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). 

All best

Christopher Skrzętnicki

On Mon, Feb 2, 2009 at 17:09, Gregg Reynolds <dev@mobileink.com> wrote:
On Mon, Feb 2, 2009 at 10:05 AM, Andrew Butterfield
<Andrew.Butterfield@cs.tcd.ie> wrote:
> Martijn van Steenbergen wrote:
>>
>>> To my naive mind this sounds
>>> suspiciously like the set of all sets, so it's too big to be a set.
>>
>> Here you're probably thinking about the distinction between countable and
>> uncountable sets. See also:
>>
>> http://en.wikipedia.org/wiki/Countable_set
>
> No - it's even bigger than those !
>
> He is thinking of proper classes, not sets.
>
> http://en.wikipedia.org/wiki/Class_(set_theory)

Yes, that's my hypothesis:  type constructors take us outside of set
theory (ZF set theory, at least).  I just can't prove it.

Thanks,

g
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