[Haskell-cafe] ordNub

[Atze van der Ploeg]

If we do not require that (a <= b) && (a >= b) ==> a == b (where <= is from
the total ordering and == is from the equality relation) then it is
trivial, take
the total ordering forall x y. x <= y that i mentioned earlier.

So the compatiblity with equality (you say field structure) is not besides
the point, in fact
antisymmetry means that the ordering corresponds to the equality relation.

Clear now or did I misunderstand?

Cheers

2015-01-01 15:39 GMT+01:00 Tom Ellis <
tom-lists-haskell-cafe-2013 at jaguarpaw.co.uk>:

> On Thu, Jan 01, 2015 at 03:37:09PM +0100, Atze van der Ploeg wrote:
> > This boils down to the question whether on each set with an equality
> > relation defined on it a total ordering (consistent with the equality
> > relation) can also be defined.
>
> I agree with the essence of this restatement.
>
> > One counterexample is the complex numbers.
>
> This is what I don't understand.  The complex numbers can be totally
> ordered.  (Not in a way compatible with the field structure, but that's
> beside the point).
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