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!

Question. Is it really possible that just one single equality $f(\sqcup_x X) = \sqcup_y f(x), x \in X$ would be enough to define continuity? Or is there a typo? I have also other doubts, more conceptual. Non-equivalent continuity notions should be perhaps in a complex correspondence to different closure operations. However, you have only asinglenotion of Scott continuity. If my doubts are founded you need to reformulate your question so that it will make sense. $\endgroup$ – Włodzimierz Holsztyński Oct 1 '14 at 21:51