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 non-empty 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 $z\le y$. 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 $e\le x$ for all $e\in D$, then $d:=\bigvee D\le x$ 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 $\mathrm{\varnothing }$ is vacuously Scott open.

• Suppose $U$ and $V$ are Scott open. Let $W=U\cap V$ and $b\in \uparrow W$. Then for some $a\in W$, $a\le b$. 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 $(\uparrow t)\cap D\subseteq (\uparrow y)\cap (\uparrow z)\cap D=((\uparrow y)\cap D)\cap ((\uparrow z)\cap D)\subseteq U\cap V=W$. 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 $a\le b$ 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 , 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.