Show that $x\mapsto \frac{1}{x}$ is not uniformly continuous for $x \in (0,1), \frac{1}{x} \in \mathbb{R}$

Question

I am meant to show that for $x \in (0,1)$, $\frac{1}{x}$ is not uniformly continuous.

I am able to come up with a proof but the given hint tells me to investigate $x_n = \frac{1}{n}$, $y_n = \frac{1}{n+1}$ and I'm not sure how to use that to construct a proof.

Is it something to do with Cauchy sequences?

Answer 1

Yes, it has to do with Cauchy sequences. A uniformly continuous function takes Cauchy sequences into Cauchy sequences.

Recalling the definition of uniformly continuous:

For each $\epsilon >0$ there exists $\delta >0$ such that for each $x,y\in D$ $$|x-y|< \delta\implies |f(x)-f(y)|<\epsilon$$

It's negation would be that there exists one epsilon such that for each $\delta >0$ there exists two points $x,y$ such that $|x-y|<\delta$ yet $|f(x)-f(y)|\geq \epsilon$. By taking $\delta=1,\frac 1 2,\frac 1 3,\dots$, we can write this as

There exists an $\epsilon >0$ and two sequence of points $x_n,y_n$ such that $$|x_n-y_n|<\frac 1 n\; ;\;n=1,2,3,\dots$$ but $$|f(x_n)-f(y_n)|\geq \epsilon \; ;\; n=1,2,3,\dots$$

In your case, $\epsilon=1$; $$x_n=\frac 1n $$ and $$y_n=\frac 1{n+1}$$ do the job.

Answer 2

Or even simpler(?): A uniformly continuos function on a bounded domain is bounded.

Answer 3

Suppose to the contrary, that f is uniformly continuous.

Fix $\epsilon$. By hypothesis, there is some $\delta$ such that, if

$$|y_n-x_n|<\delta,$$

then $|f(y_n)-f(x_n)|<\epsilon$. Can you fill in the details?