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 <d...@mobileink.com> wrote: > On Mon, Feb 2, 2009 at 10:05 AM, Andrew Butterfield > <andrew.butterfi...@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 > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe >
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