# Questions on Continuity

1. May 13, 2013

### 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.

2. May 13, 2013

### jibbles

surely you meant bounded instead of unbounded, right?

3. May 13, 2013

### Bachelier

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.

4. May 13, 2013

### Office_Shredder

Staff Emeritus
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.

5. May 14, 2013

### Bachelier

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

#### Attached Files:

• ###### Boundedness and Unif. Ctnty.pdf
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6. May 14, 2013

### lavinia

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

7. May 14, 2013

### Bachelier

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.

8. May 17, 2013

### lugita15

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.

9. May 17, 2013

### 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.