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?
Best Answer
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 $x \to 0$.