Supremum Metric on Bounded Real Sequences is Metric/Proof 2

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


 * $\ds \forall x, y \in A: \map d {x, y} := \sup_{n \mathop \in \N} \size {x_n - y_n}$

where $x = \sequence {x_i}$ and $y = \sequence {y_i}$ are bounded real sequences.

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

for all $n \in \N$.

First note that we have:

and so the exists.

Proof of
So holds for $d$.

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

Let $n \in \N$.

Thus $\map d {x, y} + \map d {y, z}$ is an upper bound for:
 * $S := \set {\size {x_n - z_n}: n \in \N}$

So:
 * $\map d {x, y} + \map d {y, z} \ge \sup S = \map d {x, z}$

So holds for $d$.

Proof of
So holds for $d$.

Proof of
As $d$ is the supremum of the absolute value of the difference of terms of the sequences $\sequence {x_i}$ and $\sequence {y_i}$:
 * $\forall x, y \in A: \map d {x, y} \ge 0$

Suppose $x, y \in A: \map d {x, y} = 0$.

Then:

So holds for $d$.