Re: Well Ordering the Reals

In article <Iq071q.6IK@xxxxxx> "*** T. Winter" <***.Winter@xxxxxx> writes:
> In article <1132069032.676676.246470@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> "Ross A. Finlayson" <raf@xxxxxxxxxxxxxxx> writes:
> > *** T. Winter wrote:
> ...
> > > No. They have it not as a set, but with the standard ordering.
> > > Gaplessness is the result of a metric, and the metric is related to the
> > > ordering. If you order the reals with a different ordering, you get a
> > > different metric, and so it is not necessarily the case that there are
> > > no gaps.
> ...
> > I would accept that the gaplessness property is the result of a metric,
> > but it's the same in this case as completeness which is a property of
> > the set.
>
> Completeness is the same as gaplessness here (depends a bit on what you
> mean with completeness).
>
> > With as you think a different ordering, that doesn't affect the
> > completeness of the reals.
>
> Are you sure that with a different ordering all Cauchy sequences converge?
> The answer is *no*. Consider the reals ordered by first the integers in
> their conventional order and next the non-integers in their conventional
> order. The sequence: 1/2^i does not converge, and so the reals (with this
> ordering) are not complete. (The sequence does not converge because
> there is no greatest lower bound for the sequence.) Of course, the above
> sequence will converge with a well-ordering. But the sequence -1/2^i may
> very well not converge. And as soon as one of the two sequences does not
> converge the reals are not complete with that ordering.

The last two sentences are nonsense, of course.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.

Quantcast