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I think the converse of this theorem is: if every continuous function over $K$ is uniformaly continuous, then $K$ is compact. To find a counterexample of it, I want to show there exist a continuous function over $s$ that is uniformaly continuous, yet the set $K$ is not compact. Correct??

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If your proposed converse is really what you want, then your proposed strategy for finding a counterexample is incorrect. You would have to give an example of a subset $K$ of $\mathbb R$ that is not compact, and such that every continuous fuction on $K$ is uniformly continuous. – Jonas Meyer Nov 12 '12 at 15:55
If you want to prove $P \rightarrow Q$, then if you can show that $\lnot Q \rightarrow \lnot P$, you are done, i.e., "if P then Q" is equivalent to "if not Q then not P". – amWhy Nov 12 '12 at 16:02
Also it seems this question does not belong to real-analysis. – Hui Yu Nov 12 '12 at 16:05
up vote 0 down vote accepted

The converse is not true in general.

First note that $X$ is compact if and only if every $f\in C(X)$ is bounded. One direction is trivial. To see the other, note that once $C(X)$ is bounded then $\|f\|=sup_X|f(x)|$ becomes a norm on $C(X)$ and it is easy that $C(X)$ is a unital abelian $C^*$-algebra under this norm. By Gelfand's theorem, $C(X)$ is $*$-isomorphic to some $C(X')$, where $X'$ is compact Hausdorff. Then one can show $X$ itself is compact since it is homeomorphic to $X'$.

Now it suffices to show 1) every $f\in C(X)$ is uniformly continuos is not equivalent to 2) every $f\in C(X)$ is bounded. But this is easy, one just take $X=\mathbb{N}$ then 1) is true but 2) is not.

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Algebra? NO yet. I did hear of isomorphism, but I am dealing with a problem in my real analysis book. Yet i would appreciate if you could find an counterexample of: every continuous function over a Set K is uniformally continuous,but the Set K is not compact(not closed and bounded) – user48601 Nov 12 '12 at 17:17
But I just gave one in the proof. Take $X=\mathbb{N}$, then all functions on $X$ are uniformly continuous (take $\delta=1/4$ if you insist on epsilon-delta language). But $\mathbb{N}$ is not compact. – Hui Yu Nov 13 '12 at 1:49
@user48601 please see my comment. $X=\mathbb{N}$ is a Counterexample. – Hui Yu Nov 14 '12 at 3:13

In general, to determine the converse, you need to formulate the expression as "$X$ implies $Y$." The converse is "$Y$ implies $X$." Then to provide a counterexample to the coverse, you need to show an example where $Y$ is true but $X$ is not.

In this case $X$ is: "$K\subset\mathbb R$ is compact."

$Y$ is: "$\forall f:K\to\mathbb R$, $f$ continuous implies $f$ is uniformly continuous."

So a counterexample would be a set $K$ for which $Y$ is true, but not $X$.

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Thanks, this is super clear to me now – user48601 Nov 12 '12 at 17:17

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