On Mon, 2009-02-02 at 17:30 +0100, Krzysztof Skrzętnicki wrote:
Do they? Haskell is a programing language. Therefore legal Haskell types has to be represented by some string. And there are countably many strings (of which only a subset is legal type representation, but that's not important).
Haskell possesses models[1] in which the carriers of all types are countable. Haskell also possesses the models[2] which do assign uncountable carriers to several Haskell types --- a -> b whenever a has an infinite carrier (and b is not a degenerate type of the form newtype B = B B ), [b] under the same conditions on b, etc. In many cases, these are the most insightful models, and I those models are what people mean when they talk about e.g. [Int] having an un-countable denotation. jcc [1] IIUC all models based on recursion theory have this property [2] IIUC most models based on domain theory have this property