Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We need to find which are uniform continuous (UC) on a) $(0,1)$ and b) $(0,\infty)$. I have done, could you confirm me, if I am wrong any where?

  1. $\frac{1}{(1-x)}$

  2. $\frac{1}{(2-x)}$

  3. $\sin x$

  4. $\sin(1/x)$

  5. $x^{1/2}$

  6. $x^3$

1) is not UC on a) because limit does not exist when $x\rightarrow 1$, on b) it is discontinous (disco) at $x=1$

2) is UC on a) and it is disco at $x=2$ so not UC on b)

3) is UC on a) and also on b) as we can show by the inequality $|\sin x-\sin y|<|x-y|$

4) is not UC on a) as limit does not exist, and also not UC on b) I am not clear enough for this one.

5) is UC on a) and not UC on b) as derivative is not bounded near $0$

6) is UC on a) and not UC on b) as derivative is not bounded.

share|cite|improve this question
@josper:Just to confirm myself. – Un Chien Andalou Jun 9 '12 at 14:29
If.. I understood several short-writings there correctly, it looks fine to me. – DonAntonio Jun 9 '12 at 14:37
Why do you need derivatives for 5. and 6., since UC functions are charaterized by $\forall \varepsilon \, \exists \delta \, \forall x \, \forall y \, ( \, |y-x|<\delta \, \to \, |f(y)-f(x)|<\varepsilon \,) $? Ok, they say: In particular, every function which is differentiable and has bounded derivative is uniformly continuous. Are these definitions interchangable? – draks ... Jun 9 '12 at 14:39
Check $5$ (a). For $4$ (b), if not UC on a set, can't be UC on bigger set. – André Nicolas Jun 9 '12 at 14:44
@Andre, thank you, I just missed the point, got it. – Un Chien Andalou Jun 9 '12 at 14:54
up vote 4 down vote accepted

I think you need to take another look at (5). If it fails to be UC on $(0,\infty)$ because the derivative is unbounded near $0$, it should fail to be UC on $(0,1)$ for the same reason. If it is UC on $(0,1)$, then the same reasoning should show that it is UC on $(0,\infty)$--since in this case, it is UC on $[1,\infty)$ and is continuous on $(0,\infty)$. I leave it to you to determine which of your arguments is correct.

As for the rest, it looks good, for the most part. One can make an even nicer argument for $\sin x$ being UC, since it is continuous (so UC) on $[0,2\pi]$ and periodic with period $2\pi$, so UC on $\mathbb{R}$, so UC on any subset of $\mathbb{R}$.

Something that may help you to clear up (4): A function $f:E\to\mathbb{R}$ (with $E\subseteq\mathbb{R}$) fails to be uniformly continuous if there is some $\varepsilon>0$ and some sequences $\{x_n\}$ and $\{y_n\}$ of points of $E$ such that $x_n-y_n\to 0$ and such that $|f(x_n)-f(y_n)|\geq\varepsilon$ for all $n$. Consider the points that cause $\sin(1/x)$ to maximize as $x\to 0$ and the points that cause $\sin(1/x)$ to minimize as $x\to 0$, and that should give you your sequences.

share|cite|improve this answer
Cameron thank you for the sum up...for $4$ we can take $x_n=\frac{1}{2n\pi}$ and $y_n=\frac{1}{(2n+1)\pi}$ – Un Chien Andalou Jun 9 '12 at 14:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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