# [Haskell-cafe] Monad laws in presence of bottoms

Miguel Mitrofanov miguelimo38 at yandex.ru
Wed Feb 22 07:45:33 CET 2012


22.02.2012, 09:30, "wren ng thornton" <wren at freegeek.org>:
> 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?



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