From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Newsgroups: sci.math.num-analysis,sci.math
Subject: Re: continuous 1-1 function from RxR to R?
Date: 10 Oct 1995 18:37:07 -0400

In article <45e623$7e7@muir.math.niu.edu>,
Dave Rusin <rusin@washington.math.niu.edu> wrote:
>In article <45doa2$432@gap.cco.caltech.edu>,
>Ilias Kastanas <ikastan@alumni.caltech.edu> wrote:
>>	Consider, then, B = R - Q (with the topology inherited from R -- no
>>   tricks).  The irrationals... so reasonable!  What have we lost?  B is a
>>   complete metric space.  
>
>What is the limit of the Cauchy sequence \pi, \pi/2, \pi/3, ..., \pi/n ?

  Precisely! Rigor fell victim to Ilias Kastanas's folksy enthusiasm.
True, the irrationals, with the topology inherited from R, are metrizable
and among the metrics there is one that makes B complete. (A general
theorem in this direction is: A metrizable space is homeomorphic to a
complete metric space iff it is a G-delta subset of a complete metric
space, i.e.  a countable intersection of open subsets.) 
  But both previous authors probably knew this anyway; I just tried to
demystify the issue. 
  To get a complete metric explicitly, we arrange the rationals into a 
sequence {r_n}_n>0 , and define the distance between two irrationals as
  d(x,y) = |x - y| + sum {|1/(x-r_n) - 1/(y-r_n)|/2^n : n>0}
 so that the sequence  {pi/n} fails to be Cauchy. (We "pushed the 
rationals to infinity".)

What is this topic doing in sci.math.num-analysis?

Cheers, ZVK (Slavek).

==============================================================================

From: ikastan@alumni.caltech.edu (Ilias Kastanas)
Newsgroups: sci.math.num-analysis,sci.math
Subject: Re: continuous 1-1 function from RxR to R?
Date: 16 Oct 1995 08:39:30 GMT

In article <45e623$7e7@muir.math.niu.edu>,
Dave Rusin <rusin@washington.math.niu.edu> wrote:
>In article <45doa2$432@gap.cco.caltech.edu>,
>Ilias Kastanas <ikastan@alumni.caltech.edu> wrote:
>>	Consider, then, B = R - Q (with the topology inherited from R -- no
>>   tricks).  The irrationals... so reasonable!  What have we lost?  B is a
>>   complete metric space.  
>
>What is the limit of the Cauchy sequence \pi, \pi/2, \pi/3, ..., \pi/n ?


	It is possible that different metrics induce a given topology, some
   of them complete and some not.

	Simple example: (0,1) with the usual topology.  The sequence {1/n} is
   Cauchy under the standard metric, and has no limit.  But (0,1) is homeomor-
   phic to R, which is complete.  Carry back that metric, and (0,1) becomes
   a complete metric space.

	B is complete if you pick the right metric.  For example, B is homeo-
   morphic to J, the irrationals in (0,1).  Using continued fraction expansi-
   ons, J is homeomorphic to the space of positive integer sequences, a com-
   plete metric space under d(x,y) = 1/(n + 1), n the least index such that
   x(n) != y(n).   And d(x,x) = 0... I'll dot the i's and cross the t's this
   time!

	Using that metric on B, the desired topology is generated; and the
   sequence {pi/n} is not Cauchy.

	Maybe I should have written "completely metrizable" to begin with,
   instead of cutting corners.  But then we wouldn't have had this bit of
   discussion!


							Ilias