Exercise 9.2 from Apostol's Mathematical Analysis book. Uniform convergence of product This is a problem (Exercise 9.2) from Apostol's Mathematical Analysis (second edition) which I couldn't solve. 
$\bullet$ Define two sequences $\{f_{n}\}$ and $\{g_{n}\}$ as follows:
$f_{n}(x) = x \Bigl(1 + \frac{1}{n}\Bigr)$ for $x \in \mathbb{R}$, $n \in \mathbb{N}$
$g_{n}(x)= \begin{cases}
 \frac{1}{n} & x=0 \  \text{or}\  x \in \mathbb{R} - \mathbb{Q} \\
 b+\frac{1}{n} & x \in \mathbb{Q},\ \text{say}\  x=\frac{a}{b}, b>0
\end{cases}$
$\mathbb{R}$ is the set of reals, $\mathbb{Q}$ is the set of rational and $\mathbb{R} - \mathbb{Q}$ is the set of irrationals.
I have to show that if $h_{n}(x)=f_{n}(x)g_{n}(x)$, then $h_{n}(x)$ does not converge uniformly on any bounded interval. 
How can I show this?
 A: What is the pointwise limit of $f_n(x)g_n(x)$ as $n\rightarrow\infty$? What is the pointwise difference between the limit and the value for a given $n$? Show that, for any $n$, this difference is unbounded on any bounded interval. This follows pretty much immediately from the fact that denominators of rational numbers are unbounded on any bounded interval.
A: Let $I$ be a bounded interval. If $x=0$ or irrational, then $h_n (x) \to 0$. If $x$ is a rational of the form $x=a/b$, $b>0$, then 
$$
h_n (x) = \frac{a}{b}\bigg(1 + \frac{1}{n}\bigg)\bigg(b + \frac{1}{n}\bigg) = a + \frac{a}{b}\frac{1}{n} + \frac{a}{n} + \frac{a}{b}\frac{1}{{n^2 }},
$$
hence $h_n (x) \to a$. Hence, $h_n$ converges pointwise on $I$; denote the limit by $h$. Since a uniform limit is a pointwise limit, it thus suffices to show that $\sup _{x \in I} |h_n (x) - h(x)| \ge 1$ for infinitely many $n$. Indeed, for any $n \in \mathbb{N}$, there exists a rational number $x_n \in I$ of the form $x_n=a/b$, $b>0$, such that $|a| \geq 2n$. Hence, if we choose $n$ sufficiently large, then
$$
|h_n (x_n) - h(x_n)| = \Big|\Big(a + \frac{a}{b}\frac{1}{n} + \frac{a}{n} + \frac{a}{b}\frac{1}{{n^2 }}\Big) - a\Big| > \frac{1}{2}\Big|\frac{a}{n}\Big| \geq 1
$$
(note that $a/b$ is uniformly bounded, since $I$ is a bounded interval).
The claim is thus established.
A: Assume that $h_n \to h$ uniformly on the domain $D$ of $\{h_n\}$, where
$$h_n(x) = 
\begin{cases} 
 \frac{x}{n}  \biggl(1 + \frac{1}{n} \biggr), & \text{if } x \text{ is   irrational} \\
 a + \frac{a}{n} + \frac{a}{b} \biggl(1 + \frac{1}{n} \biggr) \biggl(\frac{1}{n} \biggr), & \text{if } x \text{ is rational, say } x = a/b, b > 0
\end{cases}$$ 
and $$h(x) = 
\begin{cases}
 0, & \text{if }x\text{ is irrational} \\
 a, & \text{if }x\text { is rational, say }x = a/b, b > 0.
\end{cases}$$
If there is a positive rational in $D$ we can find an interval $A = [s,t]$ contained in $D$ with $s$ rational because $D$ is a bounded interval.  Then given $\epsilon = 1$ there exists a positive integer $M$ such that $n \ge M$ implies that
\begin{align}
 \biggl|h_n\biggl(\frac{a}{b}\biggr) - h\biggl(\frac{a}{b}\biggr) \biggr|        &=   \
 \biggl| \frac{a}{n} + \frac{a}{b} \biggl( \frac{1}{n} + \frac{1}{n^2} \biggr) \biggr| &< 1 
\end{align}
whenever $n \ge$ M and $a/b \in A$ where $a$ and $b$ are positive integers.
But then
\begin{align}
 \frac{a}{M} &< \frac{a}{M} + \frac{a}{b} \biggl( \frac{1}{M} + \frac{1}{M^2} \biggr) \
  &= \biggl| \frac{a}{M} + \frac{a}{b} \biggl( \frac{1}{M} + \frac{1}{M^2} \biggr) \biggr| \
  &< 1. 
\end{align}
Writing $s = c/d$ where $s$ is the left endpoint of $A$ and $c$ and $d$ are positive integers, we can choose a positive integer $k$ large enough so that both $(ck + 1)/d \in A$ and $ck + 1 > M$.  
This contradicts $(ck + 1)/M < 1$.  Therefore the assumption that $h_n \to h$ uniformly is false.  A similar argument holds if no positive rationals are in $D$
