View source for Equivalence of Cauchy sequences/Proof

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==Statement==
{{:Equivalence of Cauchy sequences/Definition}}
And that this indeed actually defines an [[equivalence relation]]
==Proof==
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'''[[Reflexive relation|Reflexivity]]''' - We must show that {{MSeq|a_n|post=\sim{{MSeq|a_n|nomath=true}}}}
* Let {{M|\epsilon>0}} be given.
** Pick {{M|1=N=1}} (any {{M|N\in\mathbb{N} }} will work)
*** Let {{M|n\in\mathbb{N} }} be given
*** There are 2 cases now, either {{M|n>N}} or {{M|n\le N}}
***# If {{M|n>N}} then by the nature of [[implies]] we require the RHS to be true, we require {{M|d(a_n,a_n)<\epsilon}} to be true.
***#* Notice {{M|1=d(a_n,a_n)=0}} by the definition of a [[metric]]
***#** As {{M|\epsilon>0}} we see {{M|1=d(a_n,a_n)=0<\epsilon}}
***#* So {{M|d(a_n,a_n)<\epsilon}} is true, as required in this case.
***# If {{M|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 {{MSeq|a_n}} is equivalent to {{MSeq|a_n}}


'''[[Transitive relation|Transitivity]]''' - we must show that {{MSeq|a_n|post=\sim {{MSeq|b_n|nomath=1}}}} and {{MSeq|b_n|post=\sim {{MSeq|c_n|nomath=1}}}} {{M|\implies}} {{MSeq|a_n|post=\sim {{MSeq|c_n|nomath=1}}}}
{{Begin Notebox}}
Workings to determine the gist of the proof
{{Begin Notebox Content}}
Let {{M|\epsilon >0}} be given, we need to show:
* {{M|d(a_n,c_n)<\epsilon}}
But by the [[metric|triangle inequality property of a metric]] we know that:
* {{M|d(a,c)\le d(a,b)+d(b,c)}} for all {{M|b}} in the space.
If we can show that {{M|d(a,b)+d(b,c)<\epsilon}} then we'd be done. 


By hypothesis:
* {{M|\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(a_n,b_n)<\epsilon]}} and:
* {{M|\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 {{M|>0}} and there is an {{M|N\in\mathbb{N} }} such that for any natural number larger than {{M|N}} the distance between either {{M|a_n}} and {{M|b_n}}, or {{M|b_n}} and {{M|c_n}} is less than that picked number.


Looking at {{M|d(a,b)+d(b,c)<\epsilon}}, we can see that if we have {{M|d(a,b)<\frac{\epsilon}{2} }} and {{M|d(b,c)<\frac{\epsilon}{2} }} then we'd have:
* {{M|1=d(a,b)+d(b,c)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon}} or
* {{M|1=d(a,b)+d(b,c)<\epsilon}} '''{{M|\longleftarrow}}this is exactly what we're looking to do'''


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


By hypothesis we see for a positive number, {{M|\frac{\epsilon}{2} }} there exists an {{M|N_1}} and {{M|N_2}} such that for all {{M|n\in\mathbb{N} }} if:
* {{M|n>N_1}} then we have {{M|d(a_n,b_n)<\frac{\epsilon}{2} }} and
* {{M|n>N_2}} then we have {{M|d(b_n,c_n)<\frac{\epsilon}{2} }}


If we pick {{M|1=N=\text{max}(N_1,N_2)}} then {{M|\forall n\in\mathbb{N} }} with {{M|n>N}} both {{M|d(a_n,b_n)}} and {{M|d(b_n,c_n)}} are {{M|<\frac{\epsilon}{2} }}
: (as {{M|n>N\implies n>N_1\text{ and }n>N_2}} - this is why we use the largest of {{M|N_1}} and {{M|N_2}})

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


'''[[Symmetric relation|Symmetry]]''' - that is that {{MSeq|a_n|post=\sim {{MSeq|b_n|nomath=1}}}} {{M|\implies}} {{MSeq|b_n|post=\sim {{MSeq|a_n|nomath=1}}}}
{{Begin Notebox}}
Workings to find the gist of the proof
{{Begin Notebox Content}}
Notice we have:
* {{M|\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:
* {{M|\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 [[metric|symmetric property of a metric]] we see that {{M|1=d(a_n,b_n)=d(b_n,a_n)}}

Thus, if {{M|d(a_n,b_n)<epsilon}} we see {{M|1=d(b_n,a_n)=d(a_n,b_n)<\epsilon}}, so {{M|d(b_n,a_n)<\epsilon}} too!
{{End Notebox Content}}{{End Notebox}}
* Let {{M|\epsilon>0}} be given. 
** By hypothesis we have:
*** {{M|\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(a_n,b_n)<\epsilon]}}
** Choose {{M|N}} to be the {{M|N\in\mathbb{N} }} which exists by hypothesis for our given {{M|\epsilon}}
*** Let {{M|n\in\mathbb{N} }} be given, there are now two cases, {{M|n>N}} and {{M|n\le N}}
***# if {{M|n>N}} then by the nature of [[implies]] we require {{M|d(b_n,a_n)<\epsilon}} to be true.
***#* Notice {{M|1=d(b_n,a_n)=d(a_n,b_n)}} [[metric|by the symmetric property of a metric]] and
***#** By our hypothesis, for our {{M|N}}, {{M|n>N\implies d(a_n,b_n)<\epsilon}} 
***#* Thus {{M|1=d(b_n,a_n)=d(a_n,b_n)<\epsilon}} and 
***#* {{M|d(b_n,a_n)<\epsilon}} as required
***# if {{M|n\le N}} then by the nature of [[implies]] the RHS can be either true or false, and the implies condition is satisfied.
***#* As {{M|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 {{MSeq|a_n|post=\sim {{MSeq|b_n|nomath=1}}}} {{M|\implies}} {{MSeq|b_n|post=\sim {{MSeq|a_n|nomath=1}}}}
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==References==
<references/>
{{Theorem Of|Metric Space|Topology|Functional Analysis|Analysis}}
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