Supremum Metric on Bounded Real Sequences is Metric/Proof 2

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


 * $\displaystyle \forall x, y \in A: d \left({x, y}\right) := \sup_{n \mathop \in \N} \left\vert{x_n - y_n}\right\vert$

where $x = \left\langle{x_i}\right\rangle$ and $y = \left\langle{y_i}\right\rangle$ are bounded real sequences.

So:
 * $\exists K, L \in \R: \left\vert{x_n}\right\vert \le K, \left\vert{y_n}\right\vert \le L$

for all $n \in \N$.

First note that we have:

and so the exists.

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

Proof of $M2$
Let $x, y, z \in A$.

Let $n \in \N$.

Thus $d \left({x, y}\right) + d \left({y, z}\right)$ is an upper bound for:
 * $S := \left\{ {\left\vert{x_n - z_n}\right\vert: n \in \N}\right\}$

So:
 * $d \left({x, y}\right) + d \left({y, z}\right) \ge \sup S = d \left({x, z}\right)$

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 difference of termselements of the sequences $\left\langle{x_i}\right\rangle$ and $\left\langle{y_i}\right\rangle$:
 * $\forall x, y \in A: d \left({x, y}\right) \ge 0$

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

Then:

So axiom $M4$ holds for $d$.