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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).
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