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

Question

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

IS THIS CORRECT ?

Thanks

Answer 1

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 $\delta$.

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Example:

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?

Answer 2

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

Hint:

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.