Sequence of continuous functions that has no convergent subsequence

Question

Let $B(\mathbb{R})$ be the space of all bounded, continuous real-valued functions in $\mathbb{R}$. Let $p$ be the metric $p(f,g) = \sup_{x\in\mathbb{R}} |f(x) - g(x)|$ for $f,g \in B(\mathbb{R})$.

How can I find a sequence of continuous functions $f_{n}$ such that for all $n \in N$ we have $\sup_{x\in\mathbb{R}} |f_{n}(x)| < 1$, and the sequence $f_{n}$ has no convergent subsequence in $B(\mathbb {R},p)$?

I've seen in many places that it is suggested to use the Arzelá-Ascoli theorem, but that theorem hasn't been covered in my course, so I'm a bit confused.

Answer 1

You could imagine something like a pulse of height $\frac{1}{2}$, which is translated more and more to the right as $n$ increases. If you make sure the pulses don't overlap, then the distance between any $f_n$ and $f_m$ is $\frac{1}{2}$, so you certainly can't have a convergent subsequence.

Answer 2

$f_n(x)=\frac 1 2 \sin (x^{n})$ is such a sequence. Note that $f_n(x) \to 0$ for $|x|<1$. If $f_{n_k}$ converges uniformly then , for any $\epsilon >0$, there exists $m$ such that $|f_{n_k}(x)| <\epsilon $ for all $k \geq m$ and for all $x\in (-1,1)$. Now let $x \to 1$ in this to get $|\sin 1| \leq 2\epsilon$. Choosing $\epsilon <\frac 1 2 \sin 1$ we get a contradiction .