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??
Prove or give a counterexample to the following converse of theorem: A continuous function on a compact set K(subset R) is uniformly continuous.
2
$\begingroup$
real-analysis
continuity
compactness
-
0Also it seems this question does not belong to *real-analysis*. – 2012-11-12
1 Answers
0
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.
-
0@user48601 please see my comment. $X=\mathbb{N}$ is a Counterexample. – 2012-11-14