Scott-continuous functions: an alternative definition

Question

I'm really struggling with this property:

Let $X,Y$ be coherence spaces and $f: Cl(X) \rightarrow Cl(Y)$ be a monotone function. $f$ is continuous if and only if $f(\bigcup_{x\in D} x)=\bigcup_{x \in D}f(x)$, for all $D \subseteq Cl(X)$ such that $D$ is a directed set.

Directed set is defined thus: $D \subseteq $ POSET$ $ is a directed set iff $ \forall x, x' \in D$ $ \exists z \in D $ such $ x \subseteq z$ and $x' \subseteq z$.
$Cl(X) $stands for cliques of X: $\{x \subseteq |X| \mid a,b \in x \Rightarrow a$ coherent $b \}$.

Many books give that as a definition of Scott-continuous functions, but unluckly not my teacher. He gave us this definition of continuous:

$f : Cl(X) \rightarrow Cl(Y)$ is continuous iff it is monotone and $\forall x \in Cl(X), \forall b \in f(x), \exists x_0 \subseteq_{fin} x, b \in f(x_0)$,
where monotone is defined as: $f$ is monotone iff $a \subseteq b \Rightarrow f(a) \subseteq f(b)$

This is the proposed proof I have, but I can't understand the last equation.

Proof of $f$ continuous implies $f(\bigcup D)=\bigcup f(D)$:
Let $b \in f(\bigcup D)$. By the definition of continuity, $\exists x_0 \subset_{fin} x \mid b \in f(x_0)$. Note that $x_0$ is the union of $\{ x_i \mid x_i \in D\}$.
If $D$ is direct then: $\exists z \in D \mid x_i \subseteq z$ then $x_0 \subseteq z$. By the definition of monotony, $f(x_0)\subseteq f(z)$ so $b \in f(z)$ (???) $\subseteq \bigcup f(D)$. And even that is true we should show that $\bigcup f(D) = f(\bigcup D)$, not just $\subseteq$.

The proof of the other implication is even worse so I can't write it here... Can you explain to me how the proof can work?

Answer 1

The definition of continuity used by your teacher is the nicer one. It tells you pretty concretely what continuity means.

Suppose $b \in f(x)$. That means that given all the information of $x$, possibly an infinite set of tokens (atoms), the function produces some element that has the atomic piece of information $b$. (It could have other information too, but we are not concerned with that at the moment.) Your teacher's definition says that it is not necessary to look at all the infinite information of $x$ in order to produce the output information $b$. Some finite subset of $x$ is enough to produce it.

(Melvin Fitting's book "Computability theory, semantics and logic programming", Oxford, 1987, calls this property compactness and defines a continuous function as being monotone and compact.)

This is the essence of continuity. To get some finite amount of information about the output of a function, you only need a finite amount of information about the input. The output produced by the function for an infinite input is obtained by piecing together the information it produces for all finite approximations of the infinite input. In other words, you don't get any magical jump in going from the finite approximations to their infinite union. Whatever you get at infinity, you should already get at some finite stage.

The standard equation $f(\bigcup_{x \in D} x) = \bigcup_{x \in D} f(x)$ is pretty to look at, but it doesn't tell you all the intuition I have explained above. However, mathematically, it is equivalent to your teacher's definition.

To show that $\bigcup_{x \in D} f(x) \subseteq f(\bigcup_{x \in D} x)$, it is enough to show that $f(x)$ is included in $f(\bigcup_{x \in D} x)$, for each $x \in D$. But that follows directly from monotonicity of $f$ because $x \subseteq \bigcup_{x \in D} x$. So, this is the "easy" direction.

The other direction, proved by your teacher, is the interesting one: $f(\bigcup_{x \in D} x) \subseteq \bigcup_{x \in D} f(x)$. To see this, use the intuition I have mentioned above. Any atomic piece of information $b$ in the left hand side comes from some finite approximation of the input: $x_0 \subseteq_{fin} \bigcup_{x \in D} x$. That is, $b \in f(x_0)$. Since $x_0$ is finite and it is included in the union of the directed set, there must be something in the directed set that is larger than $x_0$, perhaps $x_0$ itself. Call that element $z$. By monotonicity, $f(x_0) \subseteq f(z)$. So, $b \in f(z)$. Since $z \in D$, $f(z) \subseteq \bigcup_{x \in D} f(x)$. So, now $b$ is seen to be in the right hand side too. QED.

As you have noted, showing that your teacher's continuity implies the pretty equation is the easy bit. The harder bit is to show that the pretty equation, despite looking like it is not saying very much, really does say everything in your teacher's definition.

Answer 2

It occurred to me belatedly, after I wrote the last response, that the teacher's definition of continuity that I was explaining in my response is the topological notion of continuity. The algebraic formulation of continuity that is usually mentioned in Computer Science text books hides all the topological intuitions. (In fact, Dana Scott has often written that he has deliberately avoided topological formulations because Computer Scientists are not familiar with it.)

The linkage between the algebraic and topological formulations is called Stone duality, and it is now becoming increasingly clear that this linkage itself is extremely important for Computer Science.

For a conceptual exposition of these connections (and a lot more), See Abramsky's Information, processes and games.