From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: The Completeness of Metric Spaces
Date: 8 Jul 2000 06:15:40 -0400
Newsgroups: sci.math
Summary: [missing]

In article <20000708051605.02732.00003973@ng-cg1.aol.com>,
Paterices <paterices@aol.com> wrote:
:Is there a general topological method to determine whether some arbitrary
:metric space is complete

No; that is, "no, there is no general topological method ..." if we agree
what "topological" means.

"Topological" is generally understood as "invariant under homeomorphisms",
which comes down to "using only concepts defined ultimately in terms of
open sets and corresponding set-theoretical operations", in particular,
without mentioning the metric.

With this understanding, the trouble is that you can have pairs of
homeomorphic metric spaces, where one space is complete and the other
isn't. Very inconvenient but that's topology. Example: the whole line R,
and the open interval (-1,1), and a homeomorphism is x |-> x/(1+|x|) .

There is a substitute: we can call a metrizable topological space
"topologically complete" if it is homeomorphic to a complete metric space.
This sounds like cheating, but there is a nice characterization: such
spaces are "absolute G-delta spaces", that is, countable intersections of
open subsets wherever they are embedded. An equivalent condition is "being
a G-delta subset of an already complete metric space".

Example: the set of all irrationals is a topologically complete subset of
R, while the set of all rationals isn't (Baire Category Theorem prevents
it).

There are other useful generalities: a closed subset of a complete space
is complete, and a complete subset of a complete space is closed. (The
topology is understood to be defined by the restricted metric.) So, if a
subset of an already complete space is not closed then it cannot be
complete. (Sometimes it can still be topologically complete, to confuse
bystanders.)

:when there is no explicit completeness axiom, 

You don't force a space to be complete by attaching a completeness
requirement to it (that is, ultimately, it will not be complete just
because you call it complete). The misunderstanding comes perhaps from
confusing theories (such as the axiomatic theory of real numbers) and
models of theories (such as construction of the reals by completing the
rationals).

:or does
:one generally have to resort to algebraic properties (to give a specific
:example that doesn't concern me very much, can the rationals be proven
:not complete without finding specific sets that cannot include their
:closure, such as [0,sqrt2))?  

The rationals, under their usual metric, are best proved to be incomplete
by showing a Cauchy sequence of rationals which fails to have a rational
limit. Example: pick your favourite sequence of rationals that converges
in R to sqrt(2). Your example is puzzling: how do you relate the
non-closedness of that interval to completeness?

:Intuitively, I should imagine this not to
:be the case, but perhaps there is something clever that doesn't cross my
:mind.

Now I am lost: what is imagined to be or not to be the case?

Cheers, ZVK(Slavek)