Supremum Metric on Bounded Real Sequences is Metric
Theorem
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 1
By definition, a real sequence is a mapping from the natural numbers $\N$ to the real numbers $\R$.
Thus a bounded real sequence is a bounded real-valued function.
The result follows from Supremum Metric on Bounded Real-Valued Functions is Metric.
$\blacksquare$
Proof 2
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:
| \(\ds \size {x_n - y_n}\) | \(=\) | \(\ds \size {x_n + \paren {-y_n} }\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \size {x_n} + \size {\paren {-y_n} }\) | Triangle Inequality for Real Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \size {x_n} + \size {y_n}\) | Definition of Absolute Value | |||||||||||
| \(\ds \) | \(\le\) | \(\ds K + L\) |
and so the right hand side exists.
Proof of Metric Space Axiom $(\text M 1)$
| \(\ds \map d {x, x}\) | \(=\) | \(\ds \sup_{n \mathop \in \N} \size {x_n - x_n}\) | Definition of $d$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{n \mathop \in \N} \size 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
So Metric Space Axiom $(\text M 1)$ holds for $d$.
$\Box$
Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality
Let $x, y, z \in A$.
Let $n \in \N$.
| \(\ds n\) | \(\in\) | \(\ds \N\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {x_n - z_n}\) | \(\le\) | \(\ds \size {x_n - y_n} + \size {y_n - z_n}\) | Triangle Inequality for Real Numbers | ||||||||||
| \(\ds \) | \(\le\) | \(\ds \sup_{n \mathop \in \N} \size {x_n - y_n} + \sup_{n \mathop \in \N} \size {y_n - z_n}\) | Definition of Supremum of Real Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map d {x, y} + \map d {y, z}\) | Definition of $d$ |
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 $(\text M 2)$: Triangle Inequality holds for $d$.
$\Box$
Proof of Metric Space Axiom $(\text M 3)$
| \(\ds \map d {x, y}\) | \(=\) | \(\ds \sup_{n \mathop \in \N} \size {x_n - y_n}\) | Definition of $d$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{n \mathop \in \N} \size {y_n - x_n}\) | Definition of Absolute Value | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map d {y, x}\) | Definition of $d$ |
So Metric Space Axiom $(\text M 3)$ holds for $d$.
$\Box$
Proof of Metric Space Axiom $(\text M 4)$
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:
| \(\ds \map d {x, y}\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sup_{n \mathop \in \N} \size {x_n - y_n}\) | \(=\) | \(\ds 0\) | Definition of $d$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall n \mathop \in \N: \, \) | \(\ds x_n\) | \(=\) | \(\ds y_n\) | Definition of Absolute Value | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) | Equality of Mappings |
So Metric Space Axiom $(\text M 4)$ holds for $d$.
$\blacksquare$