Rudin PMA 4.20 - how can this function be unbounded ? Considering Rudin hasn't introduced "divergence" of functions yet in the chapter.

Question

Here is the very beggining of Rudin's Principles of Mathematical Analysis 4.20 theorem:

Let $E$ be a noncompact and bounded set in $\mathbb{R}^1$. Then there exists a continuous function on $E$ which is notuniformly continuous.

Proof:

"By condition there exists a limit point $x_0$ of $E$ which is not a point of $E$. Consider $$f(x)=\dfrac{1}{x-x_0}\quad (x\in E).$$ This is continuous on $E$, but evidently unbounded.

I am not able to appreciate how evidently the function $f$ is unbounded on $E$.

I have tried to suppose $f$ is bounded on $E$. As $\lim _{x\to x_0} |f(x)| = \infty$ I could possibly conclude. But, the problem is that at this point of the book, Rudin hasn't introduced the limit of a function "diverging" at a point (he has just introduced $\lim _{x\to x_0} f(x) = q \in Y$ where $Y$ is the image metric space of $f$). So, I suspect there is another way to prove this "evident" unboundedness and that Rudin had this in mind.

Answer 1

If $x_0$ is a limit point of $E$, then there is a sequence of points $x_n \in E$ such that $x_n \to x_0$ as $n \to \infty$.

This implies by the definition of a limit (already introduced by Rudin) that for any $K > 0$ there exists $m \in \mathbb{N}$ such that $|x_m - x_0| < \frac1{K}$ and $f(x_m) =\frac{1}{|x_m - x_0|} > K$. Hence $f$ is unbounded on $E$.

Answer 2

$\lim_{x\to x_0}|f(x)|=\infty$ is exactly a way to conclude that it is unbounded.

It is 'evident' because if $\left|\frac{1}{x-x_0}\right|<M$ for all $x\neq x_0$ we could contradict that with $x=x_0+\frac{1}{2M}$ (say). Rudin probably has in mind an intuitive sense of unboundedness or divergence, but honestly if you're studying uniform continuity (a more advanced concept) you should already know what is meant by that. If Rudin is supposed to teach from square one, then divergence really should have been taught earlier in the book!

Answer 3

Rudin introduced the concept of convergence in Chapter 3. But as you say, he did not define "convergence to the improper limit $\infty$".

However, the concept of convergence is not needed in the proof.

By Theorem 2.41 the non-compact set $E$ contains an infinite subset $F$ having no limit point in $E$. Since $E$ is bounded, also $F$ is bounded and by Theorem 2.42 it has a limit point $x_0 \in \mathbb R$. Clearly $x_0 \notin E$. Thus $x_0$ is a limit point of $E$ which does not belong to $E$. By Definition 2.18 we find for each $n \in \mathbb N$ an element $x_n \in E$ such that $\lvert x_n - x_0 \rvert < 1/n$. Thus $$\left\lvert \frac{1}{x_n - x_0} \right\rvert > n$$ which shows that the functon $f$ is unbounded on $E$.