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). <div>
<br></div><div>All best</div><div><br></div><div>Christopher Skrzętnicki<br><br><div class="gmail_quote">On Mon, Feb 2, 2009 at 17:09, Gregg Reynolds <span dir="ltr"><<a href="mailto:dev@mobileink.com">dev@mobileink.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><div class="Ih2E3d">On Mon, Feb 2, 2009 at 10:05 AM, Andrew Butterfield<br>
<<a href="mailto:Andrew.Butterfield@cs.tcd.ie">Andrew.Butterfield@cs.tcd.ie</a>> wrote:<br>
> Martijn van Steenbergen wrote:<br>
>><br>
>>> To my naive mind this sounds<br>
>>> suspiciously like the set of all sets, so it's too big to be a set.<br>
>><br>
>> Here you're probably thinking about the distinction between countable and<br>
>> uncountable sets. See also:<br>
>><br>
>> <a href="http://en.wikipedia.org/wiki/Countable_set" target="_blank">http://en.wikipedia.org/wiki/Countable_set</a><br>
><br>
> No - it's even bigger than those !<br>
><br>
> He is thinking of proper classes, not sets.<br>
><br>
> <a href="http://en.wikipedia.org/wiki/Class_(set_theory)" target="_blank">http://en.wikipedia.org/wiki/Class_(set_theory)</a><br>
<br>
</div>Yes, that's my hypothesis: type constructors take us outside of set<br>
theory (ZF set theory, at least). I just can't prove it.<br>
<br>
Thanks,<br>
<div><div></div><div class="Wj3C7c"><br>
g<br>
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