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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?
>
>> --
>> #191, ewill3@xxxxxxxxxxxxx
>> Windows Vista. Because it's time to refresh your hardware. Trust us.
>
--
#191, ewill3@xxxxxxxxxxxxx
Windows Vista. Because it's time to refresh your hardware. Trust us.
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