wren ng thornton wren at freegeek.org
Wed Feb 22 08:20:47 CET 2012

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