To show that $f(x)= \frac{1}{x}$ is not uniformly continuous on $(0,\infty)$

ATTEMPT Choose $\epsilon$ to be less than 1 Let $x_1=\delta$ and $x_2=\frac{\delta}{2}$

Then $|x_1-x_2|$ is less than $\delta$

$|f(x_1)-f(x_2)|= \frac{1}{\delta}$ which is bigger than 1(choosing $\delta$ to be less than 1) , and so bigger than $\epsilon$

Now by Archimedian property choose

$\frac{1}{\delta}$ greater than $\epsilon$

so this violates our condition of uniform continuity



  • $\begingroup$ What is $\varepsilon$? I suggest you write the "epsilon-delta" definition of not being uniformly continuous on $(0,\infty)$. $\endgroup$
    – Augustin
    Aug 25, 2015 at 8:44
  • $\begingroup$ @Arthur can you please provide technical details? $\endgroup$
    – Taylor Ted
    Aug 25, 2015 at 8:49
  • $\begingroup$ @Arthur Actually, you don't have $\epsilon$, you choose $\epsilon$ such that for all $\delta$... $\endgroup$
    – user110822
    Aug 25, 2015 at 8:52
  • $\begingroup$ @user110822 Yes, I was a bit sloppy there. $\endgroup$
    – Arthur
    Aug 25, 2015 at 9:02

2 Answers 2


This is mathematically incorrect, because you did not show what statement you want to disprove.

Your attempt is a very typical way a beginner approaches a mathematical proof, and that is to simply throw every possible thing they think is correct out. There is no structure, no story told in your proof.

Your attempt needs to start with:

$f$ is uniformly continuous if, for every $\epsilon > 0$, there exists a $\delta>0$ such that for all $x,y\in(0,\infty)$, if $|x-y|<\delta$, we know that $|f(x)-f(y)|<\epsilon$.

Now, since you are disproving uniform continuity, you need to prove the statement:

There exists a $\epsilon > 0$ such that for every $\delta > 0$, there exist such $x,y$ that $|x-y|<\delta$ and $|f(x)-f(y)|>\epsilon$.

You do this by actually setting a certain value of $\epsilon$ (for example, $\epsilon=1$ is possible), then taking an arbitrary value of $\delta$ and finding the correct pair of $x,y$ to prove your point.

Remember, the statement says: There exists such an $\epsilon$ that for every $\delta$, there exists a pair $x,y$, such that something is true.

You cannot, therefore, choose your $\epsilon$ after you chose your $\delta$.


here is my proof of the statement

There exists such an $\epsilon>0$ that for every $\delta>0$, it is true that $\epsilon <\delta$.

  • Let $\delta > 0$
  • Then, there exists some $n\in \mathbb N$ such that $\frac1n\leq\delta$
  • Then, I choose to set $\epsilon = \frac{1}{n+1}$.
  • Since $\frac1{n+1}<\frac1n\leq \delta$, I conclude that $\epsilon < \delta$ and the statement is proven.

Can you see the error in my argument?

  • $\begingroup$ i have proved that $|f(x_1)-f(x_2)|$ is greater than epsilon, by the help of archimedian property $\endgroup$
    – Taylor Ted
    Aug 25, 2015 at 8:51
  • $\begingroup$ @TaylorTed See the last sentence of my edited post. $\endgroup$
    – 5xum
    Aug 25, 2015 at 8:53
  • $\begingroup$ Yes i got the point. If i choose(in beginning) epsilon to be 1 and delta to be less than 1. Will that work> $\endgroup$
    – Taylor Ted
    Aug 25, 2015 at 8:54
  • $\begingroup$ @TaylorTed Edit your original post with the updated proof. Then we can talk whether it is now OK or not. And try to use the correct terminology (i.e., use phrases like "let us choose an arbitrary $\delta>0$." $\endgroup$
    – 5xum
    Aug 25, 2015 at 8:57
  • $\begingroup$ Is error is the equal to sign. Also please check edited question $\endgroup$
    – Taylor Ted
    Aug 25, 2015 at 9:07

After choosing your $\epsilon$, you have to find $x(\delta)$ and $y(\delta)$ such that $|x - y| < \delta$ and $|1/x - 1/y| \ge \epsilon$.


Find a suitable constant $c$ such that:

$$x = \delta, y = \frac{\delta}{1 + c \delta}$$

You went wrong when you mixed up the concepts. You want to disprove uniform continuity. The statement of uniform continuity is: "$\forall $ $\epsilon > 0$, $\exists$ $\delta > 0$, such that $\forall $ $x,y$, $|f(x) - f(y)| < \epsilon$.

Thus, to prove a function not uniformly continuous, you should prove that:

$$\exists \ \epsilon > 0, \forall \ \delta > 0, \exists \ x,y \text{ functions of $\delta$, such that} \ |x - y| < \delta \text{ and } |f(x) - f(y)| \ge \epsilon$$

So you choose $\epsilon$ in an independent manner of $\delta$, because you don't know the delta at that stage. In other words, the choice of $\epsilon$ is what you begin the whole process with, so you can't choose it at some later time.


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