What are some illustrative examples that demonstrate how $\succ$ can differ in behavior from $>$ and/or $\geq$?

Question

I really, really want to understand the generalization of metric spaces known as continuity spaces. Unfortunately, I always get tripped up right at the beginning. The problem is that I have little or no intuition for the meaning of $q \succ p$, which appears as Definition 2.2 in the linked article. Reworded ever-so-slightly for improved readability, it says:

Definition 2.2. Assume $V$ is a complete lattice and $q,p \in V$. Then $q$ is well-above $p$, denoted $q \succ p,$ iff for any subset $A$ of $V$, if $p \geq \mathrm{inf} \,A,$ then $q \geq a$ for some $a \in A$.

The logical structure of the definition is straightforward enough, and yet at a purely intuitive level, I don't "get" it.

Question.

1. In your own words, how do you understand the meaning of $q \succ p$?

2. What are some illustrative examples that demonstrate how $\succ$ can differ from $>$ and/or $\geq$?

Answer 1

I’ve not dealt much with the concept, but an example that I find helpful is $\Bbb R^2$ with the product partial order, $\langle x_0,y_0\rangle\le\langle x_1,y_1\rangle$ iff $x_0\le x_1$ and $y_0\le y_1$. Then it’s not hard to check that $\langle x_0,y_0\rangle\prec\langle x_1,y_1\rangle$ iff $x_0<x_1$ and $y_0<y_1$.

Consider, for instance, the set $A=\{\langle 2^{-n},2^{-n}\rangle:n\in\omega\}$:

• $\inf A=\langle 0,0\rangle$;
• $\langle 2^{-n},2^{-n}\rangle\not\le\langle 0,b\rangle$ for any $b>0$;
• $\langle 2^{-n},2^{-n}\rangle\not\le\langle b,0\rangle$ for any $b>0$;
• for any $b,c>0$ there is an $n\in\omega$ such that $\langle 2^{-n},2^{-n}\rangle\le\langle b,c\rangle$.

However, I prefer the following dual characterization of $\prec$, which is the definition that I’ve most often seen.

Let $\langle P,\le\rangle$ be a partial order. If $p,q\in P$, then $p\prec q$ iff whenever $D$ is a directed subset of $P$, and $q\le\sup D$, then $p\le d$ for some $d\in D$.

If $P$ is a complete lattice, this implies that if $p\prec q$, $A\subseteq P$, and $q\le\sup A$, then there is a finite $F\subseteq A$ such that $x\le\sup F$. If $P$ is the lattice of open sets of some space, this says that an open set $U$ is way below an open set $V$ if every open cover of $V$ has a finite subfamily covering $U$. Thus, for instance, if $P$ is the usual topology of $\Bbb R$, then $(a,b)\prec(c,d)$ whenever $c<a\le b<d$, since $[a,b]$ is then a compact subset of $(c,d)$ containing $(a,b)$.

You might find some of the examples here at least a bit illuminating, and this PDF by Grzegorz Bancerek collects a lot of facts in one place, albeit without proof.

Answer 2

To get a broader context for the well above relation you can consult any introductory text on domain theory. However, the context of continuity spaces is a bit different, and I prefer to have the intuition for Flagg's value quantales come directly from their intended role. So, the way I think about the well above relation is that it solves some nasty deficiencies that $\le$ has, even in a complete lattice. As noted, $\mathcal P(S)$, the power set of a set, is a complete lattice, $0=\emptyset$. But if $A,B>0$, it does not imply that $A\cap B>0$. This is just how sets behave. Now, this is somewhat captured by the well above relation in $\mathcal P(S)$; it may be instructive to find all the elements $A\in \mathcal P(S)$ with $A\succ 0$.

Flagg gives several more consequences of his axioms that show how the well above relation takes care of things. For instance, for any $\varepsilon \succ 0$ there exists $\delta \succ 0$ such that $\delta + \delta \prec \varepsilon $, which is extremely important for metric spaces, since it's the analogue of dividing by $2$, which is used all the time.

It is quite easy to see that if $P$ is linearly ordered, then $\le=\prec $. Again, $\mathcal P(S)$ is an example where the well above relation is very different than $\le$. An important class of examples is Flagg's $\Omega$ construction. It is basically the free locale (or frame) and it is very instructive to carry out the computations in it to see what well above means. In a sense it is the 'correct' value-quantale analogue of $\mathcal P(S)$; the latter is useless for the purposes of metric spaces, while the former is precisely what one needs (to prove e.g. that every topological spaces is metrizable). Generally, $a\le b\prec c$ implies $a\prec c$ and $a\prec b \le c$ implies $a\prec c$ (and so transitivity follows too, since $a\prec b$ implies $a\le b$).

Answer 3

I cannot in good faith answer question 1, because I didn't see enough examples of that relation to get an informal idea of it.

As for 2, first note that $q \succ p$ implies $q \geq p$ (as stated in the linked article, too), and here's an example where the two relations differ:

Consider the set $S = \{a,b,c,d\}$, so that its power set $V$ is a complete lattice wrt inclusion and the infimum is given by intersection. Then let $q = \{a,b\}, p = \{b\}$, and observe that $q \supset p$ but $q \not \succ p$. Indeed, if $A = \{\{b,c\},\{b,d\}\}$, then $$ p = \{b,c\} \cap \{b,d\} = \inf A \quad \text{but} \quad q \not \supseteq \{b,c\},\{b,d\} $$ On the other hand, $q' = \{a,b,c\}$ is well-above $p$, because any three element subset of $S$ different from $q'$ contains $d$ and $\{b,d\}$ is the only two element set containing $p$ and not contained in $q'$.

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An even more striking example: consider the set of non-negative integers ordered by divisibility, so that the infimum of a subset is given by the greatest common divisor of its elements, and fix an arbitrary positive integer $p$. Then the only integer well-above $p$ is $0$.

Indeed, let $q$ be any positive integer. Then we can always find two positive integer integers $m,n$ such that $m,n,q$ are pairwise coprime. Hence $$ \inf \{m,n\} = 1 \mid p \quad \text{but} \quad m,n \nmid q $$

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On the other hand, on $\Bbb{Z}$ with the usual order $\succ$ is indeed equivalent to $\leq$. Indeed:

• If $q \succ p$, then by definition $q \geq p = \inf \{p\}$.

• On the other hand, suppose $q \geq p$. If $A$ is unbounded, then clearly $q \geq a$ for some $a \in A$. Otherwise $m = \inf A \in A$ and $p \geq m$ implies $q \geq m$ by transitivity.