Supremum Metric on Bounded Real-Valued Functions is Metric/Proof 2

Proof
We have that the supremum metric on $A \times A$ is defined as:


 * $\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in X} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert$

where $f$ and $g$ are bounded real-valued functions.

So:
 * $\exists K, L \in \R: \left\vert{f \left({x}\right)}\right\vert \le K, \left\vert{g \left({x}\right)}\right\vert \le L$

for all $x \in X$.

First note that we have:

and so the exists.

Proof of $M1$
So axiom $M1$ holds for $d$.

Proof of $M2$
Let $f, g, h \in A$.

Let $c \in X$.

Thus $d \left({f, g}\right) + d \left({g, h}\right)$ is an upper bound for:
 * $S := \left\{ {\left\vert{f \left({c}\right) - h \left({c}\right)}\right\vert: c \in X}\right\}$

So:
 * $d \left({f, g}\right) + d \left({g, h}\right) \ge \sup S = d \left({f, h}\right)$

So axiom $M2$ holds for $d$.

Proof of $M3$
So axiom $M3$ holds for $d$.

Proof of $M4$
As $d$ is the supremum of the absolute value of the image of the pointwise sum of $f$ and $g$:
 * $\forall f, g \in A: d \left({f, g}\right) \ge 0$

Suppose $f, g \in A: d \left({f, g}\right) = 0$.

Then:

So axiom $M4$ holds for $d$.