# Equivalence of Cauchy sequences/Proof

## Statement

Given two Cauchy sequences, [ilmath](a_n)_{n=1}^\infty[/ilmath] and [ilmath](b_n)_{n=1}^\infty[/ilmath] in a metric space [ilmath](X,d)[/ilmath] we define them as equivalent if:

• $\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(a_n,b_n)<\epsilon]$

And that this indeed actually defines an equivalence relation

## Proof

Reflexivity - We must show that [ilmath] ({ a_n })_{ n = 1 }^{ \infty } \sim ({ a_n })_{ n = 1 }^{ \infty } [/ilmath]

• Let [ilmath]\epsilon>0[/ilmath] be given.
• Pick [ilmath]N=1[/ilmath] (any [ilmath]N\in\mathbb{N} [/ilmath] will work)
• Let [ilmath]n\in\mathbb{N} [/ilmath] be given
• There are 2 cases now, either [ilmath]n>N[/ilmath] or [ilmath]n\le N[/ilmath]
1. If [ilmath]n>N[/ilmath] then by the nature of implies we require the RHS to be true, we require [ilmath]d(a_n,a_n)<\epsilon[/ilmath] to be true.
• Notice [ilmath]d(a_n,a_n)=0[/ilmath] by the definition of a metric
• As [ilmath]\epsilon>0[/ilmath] we see [ilmath]d(a_n,a_n)=0<\epsilon[/ilmath]
• So [ilmath]d(a_n,a_n)<\epsilon[/ilmath] is true, as required in this case.
2. If [ilmath]n\le N[/ilmath] by the nature of implies we don't care about the RHS, it can be either true or false.
• It must be either true or false
• So we're done

This completes the proof that [ilmath] ({ a_n })_{ n = 1 }^{ \infty } [/ilmath] is equivalent to [ilmath] ({ a_n })_{ n = 1 }^{ \infty } [/ilmath]

Transitivity - we must show that [ilmath] ({ a_n })_{ n = 1 }^{ \infty } \sim ({ b_n })_{ n = 1 }^{ \infty } [/ilmath] and [ilmath] ({ b_n })_{ n = 1 }^{ \infty } \sim ({ c_n })_{ n = 1 }^{ \infty } [/ilmath] [ilmath]\implies[/ilmath] [ilmath] ({ a_n })_{ n = 1 }^{ \infty } \sim ({ c_n })_{ n = 1 }^{ \infty } [/ilmath]

Workings to determine the gist of the proof

Let [ilmath]\epsilon >0[/ilmath] be given, we need to show:

• [ilmath]d(a_n,c_n)<\epsilon[/ilmath]

But by the triangle inequality property of a metric we know that:

• [ilmath]d(a,c)\le d(a,b)+d(b,c)[/ilmath] for all [ilmath]b[/ilmath] in the space.

If we can show that [ilmath]d(a,b)+d(b,c)<\epsilon[/ilmath] then we'd be done.

By hypothesis:

• [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(a_n,b_n)<\epsilon][/ilmath] and:
• [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(b_n,c_n)<\epsilon][/ilmath]

That is we may pick any number we like, as long as it is [ilmath]>0[/ilmath] and there is an [ilmath]N\in\mathbb{N} [/ilmath] such that for any natural number larger than [ilmath]N[/ilmath] the distance between either [ilmath]a_n[/ilmath] and [ilmath]b_n[/ilmath], or [ilmath]b_n[/ilmath] and [ilmath]c_n[/ilmath] is less than that picked number.

Looking at [ilmath]d(a,b)+d(b,c)<\epsilon[/ilmath], we can see that if we have [ilmath]d(a,b)<\frac{\epsilon}{2} [/ilmath] and [ilmath]d(b,c)<\frac{\epsilon}{2} [/ilmath] then we'd have:

• [ilmath]d(a,b)+d(b,c)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon[/ilmath] or
• [ilmath]d(a,b)+d(b,c)<\epsilon[/ilmath] [ilmath]\longleftarrow[/ilmath]this is exactly what we're looking to do

Note that if [ilmath]\epsilon>0[/ilmath] then [ilmath]\frac{\epsilon}{2}>0[/ilmath] too.

By hypothesis we see for a positive number, [ilmath]\frac{\epsilon}{2} [/ilmath] there exists an [ilmath]N_1[/ilmath] and [ilmath]N_2[/ilmath] such that for all [ilmath]n\in\mathbb{N} [/ilmath] if:

• [ilmath]n>N_1[/ilmath] then we have [ilmath]d(a_n,b_n)<\frac{\epsilon}{2} [/ilmath] and
• [ilmath]n>N_2[/ilmath] then we have [ilmath]d(b_n,c_n)<\frac{\epsilon}{2} [/ilmath]

If we pick [ilmath]N=\text{max}(N_1,N_2)[/ilmath] then [ilmath]\forall n\in\mathbb{N} [/ilmath] with [ilmath]n>N[/ilmath] both [ilmath]d(a_n,b_n)[/ilmath] and [ilmath]d(b_n,c_n)[/ilmath] are [ilmath]<\frac{\epsilon}{2} [/ilmath]

(as [ilmath]n>N\implies n>N_1\text{ and }n>N_2[/ilmath] - this is why we use the largest of [ilmath]N_1[/ilmath] and [ilmath]N_2[/ilmath])

Thus we have:

• [ilmath]d(a_n,c_n)\le d(a_n,b_n)+d(b_n,c_n)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon[/ilmath], or:
• [ilmath]d(a_n,c_n)<\epsilon[/ilmath] - as required.
• Let [ilmath]\epsilon>0[/ilmath] be given.
• By hypothesis know both:
1. [ilmath]\forall\epsilon>0\exists N_1\in\mathbb{N}\forall n\in\mathbb{N}[n>N_1\implies d(a_n,b_n)<\epsilon][/ilmath] and:
2. [ilmath]\forall\epsilon>0\exists N_2\in\mathbb{N}\forall n\in\mathbb{N}[n>N_2\implies d(b_n,c_n)<\epsilon][/ilmath] to be true.
• Note that [ilmath]\epsilon>0\implies\frac{\epsilon}{2}>0[/ilmath], and in both of the hypothesised statements above, it is true for all [ilmath]\epsilon>0[/ilmath]
• Pick [ilmath]N_1\in\mathbb{N} [/ilmath] using the first statement with [ilmath]\frac{\epsilon}{2} [/ilmath] as the positive number, now:
• [ilmath]\forall n\in\mathbb{N}[n>N_1\implies d(a_n,b_n)<\frac{\epsilon}{2}][/ilmath]
• Pick [ilmath]N_2\in\mathbb{N} [/ilmath] using the second statement with [ilmath]\frac{\epsilon}{2} [/ilmath] as the positive number, now:
• [ilmath]\forall n\in\mathbb{N}[n>N_2\implies d(b_n,c_n)<\frac{\epsilon}{2}][/ilmath]
• Pick for [ilmath]N\in\mathbb{N} [/ilmath] the value [ilmath]N=\text{max}(N_1,N_2)[/ilmath]
• Now for [ilmath]n>N[/ilmath] both [ilmath]d(a_n,b_n)[/ilmath] and [ilmath]d(b_n,c_n)[/ilmath] are [ilmath]<\frac{\epsilon}{2} [/ilmath]
• Let [ilmath]n\in\mathbb{N} [/ilmath] be given, there are 2 cases now, [ilmath]n>N[/ilmath] or [ilmath]n\le N[/ilmath]
1. If [ilmath]n>N[/ilmath] then by the nature of implies we must show [ilmath]d(a_n,c_n)<\epsilon[/ilmath] to be true
• Notice: [ilmath]d(a_n,c_n)\le d(a_n,b_n)+d(b_n,c_n)[/ilmath] (by the triangle inequality property of a metric) and:
• [ilmath]d(a_n,b_n)+d(b_n,c_n)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon[/ilmath]
• Thus we have [ilmath]d(a_n,c_n)<\epsilon[/ilmath] - as required
2. If [ilmath]n\le N[/ilmath] by the nature of implies we don't actually care if [ilmath]d(a_n,c_n)<\epsilon[/ilmath] is true or false.
• As it must be either true or false, we are done.

This completes the proof that [ilmath] ({ a_n })_{ n = 1 }^{ \infty } \sim ({ b_n })_{ n = 1 }^{ \infty } [/ilmath] and [ilmath] ({ b_n })_{ n = 1 }^{ \infty } \sim ({ c_n })_{ n = 1 }^{ \infty } [/ilmath] [ilmath]\implies[/ilmath] [ilmath] ({ a_n })_{ n = 1 }^{ \infty } \sim ({ c_n })_{ n = 1 }^{ \infty } [/ilmath]

Symmetry - that is that [ilmath] ({ a_n })_{ n = 1 }^{ \infty } \sim ({ b_n })_{ n = 1 }^{ \infty } [/ilmath] [ilmath]\implies[/ilmath] [ilmath] ({ b_n })_{ n = 1 }^{ \infty } \sim ({ a_n })_{ n = 1 }^{ \infty } [/ilmath]

Workings to find the gist of the proof

Notice we have:

• [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(a_n,b_n)<\epsilon][/ilmath] and we want:
• [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(b_n,a_n)<\epsilon][/ilmath]

But by the symmetric property of a metric we see that [ilmath]d(a_n,b_n)=d(b_n,a_n)[/ilmath]

Thus, if [ilmath]d(a_n,b_n)<epsilon[/ilmath] we see [ilmath]d(b_n,a_n)=d(a_n,b_n)<\epsilon[/ilmath], so [ilmath]d(b_n,a_n)<\epsilon[/ilmath] too!

• Let [ilmath]\epsilon>0[/ilmath] be given.
• By hypothesis we have:
• [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(a_n,b_n)<\epsilon][/ilmath]
• Choose [ilmath]N[/ilmath] to be the [ilmath]N\in\mathbb{N} [/ilmath] which exists by hypothesis for our given [ilmath]\epsilon[/ilmath]
• Let [ilmath]n\in\mathbb{N} [/ilmath] be given, there are now two cases, [ilmath]n>N[/ilmath] and [ilmath]n\le N[/ilmath]
1. if [ilmath]n>N[/ilmath] then by the nature of implies we require [ilmath]d(b_n,a_n)<\epsilon[/ilmath] to be true.
• Notice [ilmath]d(b_n,a_n)=d(a_n,b_n)[/ilmath] by the symmetric property of a metric and
• By our hypothesis, for our [ilmath]N[/ilmath], [ilmath]n>N\implies d(a_n,b_n)<\epsilon[/ilmath]
• Thus [ilmath]d(b_n,a_n)=d(a_n,b_n)<\epsilon[/ilmath] and
• [ilmath]d(b_n,a_n)<\epsilon[/ilmath] as required
2. if [ilmath]n\le N[/ilmath] then by the nature of implies the RHS can be either true or false, and the implies condition is satisfied.
• As [ilmath]d(b_n,a_n)<\epsilon[/ilmath] is a statement that can only be either true or false, we see that this is satisfied

This completes the proof that [ilmath] ({ a_n })_{ n = 1 }^{ \infty } \sim ({ b_n })_{ n = 1 }^{ \infty } [/ilmath] [ilmath]\implies[/ilmath] [ilmath] ({ b_n })_{ n = 1 }^{ \infty } \sim ({ a_n })_{ n = 1 }^{ \infty } [/ilmath]