Re: Every set can be ... ordered?

In sci.logic, Rupert
<rupertmccallum@xxxxxxxxx>
wrote
on 25 Aug 2006 23:59:26 -0700
<1156575566.502150.268470@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:

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.

Hm...good point, if a bit on the thin side. :-)


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.

Is R^3 equipollent to R?


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