Uniform convergence with supremum norm Let $X$ an arbitrary set and $(f_n)$ a sequence of functions, $f_n:X\to \mathbb{R}\,\,n\in\mathbb{N}$. We say that the sequence $(f_n)$ converges uniformly to a function $f:X\to \mathbb{R}$ if given any $\epsilon>0$ there is a positive integer $N=N(\epsilon)$ such that  $$|f_n(x)−f(x)|<\epsilon \qquad \text{for every}\, n>N \,\text{and for every}\, x\in X.$$
Some authors also introduce the so called "supremum-norm": given any function $g:X\to \mathbb{R}$, we define $$\left\lVert g\right\rVert_{\infty}:=\sup_{x \in X}|g(x)|.$$ It is trivial to show that $0\leq\left\lVert g\right\rVert_{\infty}\leq +\infty$ for every $g$ and $\left\lVert g\right\rVert_{\infty}<+\infty$ if and only if $g$ is bounded. 
They also show that given $(f_n)$ a sequence of functions, $f_n:X\to \mathbb{R}\,\,n\in\mathbb{N}$, then $(f_n)$ converges uniformly to a function $f:X\to \mathbb{R}$ if, and only if, the (numerical) sequence $\{\left\lVert f_n-f\right\rVert_{\infty}\}_n$ converges to $0$.
My doubt is the following: without specifying anything about boundedness of $f_n,f,$ then  $\{\left\lVert f_n-f\right\rVert_{\infty}\}_n$ is a sequence in the extended half line $[0,+\infty]$. 
How is defined convergence to $0$ (or to any other $l\in [0,+\infty]$) in this case? I think this is important in order to understand the equivalence between the two definitions of uniform convergence.
Thanks a lot in advance.
 A: I must say that this way of trying to see uniform convergence is a bad way.
The convergence $f_n \to f$ in this norm is the same: it converges if and only if the numerical sequence $\sup_{t\in[0,\infty]} |f_n(t) - f(t)| \to 0$. However, the definition of your functions gets more convoluted, since now you need to specify the values $f_n(\infty)$ and $f(\infty)$, and check convergence of $|f_n(\infty) - f(\infty)|$.
If the functions are actually continuous, you can set $f_n(\infty) := \lim_{t\to\infty}f_n(t)$ to still have continuous functions. Note that in this way you've actually bounded your $f_n$, and it doesn't make any difference whether you're working in the usual real line or in the extended one, since:
\begin{equation}
\sup_{t\in[0,\infty]} |f_n(t) - f(t)| = \sup_{t\in[0,\infty)} |f_n(t) - f(t)| 
\end{equation}

Topological note: Your example could actually be replaced by uniform convergence in $[0,1]$, since the extended real line $[0,\infty]$ is actually homeomorphic to $[0,1]$. For example, take the function:
\begin{align}
\phi: (0,1] &\to [0,\infty)\\
t &\mapsto\phi(t) = \frac{1}{t}-1
\end{align}
and extend it to $t=0$ by continuity as $\phi(0) = \lim_{t \to 0}\phi(t)=\infty$. Similarly, we can take:
\begin{align}
\psi: [0,\infty)&\to (0,1]\\
s&\mapsto\frac{1}{s+1}
\end{align}
and again by continuity extend it to $t = \infty$ as $\psi(\infty) = \lim_{t\to\infty}\phi(t)=0$. This way, $\phi$ and $\psi$ are continuous functions that are inverses of each other, hence a homeomorphism.
A: We  can define $\|f\|=\min (1,\sup_{x\in X}|f(x)|),$ so that $d(f,g)=\|f-g\|$ is a metric on $^X\Bbb R.$
Then $(f_n)_n$ converges uniformly to $f$ iff $\|f_n-f\|\to 0$ iff $f_n\to f$ in the metric space $^X\Bbb R.$
