I'm really struggling with this property:Let $X,Y$ be coherence spaces and $f: Cl(X) \rightarrow Cl(Y)$ be a monotone function. $f$ is continuous if and only if $f(\bigcup_{x\in D} x)=\bigcup_{x \in D}f(x)$, for all $D \subseteq Cl(X)$ such that $D$ is a directed set.Directed set is defined thus: $D \subseteq $ POSET$ $ is a directed set iff $ \forall x, x' \in D$ $  \exists z \in D $ such $ x \subseteq z$ and $x' \subseteq z$.  $Cl(X) $stands for cliques of X: $\{x \subseteq |X| \mid a,b \in x \Rightarrow a$ coherent $b \}$.Many books give that as a definition of Scott-continuous functions, but unluckly not my teacher. He gave us this definition of continuous:$f : Cl(X) \rightarrow Cl(Y)$ is continuous iff it is monotone and $\forall x \in Cl(X), \forall b \in f(x), \exists x_0 \subseteq_{fin} x, b \in f(x_0)$,  where monotone is defined as:  $f$ is monotone iff $a \subseteq b \Rightarrow f(a) \subseteq f(b)$This is the proposed proof I have, but I can't understand the last equation.Proof of $f$ continuous implies $f(\bigcup D)=\bigcup f(D)$:Let $b \in f(\bigcup D)$. By the definition of continuity, $\exists x_0 \subset_{fin} x \mid b \in f(x_0)$. Note that $x_0$ is the union of  $\{ x_i \mid x_i \in D\}$.If $D$ is direct then: $\exists z \in D \mid x_i \subseteq z$ then $x_0 \subseteq z$. By the definition of monotony, $f(x_0)\subseteq f(z)$ so $b \in f(z)$ (???) $\subseteq \bigcup f(D)$. And  even that is true we should show that $\bigcup f(D) = f(\bigcup D)$, not just $\subseteq$.The proof of the other implication is even worse so I can't write it here... Can you explain to me how the proof can work?
Scott-continuous functions: an alternative definition
terminology;programming languages;semantics