I want to figure out why the Scott topology really forms a topology. In particular, I want to find out why the union of two closed sets is again closed.
So say we have a partial order $P = (P, ≤)$. A set $A ⊂ P$ is called Scott closed if it is a lower set and for each directed subset $D ⊂ A$ with a supremum $\bigvee D$ in $P$, that supremum $\bigvee D$ is already in $A$.
Why is the union $L ∪ R$ of two Scott closed sets $L$, $R$ again Scott closed?
I know that $L ∪ R$ is again a lower set, I’m struggling with the second condition.
My efforts. For any directed set $D ⊂ P$ with supremum $\bigvee D$ in $P$, I tried to argue that either in $L ∩ D$ or in $R ∩ D$ there is some directed subset with the same supremum, but to no avail. I don’t know how to construct such a subset.
Another, more general, strategy I thought of might be to construct subsets $D_L ⊂ L ∩ D$ and $D_R ⊂ R ∩ D$ with superma $d_L = \bigvee D_L$ and $d_R = \bigvee D_R$ such that $\bigvee D = d_L ∨ d_R$, but again I have no idea how to construct $D_L$ and $D_R$.
(By the way, if you can give only a hint, I’d prefer that.)
Suppose $D\subseteq P$ is directed and its supremum $\bigvee D$ is not in $L\cup R$. Since the supremum isn't in $L$, and since $L$ is closed, we have some $x\in D-L$. Similarly, we have some $y\in D-R$. Since $D$ is directed, it contains some $z$ that is $\geq$ both $x$ and $y$. $z$ can't be in $L$ or in $R$, as these sets are downward closed. So $D\not\subseteq L\cup R$.