You can enumerate all possible implementations of functions of type (Integer -> Bool). Just enumerate all strings, and give this to a Haskell compiler f :: Integer -> Bool f = <enumerated-string-goes-here> if the compiler is happy you have an implementation.
The enumerated functions do not include all mathematical functions of type (Integer -> Bool), but it does include the ones we usually mean by the type (Integer -> Bool) in Haskell. -- Lennart On Mon, Feb 2, 2009 at 4:47 PM, Martijn van Steenbergen <mart...@van.steenbergen.nl> wrote: > Lennart Augustsson wrote: >> >> The Haskell function space, A->B, is not uncountable. >> There is only a countable number of Haskell functions you can write, >> so how could there be more elements in the Haskell function space? :) >> The explanation is that the Haskell function space is not the same as >> the functions space in set theory. Most importantly Haskell functions >> have to be monotonic (in the domain theoretic sense), so that limits >> the number of possible functions. > > I was thinking about a fixed function type A -> B having uncountably many > *values* (i.e. implementations). Not about the number of function types of > the form A -> B. Is that what you meant? > > For example, fix the type to Integer -> Bool. I can't enumeratate all > possible implementations of this function. Right? > > Martijn. > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
