|
|
The Ghost In The Machine wrote:
> In sci.logic, MoeBlee
> <jazzmobe@xxxxxxxxxxx>
> wrote
> on 25 Aug 2006 16:39:22 -0700
> <1156549162.644162.249500@xxxxxxxxxxxxxxxxxxxxxxxxxxx>:
> > MoeBlee wrote:
> >> And what about "every set can be partially ordered"?
> >
> > Oops, nevermind that question. Obviously, every set is partially
> > ordered by the subset relation on the set.
> >
> > MoeBlee
> >
>
> Every *power* set, maybe. But the reals wouldn't be able
> to be ordered that way.
>
Why not? However you define them, surely the subset relation would be a
partial ordering? Remember a partial ordering doesn't have to be
connected. Even the diagonal relation is a partial ordering.
> Of course it's easy enough to total order the reals;
> if each real number r is defined by at least one Cauchy
> sequence of rational numbers, then another real number
> s can also be defined by another Cauchy sequence, and if
> it is the case that there exists an M and a rational
> d > 0 such that for every i,j > M, r_i > s_j + d, then r > s.
>
> Or something like that.
>
> Contrariwise, I am not sure if anyone's proven that one cannot
> total-order a convex R^2 or R^3 subset.
>
If R can be totally ordered, then of course every subset of a set
equipollent to R can be as well.
> --
> #191, ewill3@xxxxxxxxxxxxx
> Windows Vista. Because it's time to refresh your hardware. Trust us.
|
|