4
$\begingroup$

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!

Improve this question
$\endgroup$
1
  • $\begingroup$ I feel that there are some random errors in the formulation of the above 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 a single notion of Scott continuity. If my doubts are founded you need to reformulate your question so that it will make sense. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Oct 1, 2014 at 21:51

2 Answers 2

Reset to default
2
$\begingroup$

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 ).

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ dominiczypen: just a kind remark, you don't need to assume that $f$ is monotone in your definition of a Scott-continuous map, if one rephrases a bit the definition as follows: $f$ is Scott-continuous if for every non-empty directed subset $D$, the supremum of $f(D)$ exists and equals $f(\bigvee D)$. (Then $f$ is necessarily monotone.) $\endgroup$
    – polmath
    Commented Oct 16, 2014 at 12:40
  • $\begingroup$ Right, thanks for your remark -- and it is important to replace "monotone" by "the supremum of $f(D)$ exists" (just as you did). But I agree that your version is more economical $\endgroup$
    – Dominic van der Zypen
    Commented Oct 16, 2014 at 12:43
1
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .