We have that the supremum metric on $A \times A$ is defined as: :$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in S} d' \left({f \left({x}\right), g \left({x}\right)}\right)$ where $f$ and $g$ are bounded mappings. First note that we have: {{begin-eqn}} {{eqn | l = \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert       | r = \left\vert{f \left({x}\right) + \left({- g \left({x}\right)}\right)}\right\vert       | c =  }} {{eqn | o = \le       | r = \left\vert{f \left({x}\right)}\right\vert + \left\vert{\left({- g \left({x}\right)}\right)}\right\vert       | c = Triangle Inequality for Real Numbers }} {{eqn | r = \left\vert{f \left({x}\right)}\right\vert + \left\vert{g \left({x}\right)}\right\vert       | c = Definition of Absolute Value }} {{eqn | o = \le       | r = K + L       | c =  }} {{end-eqn}} and so the RHS exists. === Proof of $M1$ === {{begin-eqn}} {{eqn | l = d \left({f, f}\right)       | r = \sup_{x \mathop \in S} d' \left({f \left({x}\right), f \left({x}\right)}\right)       | c = Definition of $d$ }} {{eqn | r = \sup_{x \mathop \in S} \left\vert{0}\right\vert       | c = as $d'$ fulfils axiom $M1$ }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d$. {{qed|lemma}} === Proof of $M2$ === Let $f, g, h \in A$. Let $x, y \in S$. {{begin-eqn}} {{eqn | l = c       | o = \in       | r = S       | c =  }} {{eqn | ll= \implies       | l = d' \left({f \left({c}\right), h \left({c}\right)}\right)       | o = \le       | r = d' \left({f \left({c}\right), g \left({c}\right)}\right) + d' \left({g \left({c}\right), h \left({c}\right)}\right)       | c = as $d'$ fulfils axiom $M2$ }} {{eqn | o = \le       | r = \sup_{x \mathop \in S} d' \left({f \left({x}\right), g \left({x}\right)}\right) + \sup_{x \mathop \in S} d' \left({g \left({x}\right), h \left({x}\right)}\right)       | c = Definition of Supremum }} {{eqn | r = d \left({f, g}\right) + d \left({g, h}\right)       | c = Definition of $d$ }} {{end-eqn}} Thus $d \left({f, g}\right) + d \left({g, h}\right)$ is an upper bound for: :$X := \left\{ {d' \left({f \left({c}\right), g \left({c}\right)}\right): c \in X}\right\}$ So: :$d \left({f, g}\right) + d \left({g, h}\right) \ge \sup X = d \left({f, h}\right)$ So axiom $M2$ holds for $d$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = d \left({f, g}\right)       | r = \sup_{x \mathop \in S} d' \left({f \left({x}\right), g \left({x}\right)}\right)       | c = Definition of $d$ }} {{eqn | r = \sup_{x \mathop \in S} d' \left({g \left({x}\right), f \left({x}\right)}\right)       | c = as $d'$ fulfils axiom $M3$ }} {{eqn | r = d \left({g, f}\right)       | c = Definition of $d$ }} {{end-eqn}} So axiom $M3$ holds for $d$. {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | l = d \left({f, g}\right)       | r = 0       | c =  }} {{eqn | ll= \implies       | l = \sup_{x \mathop \in S} d' \left({f \left({x}\right), g \left({x}\right)}\right)       | r = 0       | c = Definition of $d$ }} {{eqn | ll= \implies       | lo= \forall x \in S:       | l = d' \left({f \left({x}\right), g \left({x}\right)}\right)       | r = 0       | c = as $d'$ fulfils axioms $M1$ and $M4$ }} {{eqn | ll= \implies       | l = f       | r = g       | c = as $d'$ fulfils axiom $M4$ }} {{end-eqn}} So axiom $M4$ holds for $d$. {{qed}}