scott continuity, sub additivity Let $(X, \sqsubseteq_x)$ and $(Y, \sqsubseteq_y)$ be two posets and let $\delta_x:X \to X$ and $\delta_y:Y \to Y$ be two closure operators (monotone, inflationary, idempotent). Then, a monotone  function $f:X \to Y$ is continuous if $f \circ \delta_x $ $\sqsubseteq_y$ $ \delta_y \circ f$.
Now, I want to show that, for dcpos/lattices, this notion of continuity coincides with the Scott's notion of continuity: that is, a monotone $f$ is continuous if $f(\sqcup_x X) = \sqcup_y f(x), x \in X$. I worked out the Scott defn. $\to$ closure defn. implication; but I don't have much handle on the other way: closure defn. $\to$ Scott defn implication, specifically the argument to prove that $f(\sqcup_x X) \sqsubseteq_y \sqcup_y f(x), x \in X$ (the dual inequality comes from the monotonicity of $f$). Am I missing something here?
Any lead would be appreciated, thanks!
 A: The two notions are equivalent if you take continuity with respect to the Scott topology (and not the one you define with the closure operators).
Let $(X,\leq_X)$ and $(Y,\leq_Y)$ be dcpos. We say that a monotone function $f: X\to Y$ is Scott-continuous if for any non-empty directed set $D\subseteq X$ we have $f(\bigvee^\uparrow D) = \bigvee^\uparrow f(D)$. (By $\bigvee^\uparrow$ we denote the supremum of a non-empty directed set which always exists in dcpos.)
Moreover the Scott topology on a dcpo $(X,\leq_X)$ is defined in the following way: $U\subseteq X$ is open if the two following conditions hold: 


*

*it is an upper set;

*whenever $D\subseteq X$ is directed with $(\bigvee^\uparrow D) \in U$ then $D\cap U \neq \emptyset$.


It turns out that for dcpos $(X,\leq_X)$ and $(Y,\leq_Y)$, a monotone function $f:X\to Y$ is Scott-continuous if and only if it is continuous with respect to the Scott topologies of $X,Y$; for more see Abramsky / Jung, "Domain Theory", http://www.cs.ox.ac.uk/samson.abramsky/handbook.ps , Proposition 2.3.4 (you can convert this to pdf by using an http://ps2pdf.com/convert.htm ).
A: Actually, the converse implication doesn't hold.
Let $X$ and $Y$ be posets with greatest elements $1_X$, $1_Y$, and let $\delta_x$ and $\delta_y$ be the constant $1_X$, $1_Y$ maps. These are clearly closure operators.
Then every function $f:X\rightarrow Y$ is continuous according to the closure definition.
However, in general not every (monotone) function $f:X\rightarrow Y$ is continuous in the Scott sense.
