Equivalence of Cauchy sequences/Proof

Statement

Given two Cauchy sequences, $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ in a metric space $(X,d)$ 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 $({ a_n })_{ n = 1 }^{ \infty } \sim ({ a_n })_{ n = 1 }^{ \infty }$

• Let $\epsilon>0$ be given.
  • Pick $N=1$ (any $N\in\mathbb{N}$ will work)
    • Let $n\in\mathbb{N}$ be given
    • There are 2 cases now, either $n>N$ or $n\le N$
      1. If $n>N$ then by the nature of implies we require the RHS to be true, we require $d(a_n,a_n)<\epsilon$ to be true.
         • Notice $d(a_n,a_n)=0$ by the definition of a metric
           • As $\epsilon>0$ we see $d(a_n,a_n)=0<\epsilon$
         • So $d(a_n,a_n)<\epsilon$ is true, as required in this case.
      2. If $n\le N$ 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 $({ a_n })_{ n = 1 }^{ \infty }$ is equivalent to $({ a_n })_{ n = 1 }^{ \infty }$

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

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Workings to determine the gist of the proof

Let $\epsilon >0$ be given, we need to show:

• $d(a_n,c_n)<\epsilon$

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

• $d(a,c)\le d(a,b)+d(b,c)$ for all $b$ in the space.

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

By hypothesis:

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

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

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

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

Note that if $\epsilon>0$ then $\frac{\epsilon}{2}>0$ too.

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

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

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

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

Thus we have:

• $d(a_n,c_n)\le d(a_n,b_n)+d(b_n,c_n)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$, or:
• $d(a_n,c_n)<\epsilon$ - as required.

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

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

Symmetry - that is that $({ a_n })_{ n = 1 }^{ \infty } \sim ({ b_n })_{ n = 1 }^{ \infty }$ $\implies$ $({ b_n })_{ n = 1 }^{ \infty } \sim ({ a_n })_{ n = 1 }^{ \infty }$

[Expand]

Workings to find the gist of the proof

Notice we have:

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

But by the symmetric property of a metric we see that $d(a_n,b_n)=d(b_n,a_n)$

Thus, if $d(a_n,b_n)<epsilon$ we see $d(b_n,a_n)=d(a_n,b_n)<\epsilon$, so $d(b_n,a_n)<\epsilon$ too!

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

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