From: bruck@pacificnet.net (Ronald Bruck)
Newsgroups: sci.math
Subject: Re: Completeness of a metric space
Date: Thu, 29 Feb 1996 21:59:58 -0700

In article <4h5ie0$ifp@sam.inforamp.net>, t762@inforamp.net (J. B.
Rainsberger) wrote:

:Everyone:
:
:I'm preparing for an exam in real analysis on Monday and one of the topics 
:that we'll be looking at on the exam is completing a space. We went over an 
:example of completing a metric space which was not already complete by 
:defining an equivalence relation on (X, d):
:
:        (x_n) ~ (y_n) <==> lim d(x_n, y_n) = 0
:
:so that if x_n and y_n converge to the same value, then (x_n) and (y_n) are 
:related. Clearly, this is an equivalence relation since, by the triangle 
:inequality, we can squeeze lim(x_n, z_n) between 0 and lim(d(x_n, y_n) + 
:d(y_n, z_n) ) = 0. And so, we note by [x] the class of equivalence containing 
:(x_n). Thus X* = { [x]:x in X }, the set of all such classes of equivalence, 
:which form a partition of X.

You're a little confused here.  This is not an equivalence relation on X;
it is an equivalence relation on the SPACE OF SEQUENCES DRAWN FROM X.  And
the sequences (x_n) DON'T have to converge in X (that's the whole point). 
The equivalence classes don't form a partition of X, but of the space of
sequences drawn from X.

:
:Now defining the injective function f from X into X* where f(x) = [x] where 
:the constant sequence (x, x, x, x, ...) is a member of [x], we have X a subset 
:of X*.
:
:We also define another distance function delta([x], [y]) = lim(d(x_n, y_n)) 
:where (x_n) is in [x] and (y_n) is in [y].
:
:Now we must show that by this series of constructions, we have completed X by 
:entending it into X*. I am somewhat unsure how to do this.

The steps are:  1.  Show that delta is well-defined on X/~; 2. Show that
delta is a metric on X/~; 3.  Show that delta([x],[y]) = d(x,y), so the
original space (X,d) can be regarded as being isometrically imbedded in
(X/~,delta); 4. Show that (X/~, delta) is complete.  It's grungy.  That's
why it's left as an exercise.
:
:Comments, please! I have a feeling that we'll be dealing with something 
:similar to the following question on the exam:
:
:Q. Let f be the collection of all real sequences which have only a finite 
:number of non-zero elements and define d(x,y) = sup{|x_n - y_n| : n natural}. 
:Show that d is a metric on f, but that f is not complete with respect to d.
:

To mix metaphors, this is a much easier kettle of fish.  Consider x_n =
(1,1/2,...,1/n,0,0,0,...), show it's Cauchy wrt d, but it doesn't
converge.

However, I have always disliked the X/~ construction.  I prefer the
following:  let (M,d) be a metric space.  Consider the space C(M) of all
bounded continuous functions from M to the reals R, with the sup norm. 
This is a Banach space, i.e. is complete in the sup norm (requires brief
proof -- essentially, it says that the uniform limit of continuous
functions is continuous).  Fix an element m_0 \in M.  For m \in M, define
f_m : M --> R by f_m(x) = d(m,x) - d(m_0,x).  Them f_m \in C(M), and the
mapping m |--> f_m is an isometry of M into C(M).  The closure (in C(M))
of the image of M under this isometry may be regarded as the completion of
M.

Every analyst should know this embedding of a metric space into C(M). 
Indeed, every analyst should know how to prove that every separable metric
space can be embedded isometrically into C[0,1].

The only disadvantage to this construction is that you can't construct the
real numbers this way (whereas they CAN be regarded as the completion of
the rationals the other way, if you're careful about what you mean by lim
d(x_n,y_n) = 0).

--Ron Bruck
  Now 100% ISDN from this address