Is every real valued continuous function on the interval $(0,1)$ is uniformly continuous?

I think the answer is no, and to reject the statement, we need to come up a continuous function probably $f(x)=\frac{1}{x}$ and follow the following link:

Coming up with an example, a function that is continuous but not uniformly continuous

But it does not work because $\delta=\min(x,1)$ cannot be applied because $x$ cannot attain $1$.

| cite | improve this question | | | | |
  • 4
    $\begingroup$ Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous $\endgroup$ – Boshu Dec 9 '18 at 6:20
  • $\begingroup$ @Boshu: That does not work as I explained in the question. $\endgroup$ – Sepide Dec 9 '18 at 6:46
  • 1
    $\begingroup$ Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $\frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it. $\endgroup$ – Boshu Dec 9 '18 at 6:52
  • $\begingroup$ @Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me? $\endgroup$ – Sepide Dec 9 '18 at 6:57
  • $\begingroup$ Will write a brief answer. $\endgroup$ – Boshu Dec 9 '18 at 7:05

The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.

Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $\dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $\dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.