# $\frac{1}{x}$ not uniformly continuous

In my textbook we saw an example of a not uniformly continuous function, $$f(x) = \frac{1}{x}$$ but i find the explanation why kinda weird. First of all, this is the definition of uniform continuity in the textbook:

Suppose we have a function $$f(x)$$ with domain $$\mathcal{D}$$ and a set $$A$$ for which $$A\subseteq \mathcal{D}$$. Then $$f$$ is uniformly continuous over $$A$$ if the following formula is true:

$$\begin{equation} (\forall \varepsilon>0)(\exists\delta_{\varepsilon}>0)(\forall x,x' \in A)(|x'-x|<\delta_{\varepsilon} \Longrightarrow \left|f(x)-f(x')\right|<\varepsilon) \end{equation}$$

Now the example was: $$f(x) = \frac{1}{x}$$ is continuous over $$]0,1]$$, but not uniformly continuous.

Because: from uniform continuity would follow that $$0<\delta<1$$ with the property that:

$$\begin{equation} (\forall x,y \in ]0,1])\left(|x-y|<\delta \Longrightarrow \left|\frac{1}{x} - \frac{1}{y}\right|<1\right) \end{equation}$$

and in case that $$y := \delta$$: $$\begin{equation} (\forall x \in ]0,\delta])\left(\left|\frac{1}{x} - \frac{1}{\delta}\right|<1\right) \end{equation}$$

which means that $$\frac{1}{x}$$ needs to be bounded over $$]0,\delta[$$ which is not true.

So first of all i don't know why they choose $$0<\delta<1$$ and why this needs to be the case. I also don't quite get why they then choose $$y := \delta$$ and how they then draw their conclusion. I think I understand the definition of uniform continuity and looked up many other examples and explanations on both this forum and other websites but I don't understand this example. Would someone be able to explain this?

In the definition of uniform continuity with $$\epsilon =1$$ you can always replace $$\delta$$ by any smaller number. For any $$x,y \in (0,1)$$ we have $$|x-y| <1$$ so taking $$\delta \geq 1$$ in $$|x-y| <\delta$$ cannot lead to any contradiction. So you take $$\delta <1$$. The reason for taking $$y=\delta$$ is $$\{x:|x-y|<\delta\}=(y-\delta,y+\delta)\cap (0,1)$$ and, in order to arrive a contradiction you want to take $$x$$ close to $$0$$. If $$y=\delta$$ then $$(y-\delta,y+\delta)\cap (0,1)=(0,2\delta)\cap (0,1)$$ and this interval contains points as close to $$0$$ as you want. You arrive at a contradiction by noting that $$|\frac 1 x -\frac 1 y| <1$$ cannot hold when $$x$$ is close to $$0$$ (because LHS $$\to \infty$$ as $$x \to 0$$.
Hint: set $$y=\frac{x}{2}$$ then $$\left|\frac{1}{x}-\frac{1}{y}\right|=\left|\frac{1}{x}-\frac{1}{\frac{x}{2}}\right|=\left|-\frac{1}{x}\right|=\frac{1}{x}>1$$ since $$x,y\in (0,1)$$ if $$\epsilon=1$$
Asserting that $$f$$ is not uniformly continuous means that there is some $$\varepsilon>0$$ such that$$(\forall\delta>0)(\exists x,y\in(0,1]):\lvert x-y\rvert<\delta\text{ and }\left\lvert\frac1x-\frac1y\right\rvert\geqslant\varepsilon.\tag1$$This holds for $$\varepsilon=1$$. Take any $$\delta>0$$ and pick $$y\leqslant\delta$$ (with $$y\in(0,1]$$) and pick $$x\in(0,y]$$ such that$$\frac1x-\frac1y\geqslant1.\tag2$$ Such an $$x$$ exists, because$$\frac1x-\frac1y\geqslant1\iff x\leqslant\frac1{1+\frac1y}=\frac y{y+1}.$$Now, it follows from $$(2)$$ that $$(1)$$ holds.