I am currently going through this pdf https://www.csie.ntu.edu.tw/~b89089/book/Apostol/ch9.pdf and in Exercise 9.2b, page 3, we have the following question.
Prove that $h_n(x)$ does not converges uniformly on any bounded interval.
\begin{align} h_n(x)=\begin{cases}\frac{x}{n}\left(1+\frac{1}{n}\right) & \text{if}\;x=0\;\text{or}\;x\in \Bbb{Q}^c,\\a+\frac{a}{n}\left(1+\frac{1}{b}+\frac{1}{bn}\right)& \text{if}\;x\in \Bbb{Q}\end{cases}\end{align}
PROOF
\begin{align} \lim\limits_{n\to \infty}h_n(x)=\begin{cases}0 & \text{if}\;x=0\;\text{or}\;x\in \Bbb{Q}^c,\\a& \text{if}\;x\in \Bbb{Q}\end{cases}\end{align} Suppose for contradiction that $h_n(x)$ converges uniformly on any bounded interval $I$, then $h_n(x)$ converges uniformly on $[c,d]\subset I$. So, given $\epsilon=\max\{|c|,|d|\}>0$, there exists a positive integer $N,$ such that as $n\geq N,$ we have \begin{align}\max\{|c|,|d|\}> |h_n(x)-h(x)|=\begin{cases}\left|\frac{x}{n}\left(1+\frac{1}{n}\right)\right| & \text{if}\;x=0\;\text{or}\;x\in \Bbb{Q}^c\cap [c,d],\\\left|\frac{a}{n}\left(1+\frac{1}{b}+\frac{1}{bn}\right)\right|& \text{if}\;x\in \Bbb{Q}\cap [c,d]\end{cases}\end{align} which implies that $(x\in \Bbb{Q}^c\cap [c,d])$ or $x=0$
\begin{align}\max\{|c|,|d|\}> \left|\frac{x}{n}\left(1+\frac{1}{n}\right)\right| \geq \frac{|x|}{n}\geq \frac{\max\{|c|,|d|\}}{n},\;\forall\;n\geq N,\end{align} which is absurd. So, $h_n(x)$ does not converges uniformly on any bounded interval.
QUESTION
What is the contradiction in, \begin{align}\max\{|c|,|d|\}>\frac{\max\{|c|,|d|\}}{n},\;\forall\;n\geq N?\end{align} Or I'm I missing something? I would be happy if the above can be explained to me. Thanks!
The solution in the linked pdf does not make any sense: it does not use at all the values of $h_n$ at rationals, while, in fact, the convergence is uniform on bounded subsets of irrationals.
The correct argument comes from the observation that for any every $M>0$, any interval $(c,d)$ (I take $c>0$ for definiteness) contains a rational point $x_M=a/b$ with $a>M$ and $a,b$ relatively prime integers. (Indeed, there are infinitely many rationals in the interval, but if $a\leq M$, then $a/b>c$ implies $b\leq M/c$, and there are only finitely many pairs of positive integers satisfying these bounds.) Therefore, $$ h_M(x_M)-h(x_M)\geq 1+\frac{1}{b}+\frac{1}{bM}\geq 1, $$ for any $M$. This is incompatible with the uniform convergence.