On 2/22/12 1:45 AM, Miguel Mitrofanov wrote:
However, there is no free ordering on:
{ (a0,b) | b<- B } \cup { (a,b0) | a<- A }
What? By definition, since, a0<= a and b0<= b, we have (a0, b0)<= (a0, b) and (a0, b0)<= (a0, b0), so, (a0, b0) is clearly the bottom of A\times B.
Sorry, the ordering relation on domain products is defined by: (a1,b1) <=_(A,B) (a2,b2) if and only if a1 <=_A a2 and b1 <=_B b2
[1] This is in the sense of domain theory. It has nothing (per se) to do with the many other uses of the term "domain" in mathematics.
Sorry, isn't the domain theory a part of mathematics?
Sure, domain theory is a part of mathematics, but the term "domain" is used to mean a bunch of different and largely unrelated things. For example: * in type theory and set theory the "domain" of a function is the set of inputs on which it's defined * in domain theory a "domain" is a partial order with a least element and the property that every non-empty countable chain has a least upper bound * in ring theory a "domain" is is a ring which has no left nor right zero-divisors * in analysis a "domain" is any connected open subset of a finite-dimensional vector space etc. -- Live well, ~wren