If $E$ is a non-compact and bounded set in $\mathbb{R}$, then there exists a continuous function on $E$ which is not uniformly continuous.
In the proof, the book says that if $E$ is a non-compact and unbounded, there doesn't exist a non-uniformly continuous, by suggesting the example of $E=\mathbb{Z}$. But I can't understand this, because if $E$ is $\mathbb{R}$, which is unbounded, and $f(x)=x^2$, then $f(x)$ is a non-uniformly continuous function.
Please help me ;(
Since $E$ is bounded and non-compact, it fails to be closed (by the Bolzano-Weierstrass Theorem).
In other words, $E$ has a limit point $a$ that is not in $E$. The function $$ f(x) = {1 \over x - a} $$ is continous at all points in $\mathbb{R}$ except at $x=a$. What is you opinion about whether $f(x)$ is uniformly continuous on $E$?
And use a good book on analysis; e.g., Kolmogorov and Fomin's.:)
I think its from Rudin. The result is
If $E$ is a non compact set and bounded , then there exists a continuous function on E which is not uniformly continuous.
If we assume $E$ is not bounded then this result is false, since every continuous function on $\Bbb Z$ is uniformly continuous( because every point of $\Bbb Z$ is isolated). That's what Rudin say.
The result is false means we provide an unbounded set inwhich every function is uniformly continuous.
Consider this result: Every cyclic group is Abelian.
If we remove cyclic hypothesis, then this result is false.
The false stetement is: Every group is Abelian.
In this place , we cannot do "consider $V_4$, then the result is true"
This situation is exactly what we do with $x \mapsto x^2$ in your example
