22.02.2012, 09:30, "wren ng thornton" <[email protected]>: > On 2/21/12 11:27 AM, MigMit wrote: > >> Ehm... why exactly don't domain products form domains? > > One important property of domains[1] is that they have a unique bottom > element. Given domains A and B, let us denote the domain product as: > > (A,B) def= { (a,b) | a <- A, b <- B } > > Which will inherit an ordering in the obvious/free way from the domain > orderings on A and B. Since both A and B are domains, they have bottom > elements: > > exists a0:A. forall a:A. (a0 <=_A a) > exists b0:B. forall b:B. (b0 <=_B b) > > 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. > > So all of those are minimal elements of (A,B) but none of them is a > unique minimum; hence (A,B) is not a domain. > > The smash product gets around this because it takes all those elements > and makes them equal, just like a strict tuple would in Haskell. > > [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? _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
