How to show a set of functions is or is not an open set on the sup-metric? This is an excerpt from my text:

The set $G$ of functions $g:\mathbb{R}\rightarrow\mathbb{R}$ such that $|g(x)|\le 1$ for all $x$ is not an open set in the sup-metric space. For instance, consider the function $g(x)=\frac{2}{\pi}\tan^{-1}(x)$: There is no positive $\epsilon$ such that $B_\epsilon(g)\subseteq G$.

I don't understand why the last statement is true. So I am attempting to prove it myself.
I am able to show that $g(x)=\frac{2}{\pi}\tan^{-1}(x)$ is in $G$. So, the next step is to attempt to construct an open ball around $g(x)$ and realise that no matter what radius, $\epsilon$ I choose, that open ball will not be in $G$.
So, for some $\epsilon>0$,
$B_\epsilon(g)=\{f(x)\in G  :\, d_\infty(f(x),g(x))<\epsilon \}$.
Where $d_\infty(x,y)$ is the sup-metric.
I don't see how to get a contradiction from the definition of $B_\epsilon(g)$. Please help.
In fact, I am not even sure what an open ball of a set of functions looks like. I feel that the method to show that the open ball is a or is not an open set should be analogous to doing the same in $\mathbb{R}^n$, but I just don't see a connection.
So, can I also have an example when the open ball of a set of functions is open and how to show it is true? The treatment of my text on topic is very scarce.
 A: Suppose $\lim_{x \to \infty} |g(x)|  =1$,  and let $\epsilon>0$.
Let $f(x) = (1+ {1 \over 2}\epsilon\operatorname{sgn} g(x))g(x)$. Then $f \in B(g,\epsilon)$, but since $\lim_{x \to \infty} |f(x)|  =1+ { 1\over 2} \epsilon$, we see that $f \notin G$ (since for some finite $x$, we have $|f(x)|>1$).
We see that $\lim_{x \to \infty} {2 \over \pi} \arctan x = 1$
A: Hint: The function $f(x) \pm \epsilon/2$ is inside the ball centered around $f$ with radius $\epsilon$ according to the definition of sup norm. So what happens if your function either takes on the value $\pm 1$, or approaches the value $\pm 1$ as $x \to  \pm \infty$ or $x \to c$ for some $c$? Or more generally, what if you have a function that takes on various values $\pm (1 - \delta_i)$ for a sequence $\delta_i$ that goes to 0?
A: An $\epsilon$ neighbourhood around a function $f$ is the set of all functions $g$ such that $|f(x)-g(x)|<\epsilon$ for ALL $x\in\mathbb{R}$.
Observe that, geometrically, what happens is that $f=(2/\pi)\tan^{-1}(x)$ tends towards $\pm 1$ in the extremes. Consider the function $g$ define by 
$$ g(x)=f(x)+\epsilon/2.$$
By our definition, $g$ is in the $\epsilon$ open neighbourhood of $f$. However, one can always find $x_0$ such that $1-f(x_0)<\epsilon/2$ (since $\lim_{x\to\infty}f(x)=1$. But in this which case we have $g(x_0) > 1$.
And this is true for ALL $\epsilon$. Which means that there cannot be any open $\epsilon$-neighbourhood around $f$ contained in the set of $\{f\  \mid |f(x)|\leq 1\; \text{for all }x\in \mathbb{R}\}$.
Picture:

The key point of an $\epsilon$-nbhd is that the all the functions whose graphs lie inside the shaded area, are a part of this nbhd. As you can see, since $f$ aproaches $\pm 1$, you can always find a function whose graphs is in this shaded area but it leaves the $\pm 1$ bound.
Any open set in the sup metric can be express as a union of these $\epsilon$-nbhds. In other words, they are a basis for the sup topology.
A: When I imagine an open ball in this space, I try to have in mind the graphic of the center function and a strip $2\epsilon$ width around it. Any function whose graph is strictly contained in this strip is in the ball.
Now try to imagine the strip around $\tan^{-1}$ graph. Must this strip (no matter how narrow is) go out the limit at $y=1$? Does the fact that
$$\lim_{x\to\infty}\tan^{-1}(x)=1$$
have anything to do with it?
