Limit function of uniform convergent bounded sequences is bounded 
TRY:
Let $(f_n)$ be given sequence of bounded functions: $|f_n(x) | \leq M$ for all $n$ and for all $x \in A$. 
Choose $\epsilon = 1$ and select $N$ such that for all $n \geq N$ one has $|f_n-f|  <1$. Then 
$$ |f(x)| = |f-f_n+f_n| \leq 1 + |f_n(x) | \leq 1+ M $$
and this is true for every $x \in A$. Thus $f(x)$ is bounded as desired.!
For point wise convergence, isn't just basically the same argument as above? Can someone clarify this to me? Im still trying to completely understand the difference between point wise and uniform 
 A: You're assuming what you need to prove. 
A sequence of bounded functions means that for each $n$, there is a $M_n>0$ such that $|f_n(x)| \le M_n$, for each $x \in A$.
Now, as you said, there is an $N$ such that $|f(x) - f_n(x)| < 1$ for $n > N$ and $x \in A$.
By Cauchy's criterion for uniform convergence, there is also an $N_1$ such that $m, n > N_1$ implies $|f_m(x) - f_n(x)| < 1$, for $x \in A$.
Then, $m, n > N + N_1$ implies $|f(x)| \le |f(x) - f_m(x)| + |f_m(x) - f_n(x)| + |f_n(x)|$ < $2 + M_n$, for $x \in A$.
As for pointwise convergence, consider $f_n(x) = \frac{1}{x} \cdot 1_{\left [\frac{1}{n}, 1 \right ]}$, for $x \in (0, 1]$.
A: I think you have to be careful with 2 facts:


*

*Not all $f_n$ must share the same bound. i.e. $\exists\; M_n \in \mathbb{R}$ s.t. $|f_n(x)| \leq M_n$ $\forall x\in A$. Your proof still works just change this fact.

*If $f_n$ converges pointwise to $f$, it is not true that given $\epsilon =1 $ you get a single $N$ that makes $|f(x) - f_n(x)| < 1$ $\forall n>N$ and $\forall x\in A$ . This is the big difference, in pointwise convergence you get an $N$ for each $x$ which need not be the same for all $x$.


As for why is it not true for pointwise, check the following counterexample:
$f_n(x): (0,1) \rightarrow \mathbb{R}$
$$f_n(x) = \frac{n}{nx+1}$$
It converges pointwise to $ \frac{1}{x}$ which is not bounded in $(0,1)$.
