Theorem: Supremum Metric on Bounded Real Sequences is Metric

Let $A$ be the set of all bounded real sequences.
Let $d: A \times A \to \R$ be the supremum metric on $A$.
Then $d$ is a metric.

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:
{{begin-eqn}}
{{eqn | l = \size {x_n - y_n}
      | r = \size {x_n + \paren {-y_n} }
      | c = 
}}
{{eqn | o = \le
      | r = \size {x_n} + \size {\paren {-y_n} }
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | r = \size {x_n} + \size {y_n}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | o = \le
      | r = K + L
      | c = 
}}
{{end-eqn}}
and so the {{RHS}} exists.
=== Proof of {{Metric-space-axiom|1|nolink}} ===
{{begin-eqn}}
{{eqn | l = \map d {x, x}
      | r = \sup_{n \mathop \in \N} \size {x_n - x_n}
      | c = Definition of $d$
}}
{{eqn | r = \sup_{n \mathop \in \N} \size 0
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
So {{Metric-space-axiom|1}} holds for $d$.
{{qed|lemma}}
=== Proof of {{Metric-space-axiom|2|nolink}} ===
Let $x, y, z \in A$.
Let $n \in \N$.
{{begin-eqn}}
{{eqn | l = n
      | o = \in
      | r = \N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {x_n - z_n}
      | o = \le
      | r = \size {x_n - y_n} + \size {y_n - z_n}
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | o = \le
      | r = \sup_{n \mathop \in \N} \size {x_n - y_n} + \sup_{n \mathop \in \N} \size {y_n - z_n}
      | c = {{Defof|Supremum of Real Sequence}}
}}
{{eqn | r = \map d {x, y} + \map d {y, z}
      | c = Definition of $d$
}}
{{end-eqn}}
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 {{Metric-space-axiom|2}} holds for $d$.
{{qed|lemma}}
=== Proof of {{Metric-space-axiom|3|nolink}} ===
{{begin-eqn}}
{{eqn | l = \map d {x, y}
      | r = \sup_{n \mathop \in \N} \size {x_n - y_n}
      | c = Definition of $d$
}}
{{eqn | r = \sup_{n \mathop \in \N} \size {y_n - x_n}
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = \map d {y, x}
      | c = Definition of $d$
}}
{{end-eqn}}
So {{Metric-space-axiom|3}} holds for $d$.
{{qed|lemma}}
=== Proof of {{Metric-space-axiom|4|nolink}} ===
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:
{{begin-eqn}}
{{eqn | l = \map d {x, y}
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sup_{n \mathop \in \N} \size {x_n - y_n}
      | r = 0
      | c = Definition of $d$
}}
{{eqn | ll= \leadsto
      | q = \forall n \mathop \in \N
      | l = x_n 
      | r = y_n
      | c = {{Defof|Absolute Value}}
}}
{{eqn | ll= \leadsto
      | l = x
      | r = y
      | c = Equality of Mappings
}}
{{end-eqn}}
So {{Metric-space-axiom|4}} holds for $d$.
{{qed}}