On the definition of Scott Continuity I somewhere encountered the concept of "Scott Continuity" as follows.
Let $P,Q$ be partially ordered sets; a function $f:P\to Q$ is Scott continuous if it preserves directed suprema, i.e. for all directed $D\subseteq P$ that have a supremum there holds: $$\sup(f(D))=f(\sup(D)).$$
I have also found that this definition can be restated just requiring that $f$ preserves suprema of non empty chains. 
It is clear that the first definition implies the second one (as chains are just special directed sets) and I really find it difficult to even imagine a directed set whose supremum is not the supremum of one of its subchains.
I am therefore asking you: are the two definitions equivalent?
Is there a simple proof?   
 A: I am answering to your remark "I really find it difficult to even imagine a directed set whose supremum is not the supremum of one of its subchains."
Actually, consider the poset $P = 2^{\mathbb{R}}$ ordered by inclusion, and the directed subset $\mathcal{D} = \{ F \subset \mathbb{R} : F \mbox{ finite } \}$. Then the supremum of $\mathcal{D}$ in $P$ is $\bigcup \mathcal{D} = \mathbb{R}$. 
However, one cannot write $\mathbb{R}$ under the form $\bigcup \mathcal{C}$ for some subchain $\mathcal{C}$ of $\mathcal{D}$. 
Indeed, the map $\mathcal{C} \rightarrow \mathbb{N}, F \mapsto \# F$ is injective, hence $\mathcal{C}$ is countable, so that $\bigcup \mathcal{C} = \bigcup_{F \in \mathcal{C}} F$ is countable. 
A: I've found that your question finds a positive answer in the paper by Markowsky "Chain-complete posets and directed sets with applications", Corollary 3 (see here), provided that the posets $P$ and $Q$ are chain-complete or directed-complete. Now the question becomes: is there a more direct proof? 
