Equivalence Relation on Cauchy Sequences

Lemma
Let $\struct {X, d}$ be a metric space.

Let $\CC \sqbrk X$ denote the set of all Cauchy sequences of elements from $X$.

Let a relation $\sim$ be defined on $\CC \sqbrk X$ by:


 * $\displaystyle \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$

Then $\sim$ is an equivalence relation on $\CC \sqbrk X$.

Proof
We must show that $\sim$ is
 * reflexive,
 * symmetric and
 * transitive

on $\CC \sqbrk X$.

To this end, let $\sequence {x_n}, \sequence {y_n}, \sequence {z_n} \in \CC \sqbrk X$ be arbitrary.

For each $n \in \N$ we have that $\map d {x_n, x_n} = 0$ by metric space axiom M1.

Therefore $\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, x_n} = 0$.

This shows that $\sequence {x_n} \sim \sequence {x_n}$.

Thus $\sim$ is reflexive.

By metric space axiom M3, $\map d {x_n, y_n} = \map d {y_n, x_n}$ for each $n \in \N$.

Therefore:
 * $\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \lim_{n \mathop \to \infty} \map d {y_n, x_n}$

So $\sequence {x_n} \sim \sequence {y_n}$ implies that $\sequence {y_n} \sim \sequence {x_n}$.

Thus $\sim$ is symmetric.

Finally, by metric space axiom M2, $\map d {x_n, z_n} \le \map d {x_n, y_n} + \map d {y_n, z_n}$ for each $n \in \N$.

Therefore, by the sum rule for limits of sequences:


 * $\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, z_n} \le \lim_{n \mathop \to \infty} \map d {x_n, y_n} + \lim_{n \mathop \to \infty} \map d {y_n, z_n}$

Thus $\sequence {x_n} \sim \sequence {y_n}$ and $\sequence {y_n} \sim \sequence {z_n}$ together imply that $\sequence {x_n} \sim \sequence {z_n}$.

Thus $\sim$ is transitive.

So $\sim$ is shown to be reflexive, symmetric and transitive, and therefore an equivalence relation on $\CC \sqbrk X$.