I am trying to understand the Scott Topology. I am looking at the page here: http://planetmath.org/scotttopology and for the definition of "upper set": http://planetmath.org/node/37801

I am having a difficult time coming up with a working example (with actual elements of a set) to see how these "Scott open sets" work.

In starting: Let P be a directed complete partially ordered set. A subset $U$ of $P$ is said to be "Scott open" if it satisfies the following two conditions:

  1. U is an upper set: $\uparrow U= U$
  2. if D is a directed set with $\bigvee D \in U$, then there is a $y \in D$ such that $(\uparrow y) \cap D \subseteq U$.

I am trying to figure this out with an example by letting $P = \mathbb{Z}_7 = \{0,1,2,3,4,5,6\}$ and $U = \{5,6\}$, as if I am correct with the calculations (could someone please let me know) we would have to have that $\uparrow U = U$ as 5 and 6 are the only elements $p \in P$ such that $p \leq x \in U$and thus the first condition above would be satisfied.

I am now stuck. I see that we can find the supremum which is the join of the sets of D - I don't understand what D is supposed to be, both in an example that works with the above - or what the point is of finding the join in the first place and so on and so forth (why we are needing the intersection of $\bigvee \cap \uparrow \{y\}$ to be a subset of U), why this is all important.

If someone could help me verify that I am going about finding $\uparrow U$ and if my calculations are correct, and perhaps some values for D, it would be much appreciated.

thank you,



The idea of domain representation is that the elements of a dcpo $P$ can be viewed as approximations to the maximal elements of $P$, which we view as "complete" points rather than just approximations. The Scott topology is put on the entire poset $P$, so we can not only say what it means for a sequence of approximations to converge to a maximal point, we can say what it means for a sequence of approximations to converge to another approximation.

The classical example lets $P$ be the poset of nonempty closed intervals of $\mathbb{R}$, ordered by reverse inclusion. In fact $P$ is a dcpo whose maximal elements are one-point closed intervals. The Scott topology, when restricted to the maximal elements, will give a space homeomorphic to $\mathbb{R}$ with its standard topology.

A directed set "converges" to its supremum, which is its join. Recall that, for $p, q \in P$, $p$ is way below $q$, written $p \ll q$, if every directed set that converges to $q$ must go above $p$: if $\sup D = q$ there is an $r \in D$ with $p \leq r$.

In my example $P$, $[1,2] \leq [1,1]$, but $[1,2]$ is not way below $[1,1])$, because there is a sequence of open intervals that converges to $[1,1]$ without ever being above $[1,2]$, namely $[1 - 1/n,1]$. So $[1,2]$ is some sort of approximation to $[1,1]$, but not a good one.

The set of $a$ such that the interval $[a,a]$ is way above $[1,2]$ is exactly the set of $a$ in the open interval $(1,2)$. These are the points for which $[1,2]$ is a good approximation, in the sense that any directed set that converges to one of these points has to actually refine the approximation $[1,2]$. Notice that although $[1,2]$ is a closed interval, it determines an open-in-$\mathbb{R}$ set of maximal elements that are way above it.

So that is how the Scott topology works for the maximal elements of $P$. The interesting thing is that the topology is on the entire poset $P$, not just on the set of maximal elements.

The set $U = \{[a,b] : 1 < a \text{ and } b< 2\}$ is Scott open in $P$. If $D$ is an ideal on $P$ with $\sup D \in U$ then, letting $\sup D = [a,b]$, we know there must be some $[c,d] \in D$ with $1 < c < a$ and some $[e,f] \in D$ with $b < f < 2$. Then $[c,d] \vee [e,f] = [c,f] \in D \cap U$.

In general a directed set $D$ corresponds to an ideal, but it is easier if we temporarily pretend $D$ is just an increasing sequence in $P$ that converges to some element in $U$. Then we want to know that $D$ itself is eventually in $U$. That is the naive topological meaning of an "open set" - a sequence that converges to a point in an open set must eventually lie in the open set. Now, since $D$ is increasing and $U$ is upward closed, saying that $D$ is eventually in $U$ is the same as saying $D \cap U $ is nonempty. So, intuitively, the Scott open sets are indeed the ones that, if an increasing sequence converges to an element of the set, the sequence must actually contain an element of the set.

Another way to understand the way below relation is that if $p$ is way below $q$, then $p$ is an "observable requirement" for $q$ in the sense that if we have an increasing sequence that converges to $q$ we will eventually observe that the sequence gets above $p$ in the order on $P$. We define the Scott open set determined by $p$ to be the set of all $q$ for which $p$ does serve as a requirement in this way. There is one such set for each $p \in P$, and the Scott topology is generated by this subbasis.

  • $\begingroup$ Thank you Carl. It is hard to choose an accepted answer as both you and Niels have helped a lot. Thanks again though. $\endgroup$ – Relative0 Oct 25 '13 at 18:07

The example you have chosen is not a good way to demonstrate the effect of rule 2. For any finite poset that rule is trivially satisfied, because every directed subset contains its supremum. Better examples are suggested on the Planet Math page you linked to.

The basic idea of rule 2 is a variation on a principle that holds for open sets in complete metric spaces: a Cauchy sequence of points outside the set has its limit outside the set. Just replace Cauchy sequence with directed set, and replace limit with supremum.

Let's look at how this works out for the first example: the unit interval. An upper set is here either a closed interval $[ a, 1 ]$, or a half-open interval $(a, 1]$. As the order is linear, every subset is directed. For upper sets of the form $U = [a, 1]$ with $0 < a < 1$, we have the directed set $D = [0, a)$ that does not intersect $U$, but $\sup D = a \in U$, so these sets are not open. On the other hand, if $U = (a, 1]$, then for any set $D$ that does not meet $U$ we have $\sup D \le a$, hence $\sup D \notin U$, so these sets are open.

We find then that the Scott topology on $[0, 1]$ is $\{ \emptyset, [0, 1] \} \cup \{ (a, 1] \mid 0 \le a \lt 1 \}$.


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