# Real analysis; uniformly continuous function.

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.

• I think the author is trying to say that unboundedness of $E$ alone is not sufficient to guarantee the existence of non-uniform continuous function $f : E \to \mathbb{R}$. So, although some unbounded $E$ may actually work to give a non-uniform one, some choices such as $E = \mathbb{Z}$ do not. – Sangchul Lee Apr 9 '19 at 7:04

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

• I understand why Rudin suggested some unbounded and non-compact set where all functions are uniformly continuous. However, I can't understand why the function defined on Z is all uniformly continuous. As you say, isn't the value of the function defined on Z discrete? Then it cannot be "continuous". Could you explain more about it? :-) – 주혜민 Apr 9 '19 at 9:01
• Take any $\varepsilon >0$. Then choose $\delta <1$, Then we prove $$\vert x-y |< \delta \implies |f(x)-f(y)|<\varepsilon$$ But $|x-y|<\delta$ means ,there is only one point in $(y-\delta,y+\delta)$, so $f(x)=f(y)$ and hence result follows – Chinnapparaj R Apr 9 '19 at 9:20