Scott topology on $\mathcal{O}X$ In Wikipedya  Scott-open set is defined as follows:
Def. A subset $O$ of a partially ordered set $P$ is called Scott-open if 


*

*It is an up set.

*All directed sets $D$ with supremum in $O$ have non-empty intersection with $O$.


Now consider a topological space $X$ with set of opens $\mathcal{O}X$ partially ordered by $\subseteq $. Martin Escardo and Reinhold Heckmann in their paper Topologies on spaces of continuous functions for case of a $\mathcal{O}X$ define Scott-open as follows:
An upset $O\subseteq \mathcal{O}X$ is called Scott-open if every open cover of a member of $O$ has finite subcover of a member of $O$. 
They then state that for any subset $Q$ of $X$, the set $\mathcal{V}_Q=\{V\in\mathcal{O}X| ~Q\subseteq V\}$ the set $\mathcal{V}_Q$ Scott-open  iff $Q$ is compact. 
Are those the same definitions? I can not see why it is if it is so. How we can get the same result with Wiki definition?  
 A: I figured out the answer and forgot to write here. Really it turned out very easy. 
Our Poset is $(\mathcal{O}X, \subseteq)$
$\Rightarrow $
Suppose $O$ is an up-set and all directed sets $D$ with supremum in $O$ have non-empty intersection with $O$. Then $O$ is still an upset; For every open cover add finite unions to the cover, the supremum, i.e. covered set a member of $O$ remains the same, and by assumption there is a member of a new cover, that is finite union of old members which is in $O$.
$\Leftarrow$ 
Suppose an upset $O\subseteq \mathcal{O}X$ is called such that every open cover of a member of $O$ has finite subcover of a member of $O$. Then we do the same trick, we add finite unions of members to the covers, the covers remin the same and since our new covers are directed, consist of finite unions of old members and all directed families are the mentioned form. By assumption we get that the directed families with supremum in $O$ have nonempty intersection with $O$. 
That way the definitions are equivalent in the case of poset of opens.     
