Prove that $f_n$ is uniformly bounded on $E$ and $f$ is a bounded function on $E$

Question

I have the solution for the following problem, but I don't understand most of it.

Question: A sequence of functions $f_n$ is said to be uniformly bounded on a set $E$ iff there exists $M>0$ such that $|f_n(x)\le M$ for all $x\in E$ and all $n\in \mathbb{N}$.

Suppose that for each $n\in\mathbb{N}$, $f_n:E\to\mathbb{R}$ is bounded. If $f_n\to f$ uniformly on $E$, as $n\to\infty$, prove that ${f_n}$ is uniformly bounded on $E$ and $f$ is a bounded function on $E$.

Solution:

Choose $N$ so large that $|f(x)-f_n(x)|<1$ for all $x\in E$ and $n\ge N$.

Why does it have to be less than $1$?

Set $M:=\sup_{x\in E}|f_N(x)|$ and observe by the Triangle Inequality that $$|f(x)|\le|f(x)-f_N(x)|+|f_N(x)|<1+M \text{ for all }x\in E\text{.}$$

Therefore, $|f_n(x)|\le|f(x)|+1\le(1+M)+1=2+M$ for all $n\ge N$ and $x\in E$, i.e., $\{f_n\}_{n\ge N}$ is uniformly bounded on $E$.

I understand this, but not what is below. What exactly is it trying to show? And how does that prove that $f$ is a bounded function on $E$?

In particular, $$|f_n(x)\le\hat{M}:=\max\{2+M,\sup_{x\in[a,b]}|f_1(x)|,...,\sup_{x\in[a,b]}| f_{N-1}(x)|\}<\infty$$ for all $n\in\mathbb{N}$ and $x\in E$.

Answer 1

Answers, in order:

• the choice of $1$ was arbitrary; this would work for any $\epsilon > 0$, so we might as well choose $\epsilon = 1$. As Daniel Fischer notes, $\epsilon = \pi$ and $\epsilon = 10^{10^{1000}}$ would work just as well, for the purposes of the proof.
• in the above, we've shown that $|f(x)| \leq 1 + M$, so we know that $f$ is bounded. We also know that for any $n \geq N$, $f_n$ is also bounded by $2+M$ so that $\{f_n\}_{n \geq N}$ is uniformly bounded. What remains to be shown is that the first elements $\{f_1,\dots,f_{N-1}\}$ have a common bound, so that the entire sequence $\{f_n\}_{n \geq 1}$ can be shown to be bounded. In the last line, we establish that the entire sequence is in fact uniformly bounded.