Supremum Metric on Bounded Real Functions on Closed Interval is Metric/Proof 2

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


 * $\ds \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$

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

So:
 * $\exists K, L \in \R: \size {\map f x} \le K, \size {\map g x} \le L$

for all $x \in \closedint a b$.

First note that we have:

and so the exists.

Proof of
So holds for $d$.

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

Let $c \in \closedint a b$.

Thus $\map d {f, g} + \map d {g, h}$ is an upper bound for:
 * $S := \set {\size {\map f c - \map h c}: c \in \closedint a b}$

So:
 * $\map d {f, g} + \map d {g, h} \ge \sup S = \map d {f, h}$

So holds for $d$.

Proof of
So holds for $d$.

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

Suppose $f, g \in A: \map d {f, g} = 0$.

Then:

So holds for $d$.