Prove for the space of bounded functions $B(A)$ on a set $A$, that
$$d(x,y) = \sup_{t \in A} |x(t) - y(t)|$$
defines a metric. Below is my attempt.
Other properties aside, here I'm just concerned with showing that
$$d(x,y) \leq d(x,z) + d(z,y)$$
for any $z$ that's also a bounded function on $A$.
Proof.
\begin{align} d(x,y) = \sup_{t \in A}{|x(t) - y(t)|} &= \sup_{t \in A}{|x(t) -z(t) + z(t)- y(t)|} \\ \\ &\leq \sup_{t \in A}\{|x(t) -z(t)| + |z(t)- y(t)|\} \tag{1}\\ \\ &= \sup_{t \in A}\{|x(t) -z(t)|\} + \sup_{t\in A}\{|z(t)- y(t)|\} \tag{2}\\ \\ &= d(x,z) + d(z,y) \end{align}
I'm concerned with the supremum properties I used in (1) and (2). I assert (1), because the set $S_1 = \{|x(t) - z(t) + z(t)- y(t)|\}$ will be bounded above by any element of $S_2 = \{|x(t) - z(t)| + |z(t)- y(t)|\}$ for all $t\in A$, which implies the least upper bound of $S_2$ must be greater than or equal to the least upper bound of $S_1$. I assert (2) because
$$\sup\{D + E\} = \sup D + \sup E$$
if the largest elements of $D$ and $E$ are nonnegative. But that seems kind of like a leap since in (2) $t$ isn't, for example, an element of the the set $|x(t) - z(t)|$. So, what I actually was doing is taking the supremum of the set the elements $t\in A$ map to by the functions $f(t) = |x(t) - z(t)|$ and $g(t) = |z(t) - y(t)|$. In other words I'm claiming that for $g,f \in B(A)$, if $g,f \geq 0$, then
$$\sup_{t\in A}\{f(t) + g(t)\} = \sup_{t\in A}f(t) + \sup_{t\in A} g(t)$$
if we view $f$ and $g$ as sets then this should make sense I think.
I'm skeptical of my proof because it seems straight forward but the author (Kreyszig) in my text proved the result above in a different way:
For all $t\in A$ we have
\begin{align} |x(t) - y(t)| &\leq |x(t) - z(t)| + |z(t) - y(t)| \\ \\ &\leq \sup_{t \in A}|x(t) - z(t)| + \sup_{t\in A}|z(t) - y(t)| \end{align}
This shows $x-y$ is bounded on $A$. Since the bound given by the expression in the second line does not depend on $t$, we make the supremum on the left to obtain the desired result.
So part of my question, is why did he do it like that?
You are claiming that for $g,f \in B(A)$, if $g,f \geq 0$, then
$$\sup_{t\in A}\{f(t) + g(t)\} = \sup_{t\in A}f(t) + \sup_{t\in A} g(t)$$
and you are right to be sceptical about it, since this is not true. However
$$\sup_{t\in A}\{f(t) + g(t)\} \le \sup_{t\in A}f(t) + \sup_{t\in A} g(t)$$
is true and this is sufficient for your proof.