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

Question

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?

Answer 1

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 $x \to 0$.

Answer 2

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 $$\epsilon=1$$

Answer 3

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.