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 ;(

  • 1
    $\begingroup$ 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. $\endgroup$ – 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

  • $\begingroup$ 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? :-) $\endgroup$ – 주혜민 Apr 9 '19 at 9:01
  • $\begingroup$ 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 $\endgroup$ – Chinnapparaj R Apr 9 '19 at 9:20

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