Proof that the following set of bounded, continuous functions comprises a metric space? I need to prove that $X=C[0,T]$ is a metric space (i.e. the set of continuous functions in the interval $0$ to $T$).
The metric in use is
\begin{equation}
d(f,g)=\max_{{}t\in[0,T]}|f(t)-g(t)|
\end{equation}
I'm just making sure that I have everything needed to prove the conditions of a metric space. So looking at the three conditions:

*

*$d(f,g)\gt0 \; (f\ne g)$:

Does this follow immediately from the fact that we are taking the absolute value of the distance between the two distinct function values? Why is it not possible that $f(t)$ and $g(t)$ have the same values in the range of $t$, to give $|f-g|=0$?


*$d(f,g)=d(g,f)$:

Again, is this just a case of saying $|f-g| = |g-f|$?


*$d(f,g)\le d(f,h)+d(h,g)$ for $h \in X$:

So in other words, \begin{equation}\max_{{}t\in[0,T]}|f(t)-g(t)|\le\max_{{}t\in[0,T]}|f(t)-h(t)|+\max_{{}t\in[0,T]}|h(t)-g(t)|\end{equation}
This is the condition I'm not entirely sure how to prove.
If anyone could help me to consolidate my understanding of how this proof goes, I would be very thankful!
 A: For a specific $x\in [0,T]$, it will always be the case that
$|f(x)-g(x)|$
$\leq|f(x)-h(x)|+|h(x)-g(x)|$
$\leq \max_{{}t\in[0,T]}|f(t)-h(t)|+\max_{{}t\in[0,T]}|h(t)-g(t)|=K$.
If $|f(x)-g(x)|\leq K$ for all $x\in [0,T]$,
then also $\max_{{}t\in[0,T]}|f(t)-g(t)|\leq K$.
A: For 1, if $f \neq g$, then there is a $t \in [0, T]$, such that $f(t) \neq g(t)$, so by its definition, $d(f, g) \ge |f(t) - g(t)| > 0$. So $d(f, g) \neq 0$ if $f \neq g$.
For 2, your reasoning is fine.
For 3, note that by the triangle inequality:
$$|f(t) - g(t)| \le |f(t) - h(t)| + |h(t) - g(t)|$$
So
\begin{align}
d(f, g) &= \max_{t \in [0, T]}|f(t) - g(t)| \\&\le \max_{t \in [0, T]}(|f(t) - h(t)| + |h(t) - g(t)|) \\
 &\le\max_{t_1 \in [0, T]}(|f(t_1) - h(t_1)|) + \max_{t_2 \in [0, T]}(|h(t_2) - g(t_2)|) \\
 &= d(f, h) + d(g, h)
\end{align}
where, informally, the second inequality holds because on the right-hand side you can always do at a least as well as on the left-hand side, by taking $t_1$ and $t_2$ to be the same and you may be able to do better by taking them to be different. (I've changed $t$ to $t_1$ and $t_2$ for clarity, you don't actually need to do that.)
