Theorem 4.20 in Baby Rudin: How is this map not uniformly continuous?

Question

Let $E$ be a bounded, non-compact subset of $\mathbb{R}$, let $x_0$ be a limit point of $E$ such that $x_0 \not\in E$, and let $f \colon E \to \mathbb{R}$ be defined by $$f(x) \colon= \frac{1}{x-x_0} \ \ \forall x \in E.$$ Then how to explicitly and rigorously show that the map $f$ is not uniformly continuous on $E$? Of course, $f$ is continuous on all of $E$!

I've not been able to fully convince myself as to the soundness of the argument adduced by Rudin.

My effort:

For any points $x$, $y \in E$, we have $$\vert f(x) - f(y) \vert = \frac{\vert x - y \vert}{\vert x-x_0 \vert \cdot \vert y-x_0\vert}.$$ Let $\delta > 0$ be arbitrary. Take $x$, $y \in E$ such that $\vert x - x_0 \vert < \min(1, \frac{\delta}{4})$, $\vert y - x_0 \vert < \min(1, \frac{\delta}{4})$. Then we have $$\frac{1}{\vert x-x_0\vert } > 1,$$ $$\frac{1}{\vert y-x_0\vert } > 1,$$ $$\vert x - y \vert \leq \vert x - x_0 \vert + \vert x_0 - y \vert < \frac{\delta}{4} + \frac{\delta}{4} = \frac{\delta}{2} < \delta.$$ Howvever, $$\vert f(x) -f(y) \vert = \frac{\vert x - y \vert}{\vert x-x_0 \vert \cdot \vert y-x_0\vert} > \frac{\vert x -y \vert}{1} \geq \vert x-x_0 \vert - \vert y-x_0\vert.$$

What next? How to proceed from here?

Or, is there some other way of stipulating the bounds on the distances of $x$ and $y$ from $x_0$?

We have to find some real number $\epsilon > 0$ such that there are points, say, $x_\epsilon$, $y_\epsilon \in E$ for which $$\vert x_\epsilon - y_\epsilon \vert < \delta,$$ but $$\left\vert f(x_\epsilon) - f(y_\epsilon) \right\vert \not< \epsilon.$$

Answer 1

$\newcommand{\Reals}{\mathbf{R}}$You've got what appears to be a good proof. Here's a sketch of a rigorous alternative (possibly more conceptual) approach:

Lemma 1: If $f:E \to \Reals$ is uniformly continuous, and if $(x_{n})$ is a Cauchy sequence in $E$, then $\bigl(f(x_{n})\bigr)$ is a Cauchy sequence.

Lemma 2: If $f:E \to \Reals$ is uniformly continuous, then $|f|:E \to \Reals$ is uniformly continuous. (Generally, a composition of uniformly continuous functions is uniformly continuous.)

(The proof of each involves little more than walking through the definitions.)

Now, since $x_{0}$ is a limit point of $E$ but is not an element of $E$, for each positive integer $n$ there exists an $x_{n}$ in $E$ such that $0 < |x_{n} - x_{0}| < 1/n$. The sequence $(x_{n})_{n=1}^{\infty}$ is (almost obviously) Cauchy, but $$ |f|(x_{n}) = \frac{1}{|x_{n} - x_{0}|} > n\quad\text{for each $n$.} $$ Consequently, $\bigl(|f|(x_{n})\bigr)_{n=1}^{\infty}$ is not Cauchy. Contrapositively, Lemma 1 implies $|f|$ is not uniformly continuous, and Lemma 2 implies $f$ itself is not uniformly continuous.

Answer 2

I've finally come up with a proof. Here it is:

Let's choose a real number $\epsilon$ such that $0 < \epsilon < 1$. Let $\delta > 0$ be any real number. Now we choose points $x$, $y \in E$ as follows:

Choose $x \in E$ such that $0 < \vert x - x_0 \vert < \min(1, \frac{\delta}{3})$, and choose $y \in E$ such that $0 < \vert y-x_0 \vert < \frac{1}{2} \vert x - x_0 \vert = \vert \frac{x+x_0}{2} - x_0 \vert$. Such points $x$ and $y$ do exist in $E$ since $x_0$ is a limit point of $E$ (but not an element of $E$).

Then we find that $$\vert x - y \vert \leq \vert x - x_0 \vert + \vert y - x_0 \vert \leq \frac{3}{2} \vert x - x_0 \vert < \frac{3}{2} \frac{\delta}{3} = \frac{\delta}{2} < \delta. $$ However, \begin{eqnarray*} \vert f(x) - f(y) \vert &=& \left\vert \frac{1}{x-x_0} - \frac{1}{y-x_0} \right\vert \\ &>& \left\vert \frac{1}{x-x_0} - \frac{1}{\frac{x+x_0}{2} - x_0 } \right\vert \text{ [Is this step correct?] } \\ &=& \frac{1}{\vert x - x_0 \vert} \\ &>& 1 \\ &>& \epsilon. \end{eqnarray*}

Is this proof correct?