You are here
HomeScott topology
Primary tabs
Scott topology
Let $P$ be a dcpo. A subset $U$ of $P$ is said to be Scott open if it satisfies the following two conditions:
1. $U$ an upper set: $\uparrow\!\!U=U$, and
2. if $D$ is a directed set with $\bigvee D\in U$, then there is a $y\in D$ such that $(\ \uparrow\!\!y\ )\cap D\subseteq U$.
Condition 2 is equivalent to saying that $U$ has nonempty intersection with $D$ whenever $D$ is directed and its supremum is in $U$.
For example, for any $x\in P$, the set $U(x):=P(\ \downarrow\!\!x\ )$ is Scott open: if $y\in\uparrow\!\!U(x)$, then there is $z\in U(x)$ with $z\leq y$. Since $z\notin\downarrow\!\!x$, $y\notin\downarrow\!\!x$. So $y\in U(x)$, or that $U(x)$ is upper. If $D$ is directed and $e\leq x$ for all $e\in D$, then $d:=\bigvee D\leq x$ as well. Therefore, $d\in U(x)$ implies $e\in U(x)$ for some $e\in D$. Hence $U(x)$ is Scott open.
The collection $\sigma(P)$ of all Scott open sets of $P$ is a topology, called the Scott topology of $P$, named after its inventor Dana Scott. Let us prove that $\sigma(P)$ is indeed a topology:
Proof.
We verify each of the axioms of an open set:

Clearly $P$ itself is Scott open, and $\varnothing$ is vacuously Scott open.

Suppose $U$ and $V$ are Scott open. Let $W=U\cap V$ and $b\in\uparrow\!\!W$. Then for some $a\in W$, $a\leq b$. Since $a\in U\cap V$, $b\in\ \uparrow\!\!U=U$ and $b\in\ \uparrow\!\!V=V$. This means $b\in W$, so $W$ is an upper set. Next, if $D$ is directed with $\bigvee D\in W$, then, $\bigvee D\in U\cap V$. So there are $y,z\in D$ with $(\ \uparrow\!\!y\ )\cap D\subseteq U$ and $(\ \uparrow\!\!z\ )\cap D\subseteq V$. Since $D$ is directed, there is $t\in D$ such that $t\in(\ \uparrow\!\!y\ )\cap(\ \uparrow\!\!z\ )$. So $(\ \uparrow\!\!t\ )\cap D\subseteq(\ \uparrow\!\!y\ )\cap(\ \uparrow\!\!z\ )% \cap D=\big((\ \uparrow\!\!y\ )\cap D\big)\cap\big((\ \uparrow\!\!z\ )\cap D% \big)\subseteq U\cap V=W$. This means that $W$ is Scott open.

Suppose $U_{i}$ are open and $i\in I$ an index set. Let $U=\bigcup\{U_{i}\mid i\in I\}$ and $b\in\uparrow\!\!U$. So $a\leq b$ for some $a\in U$. Since $a\in U_{i}$ for some $i\in I$, $b\in\uparrow\!\!U_{i}=U_{i}$ as $U_{i}$ is upper. Hence $b\in U_{i}\subseteq U$, or that $U$ is upper. Next, suppose $D$ is directed with $\bigvee D\in U$. Then $\bigvee D\in U_{i}$ for some $i\in I$. Since $U_{i}$ is Scott open, there is $y\in D$ with $(\ \uparrow\!\!y\ )\cap D\subseteq U_{i}\subseteq U$, so $U$ is Scott open.
Since the Scott open sets satisfy the axioms of a topology, $\sigma(P)$ is a topology on $P$. ∎
Examples. If $P$ is the unit interval: $P=[0,1]$, then $P$ is a complete chain, hence a dcpo. Any Scott open set has the form $(a,1]$ if $0<a\leq 1$, or $[0,1]$. If $P=[0,1]\times[0,1]$, the unit square, then $P$ is a dcpo as it is already a continuous lattice. The Scott open sets of $P$ are any upper subset of $P$ that is also an open set in the usual sense.
References
 1 G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. W. Mislove, D. S. Scott, Continuous Lattices and Domains, Cambridge University Press, Cambridge (2003).
Mathematics Subject Classification
06B35 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections