Difference between revisions of "Equivalence of Cauchy sequences/Proof"

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(Created page with "<noinclude> ==Statement== {{:Equivalence of Cauchy sequences/Definition}} And that this indeed actually defines an equivalence relation ==Proof== </noinclude> '''Reflexivi...")
 
(Added proofs for all 3 statements in a first-year friendly way. Will mark parent page as exemplary/first-year friendly.)
 
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==Proof==
 
==Proof==
 
</noinclude>
 
</noinclude>
−
'''Reflexivity'''
+
'''[[Reflexive relation|Reflexivity]]''' - We must show that {{MSeq|a_n|post=\sim{{MSeq|a_n|nomath=true}}}}
−
{{Requires proof}}
+
* Let {{M|\epsilon>0}} be given.
−
'''Transitivity'''
+
** Pick {{M|1=N=1}} (any {{M|N\in\mathbb{N} }} will work)
−
{{Requires proof}}
+
*** Let {{M|n\in\mathbb{N} }} be given
−
'''Symmetry'''
+
*** There are 2 cases now, either {{M|n>N}} or {{M|n\le N}}
−
{{Requires proof}}
+
***# 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}}}}
 
<noinclude>
 
<noinclude>
 
==References==
 
==References==

Latest revision as of 21:02, 20 April 2016

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[1]:

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

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]

References

  1. ↑ Analysis - Part 1: Elements - Krzysztof Maurin
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