\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