis of quasicompact
neighborhoods.  By 1.10 the sets $\doubleuparrow y$ with $y\ll x$ form a basis for the
neighborhoods of the point.  But as we know, if $x\in U\in\sigma(L)$, then actually we
have a $y\in U$ with $y\ll x$; hence, $\uparrow y\subseteq U$, and so the sets $\uparrow y$ can be used as
neighborhoods.  Since $\uparrow y$ (and hence, in particular $L = \uparrow 0$) is trivially
quasicompact with respect to any topology whose open sets are upper sets, the
assertion is proved.  That $(L,\sigma(L))$ is a Baire space follows from I-3.43.10.  $\Box$

\medskip
We wish to warn the reader about a subtlety concerning the joint con%-
tinuity of the $\sup$-operation above.  We cannot be satisfied by saying that the
$\sup$-operation is a continuous function $(L\times L,\sigma(L\times L))\to(L,\sigma(L))$; this con%-
tinuity is weaker, since in general we have a proper containment of topologies:
$\sigma(L\times L)\supset\sigma(L)\times\sigma(L)$.  We will return to this question at greater length in
Section 4 below (see 4.11 ff).

%107
We know enough about the Scott topology now to use it for yet another
characterization theorem for continuous lattices.

\medskip
{\bf 1.14. Theorem.}  {\it For any complete lattice, the following conditions are equivalent:

{\rm (1)} $L$ is continuous;

{\rm (2)} If $U\in\sigma(L)$, then $U = \bigcup\{\doubleuparrow x : x\in U\}$;

{\rm (3)} Each point has a neighborhood basis (in the Scott topology) of
Scott-open filters, and $\sigma(L)$ is a continuous lattice;

{\rm (4)} For each point $x\in L$ we have $x = \sup \{\inf U : x\in U\in\sigma(L)\}$

{\rm (5)} $\sigma(L)$ has enough co-primes and is a continuous lattice;

{\rm (6)} $\sigma(L)$ is completely distributive;

{\rm (7)} Both $\sigma(L)$ and $\sigma(L)^{\hbox{\sevenrm{op}}}$ are continuous.}

\medskip
{\bf Proof.}  (1) implies (2): Use 1.10.

(2) implies (1):  Let $x\in L$ and set $y = \sup \doubledownarrow x\le x$.  If $y<x$, then $L\setminus\downarrow y$ is a
Scott-open neighborhood of $x$; hence by (2) it contains an open neighborhood
$\doubleuparrow z$ of $x$ with $z\in L\setminus\downarrow y$.  But then $z\ll x$, and thus $z\le\sup \doubledownarrow x = y$, a contra%-
diction.

(1) implies (3):  By (1) iff (2), $x$ has arbitrarily small neighborhoods of
the form $\doubleuparrow y$ with $y\ll x$.  By 1.10 and I-3.3, we know then that $x$ has arbitrarily
small Scott-open neighborhoods which are filters.  In order to prove the
continuity of $\sigma(L)$, we let $U$ be Scott open.  For any $x\in U$ we find a $y\in U$ with
$y\ll x$ by (2).  Then $x\in\doubleuparrow y\in\sigma(L)$, and we claim that $\doubleuparrow y\ll U$: Indeed, if $D$ is a
directed family of Scott-open sets covering $U$, then one of its members must
contain $y$, hence $\uparrow y$, since Scott-open sets are upper sets, and thus it contains
$\doubleuparrow y$.  We have  shown $U = \bigcup\{V : V\ll U\}$.

(3) implies (4):  For $x\in L$ set
$$y = \sup\{\inf U : x\in U\in\sigma(L)\}\le x.$$
If $y<x$, then $L\setminus\downarrow y$ is a Scott-open neighborhood of $x$.  Let $V$ be a Scott-open
neighborhood of $x$ with $V\ll L\setminus\downarrow y$, which exists since $\sigma(L)$ is continuous by
(3).  Now use (3) to find a Scott-open filter neighborhood $U$ of $x$ within $V$.  By
the definition of $y$ we have $\inf U\le y$.  Then
$$L\setminus\downarrow y\subseteq L\setminus\downarrow \inf U = L\setminus\bigcap\{\downarrow u : u\in U\} = \bigcup\{L\setminus\downarrow u : u\in U\}.$$
Since $U$ is a filter, the $L\setminus\downarrow u$ for $u\in U$ form  a directed family of Scott-open
sets.  Since $V\ll L\setminus\downarrow y$, there must be a $u\in U$ such that $V\subseteq L\setminus\downarrow u$ and so $u\not\in V$.
This is a contradiction to $U\subseteq V$.  Thus $x = y$.

(4) implies (1):  Clear since for every Scott-open neighborhood $U$ of $x$
one has $\inf U\ll x$.

(3) iff (5):  Clear from 1.11.

(5) iff (6) iff (7):  Consequence of I-3.15.  $\Box$

\medskip
Note that condition (2) is stronger than saying that $\sigma(L)$ has a basis of
%108
sets $\doubleuparrow x$.  A parallel result to 1.14 for algebraic lattices reads as follows:

\medskip
{\bf 1.15. Corollary.}  {\it For any complete lattice $L$ the following conditions are
equivalent:

{\rm (1)}  $L$ is algebraic.

{\rm (2)}  The Scott topology has a basis of sets $\uparrow k$ where $k\in K(L)$.

{\rm (3)}  $\sigma(L)$ is algebraic and has enough co-primes.}

\medskip
{\bf Proof.}  (1) implies (2): Let $U$ be a Scott-open neighborhood of $x$.  We
recall that $x = \sup (\downarrow x\cap K(L))$ and $\downarrow x\cap K(L)$ is deirected.  Hence by 1.2(ii), we
find a $k\in\downarrow x\cap K(L)\cap U$.  Then $\uparrow k = \doubleuparrow k$ is a Scott-open neighborhood of $k$,
hence of $x$, with $\uparrow k\subseteq U$.

(2) implies (3):  If $k\in K(L)$, then $\uparrow k\in\sigma(L)$ since $\doubleuparrow k = \uparrow k$.  Now $\uparrow k$ is a
quasicompact set (if $\uparrow k$ is covered by Scott-open sets, then one of them must
contain $k$, hence $\uparrow k$ by 1.2(i)).  Thus $\sigma(L)$ is algebraic by I-4.4.  Since all $\uparrow k$ are
filters, hence co-primes by 1.11, we are done.

(3) implies (1):  Since $\sigma(L)$ is algebraic, the Scott topology has a basis of
quasicompact sets $U$.  Since there are enough co-primes, $U$ is a union of open
filters by 1.11, and thus, by quasicompactness, $U = U_1\cup\ldots\cup U_n$ with open
filters $U_k$.  It is no loss of generality to assume that none of the $U_k$ is contained
in the union of the others.  Then $V = U_1\setminus(U_2\cup\ldots\cup U_n)$ is quasicompact and
filtered.  We claim that $V$ has a smallest element $u_1$.

For if not, then  $V\subseteq\bigcup\{L\setminus\downarrow v : v\in V\}$; and by quasicompactness and the
fact that the $L\setminus\downarrow v$ form a directed family, there would be a $v\in V$ with
$V\subseteq L\setminus\downarrow v$, notably $v\not\in V$, which is impossible.  Since $U_2\cup\ldots\cup U_n$ is an upper set,
it cannot contain $\inf U_1$.  Hence $u_1 = \min U_1$.  Since $U_1$ is an upper set, $U_1 =
\uparrow u_1$.  Since $U_1$ is Scott open, $u_1\in K(L)$.

We have shown that $\sigma(L)$ has a basis of sets $\uparrow k$ where $k\in K(L)$.  Now let
$x\in L$, set $y = \sup (\downarrow x\cap K(L))\le x$.  If $y<x$, then the Scott-open neighborhood
$L\setminus\downarrow y$ of $x$ would contain a basic neighborhood $\uparrow k$ of $x$.  Then there would be
a $k\in K(L)$ with $k\le x$ and $k\not\le y$ which contradicts the definition of $y$.  $\Box$

\bye