Is Every Continuous Function on an Unbounded Set Uniformly Continuous?

[Bachelier]

$prop:$ let set $E \subset \mathbb{R}$ be unbounded, then $\forall f$ well-defined on $E$, if $f$ is continuous, then $f$ is uniformly continuous.

First am I reading this correctly, and second, I am having a hard time seeing this. Could someone please shed some light on this?

Thanks.

 

[jibbles]

surely you meant bounded instead of unbounded, right?

 

[Bachelier]

surely you meant bounded instead of unbounded, right?

If it was bounded, then $E$ must be closed as well for $f$ to be uniform continuous. I am citing a case where $E$ is

not bounded. Rudin gives the example of $\mathbb{Z}$ and states that ANY function defined on $\mathbb{Z}$ is

indeed uniformly continuous.

 

[Office_Shredder]

Is your definition of unbounded weird? Because if E=R then it's saying that all continuous functions are uniformly continuous.

The example of Z isn't because Z is Z is unbounded, the key property is that Z is discrete - any discrete set has that all functions on them are uniformly continuous.

 

[Bachelier]

Is your definition of unbounded weird? Because if E=R then it's saying that all continuous functions are uniformly continuous.

The example of Z isn't because Z is Z is unbounded, the key property is that Z is discrete - any discrete set has that all functions on them are uniformly continuous.

Yea I agree. That is why I am asking. Please see the attached theorem 4.20. The assumption on the boundedness is at the end of page 2

 

[lavinia]

Yea I agree. That is why I am asking. Please see the attached theorem 4.20. The assumption on the boundedness is at the end of page 2

the theorem says there exists a continuous function on E that is notuniformly continuous

 

[Bachelier]

the theorem says there exists a continuous function on E that is notuniformly continuous

Maybe I am reading too much into this.

After equation (23), Rudin writes:

"...Assertion (c) would be false if boundedness were omitted from the hypotheses."

Can you explain this further? especially via an example without using the set of integers.

 

[lugita15]

Maybe I am reading too much into this.

After equation (23), Rudin writes:

"...Assertion (c) would be false if boundedness were omitted from the hypotheses."

Can you explain this further? especially via an example without using the set of integers.

He means there are some noncompact unbounded sets E for which all continuous functions on E are uniformly continuous. Of course, any unbounded set is noncompact, so he is saying that there are some unbounded sets E for which all continuous functions on E are uniformly conitnuous.

By the way, note the word "some". That's why the proposition in your OP isn't stated correctly.

 

[Bacle2]

It is definitely not always true, if I understood correctly ( or, If I have not jumped the gun, like I have sometimes done, embarrassingly).

Take f: Q<ℝ → Q , with f(x)=1/(x-√2) .

Q is unbounded in ℝ , but f is not uniformly-continuous (fails near √2 ; if you want it to fail
at more points, you can repeat the idea.