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From: William Elliot on 8 Jul 2010 05:33
On Wed, 7 Jul 2010, Edson 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.

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.
From: David C. Ullrich on 8 Jul 2010 09:35
On Wed, 07 Jul 2010 18:59:10 EDT, cwldoc <cwldoc(a)aol.com> wrote:

>> 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.


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).



>
>>
>> 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: W^3 on 9 Jul 2010 00:59
In article
<2069423442.83636.1278514688866.JavaMail.root(a)gallium.mathforum.org>,
Edson <edsonm37(a)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.
From: Robert Israel on 9 Jul 2010 14:02
W^3 <aderamey.addw(a)comcast.net> writes:

> In article
> <2069423442.83636.1278514688866.JavaMail.root(a)gallium.mathforum.org>,
> Edson <edsonm37(a)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.

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 israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: W^3 on 10 Jul 2010 16:32
In article <aderamey.addw-B58437.21592008072010(a)News.Individual.NET>,
W^3 <aderamey.addw(a)comcast.net> wrote:

> In article
> <2069423442.83636.1278514688866.JavaMail.root(a)gallium.mathforum.org>,
> Edson <edsonm37(a)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."

> 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.
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