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From: Edson on 7 Jul 2010 06:57
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.

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
From: Ronald Benedik on 7 Jul 2010 11:35


"Edson" <edsonm37(a)yahoo.com> schrieb im Newsbeitrag
news:2069423442.83636.1278514688866.JavaMail.root(a)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.


From: David C. Ullrich on 7 Jul 2010 12:37
On Wed, 07 Jul 2010 10:57:38 EDT, Edson <edsonm37(a)yahoo.com> wrote:

>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

From: Dave L. Renfro on 7 Jul 2010 13:07
Edson wrote (in part):

> 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
From: cwldoc on 7 Jul 2010 14:59
> On Wed, 07 Jul 2010 10:57:38 EDT, Edson
> <edsonm37(a)yahoo.com> wrote:
>
> >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

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.

>
> 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
>
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