Compactness in metric spaces
Edson
Since Heine Borel theorem holds in R^n, this is not hard to prove. If S is not compact, then it's unbounded or not closed. If it's unbounded, put f(x) = |x|; If it's not closed, choose a limit point a of S that is not in S and define f(x) = 1/|x - a|. It's easy to see these are continuous but unbounded functions from S to R. By contraposition, S is compact.
A similar reasoning shows this is true in any metric space where Heine Borel relation holds. But what if we have a metric space where Heine Borel condition doesn't hold? We have the following general question.
Let S be a subset of a metric space (M, d) with the property that every continuous function from S to R (R with the euclidean metric) is bounded. The, does S need to be compact.
I tried to come to a conclusion using the general fact that a metric space is compact if, and only if, it's complete and totally bounded. But I'd like some help, please.
Edson
Ronald Benedik
"Edson" <edso...@yahoo.com> schrieb im Newsbeitrag
news:2069423442.83636.1278514688866.JavaMail.root@gallium.mathforum.org...
> This question was motivated by a the following problem: Let S be a subset
> of R^n such that every continuous function from S to R is bounded. Show
> that S is compact.
>
> Since Heine Borel theorem holds in R^n, this is not hard to prove. If S is
> not compact, then it's unbounded or not closed. If it's unbounded, put
> f(x) = |x|; If it's not closed, choose a limit point a of S that is not in
> S and define f(x) = 1/|x - a|. It's easy to see these are continuous but
> unbounded functions from S to R. By contraposition, S is compact.
>
> A similar reasoning shows this is true in any metric space where Heine
> Borel relation holds. But what if we have a metric space where Heine Borel
> condition doesn't hold? We have the following general question.
>
> Let S be a subset of a metric space (M, d) with the property that every
> continuous function from S to R (R with the euclidean metric) is bounded.
> The, does S need to be compact.
The equivalent indirect formulation would be:
S is not compact, so there exists a continuous function which is not
bounded.
Correct me if I'm wrong.
David C. Ullrich
>This question was motivated by a the following problem: Let S be a subset of R^n such that every continuous function from S to R is bounded. Show that S is compact.
>
>Since Heine Borel theorem holds in R^n, this is not hard to prove. If S is not compact, then it's unbounded or not closed. If it's unbounded, put f(x) = |x|; If it's not closed, choose a limit point a of S that is not in S and define f(x) = 1/|x - a|. It's easy to see these are continuous but unbounded functions from S to R. By contraposition, S is compact.
>
>A similar reasoning shows this is true in any metric space where Heine Borel relation holds. But what if we have a metric space where Heine Borel condition doesn't hold? We have the following general question.
>
>Let S be a subset of a metric space (M, d) with the property that every continuous function from S to R (R with the euclidean metric) is bounded. The, does S need to be compact.
There's no reason to talk about subsets here; this is the same as the
question of whether a metric space is compact if every countinuous
real-valued function is bounded.
>I tried to come to a conclusion using the general fact that a metric space is compact if, and only if, it's complete and totally bounded. But I'd like some help, please.
Well, if M is not complete then there exists an unbounded continuous
function. Possibly the simplest way to see that is to use the fact
that every metric space has a completion. Say N is the completion
of M and N <> M. Say p is a point of N not in M. Then the function
1/d(x,p) is countinuous and unbounded on M.
And I believe that if M is not totally bounded it also folllows
that there is an unbounded continuous real-valued function,
although I haven't worked out every detail.
Say r > 0 and the closed balls B(x_n, r) are
oairwise disjoint. Let B_n = B(x_n, r/3). Or r/10
or somethng, whatever works. Letting
f_n(x) = g_n(f(x, x_n))
for an appropriate g_n you get continuous functions
f_n with f_n > 0, f_n(x_n) > n, and
f_n(x) < 2^{-n} for all x in M \ B_n. It seems to
me that the sum of the f_n is then continuous
and unbounded.
Ah, here we go. Let x be any point of M. There
exists at most one n such that B(x, /3r) intersects
B_n (this follows from the triangle inequality
plus the disjointness of B(x_n, r)). Say x is in
M and x is not in B_n for n <> m. Then the sum
of the f_n for n <> m converges uniformly on
B(x, r/3), and hence the sum of _all_ the f_n is
continuous on B(x, r/3), hence continuous at x.
I lied, I _have_ worked out the details.
>Edson
Dave L. Renfro
> Let S be a subset of a metric space (M, d) with the property that
> every continuous function from S to R (R with the euclidean metric)
> is bounded. The, does S need to be compact.
It looks like David C. Ullrich has already answered your question
(I say "looks like" only because I haven't read over Ullrich's
post carefully), but you might also be interested in the eomments
about this problem in these March 1999 sci.math posts:
http://www.math.niu.edu/~rusin/known-math/99/cpt_metric
Dave L. Renfro
cwldoc
Maybe I'm missing something here, but I do not follow this part of your proof.
[M is not totally bounded] means there exists an r > 0 such that any finite collection of r-balls fails to cover M (or equivalently that any collection of r-balls that covers M is infinite). I don't see where you are getting the collection of balls referred to above.
William Elliot
> This question was motivated by a the following problem: Let S be a
> subset of R^n such that every continuous function from S to R is
> bounded. Show that S is compact.
A space S is pseudo-compact when for all f in C(S,R), f is bounded.
If S is a normal T1 space, then
S is pseudo-compact iff S is countably compact.
This uses Tietze extension theorem.
If S is a metric space, then
S is countably compact iff S is compact.
> But what if we have a metric space where Heine
> Borel condition doesn't hold?
It's true of all metric spaces and
almost true of all normal T1 spaces.
David C. Ullrich
Say R > 0 has the property that no finite collection of balls of
radius R covers M. Recursively choose a sequence x_1, x_2, ...
such that x_n is not in the union of B(x_j, R) for j < n.
That says that d(x_j, x_n) >= R whenever j < n. Hence
d(x_n, x_m) >= R whenever n <> m. Let r = R/3. Then
the balls B(x_n, r) are pairwise disjoint (if p is in B(x_n, r)
intersect B(x_m, r) wth n <> m then the triangle
inequality shows that d(x_n, x_m) <= 2r < R).
W^3
<2069423442.83636.12785...@gallium.mathforum.org>,
Edson <edso...@yahoo.com> wrote:
> Let S be a subset of a metric space (M, d) with the property that every
> continuous function from S to R (R with the euclidean metric) is bounded.
> The, does S need to be compact.
Yes. Let's work in the metric space S; no need to worry about any
containing space. If S is not compact, then there is a sequence x_n of
distinct points in S having no convergent subsequence. It follows that
each x_n is an isolated point of S, and that E= {x_n :n in N} is
closed in S. The function f : E -> R given by f(x_n) = n is continuous
on E. By Tietze's extension theorem, f extends to a continuous
function on S, which is obviously unbounded on S, contradiction.
Robert Israel
You can also do it without Tietze. Consider a sequence x_n with no convergent
subsequence. Let f_n(x) = max(0, n - n^2 d(x, x_n)), so f_n(x_n) = n and
f_n(x) = 0 if d(x, x_n) >= 1/n, and let f(x) = sup_n f_n(x). For every x,
there is some delta > 0 such that (with at most one exception) all
d(x, x_n) > delta, and in particular all but finitely many f_n are 0 in
the ball of radius delta/2 about x. Therefore f is continuous and unbounded.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
W^3
W^3 <aderam...@comcast.net> wrote:
> In article
> <2069423442.83636.12785...@gallium.mathforum.org>,
> Edson <edso...@yahoo.com> wrote:
>
> > Let S be a subset of a metric space (M, d) with the property that every
> > continuous function from S to R (R with the euclidean metric) is bounded.
> > The, does S need to be compact.
>
> Yes. Let's work in the metric space S; no need to worry about any
> containing space. If S is not compact, then there is a sequence x_n of
> distinct points in S having no convergent subsequence. It follows that
> each x_n is an isolated point of S, and that E= {x_n :n in N} is
> closed in S.
I should have said "It follows that E = {x_n :n in N} is closed in S,
and each x_n is an isolated point of E."