Recall that a dcpo is a partially ordered set $(X,\leqslant)$ possessing suprema for all directed subsets.
The Scott topology of a dcpo $(X,\leqslant)$ has as closed sets those sub-dcpos which are also downwards closed, i. e. closed in the Alexandroff topology.
In other words, $C\subseteq X$ is closed in the Scott topology iff
- $x\leqslant y$, $y\in C$ $\Rightarrow$ $x\in C$;
- $S\subseteq C$, $S$ directed $\Rightarrow$ $\sup(S)\in C$.
My question is: if $Y\subseteq X$ is a sub-dcpo, i. e. satisfies 2., then is its down-closure, i. e. $\{x\in X\mid\exists\ y\in Y\ x\leqslant y\}$, also a sub-dcpo?
No, not necessarily: a standard example:
$X = \{a_0,a_1,a_2,\ldots\} \cup \{b_0, b_1, b_2, \ldots\} \cup \{\infty\}$
with the partial order $a_0 \le a_1 \le a_2 \le a_3 \ldots$ with supremum $\infty$ while the $b_i$ are incomparable and we have $a_i \le b_i$ for all $i$.
Then $X$ is a dcpo (there is essentially one directed set and it has a sup). and $Y=\{b_0,b_1,\ldots\}$ is a sub-dcpo (voidly; there are no directed subfamilies that need a sup). But the down-closure of $Y$ is $X\setminus \{\infty\}$ which is not a dcpo, as we now miss our essential supremum for the $a_i$.