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 $P$. ∎