Difference between revisions of "Subsequence/Definition"

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(Created page with "<noinclude> ==Definition== </noinclude>Given a sequence {{M|1=(x_n)_{n=1}^\infty}} we define a ''subsequence of {{M|1=(x_n)^\infty_{n=1} }}''{{rAPIKM}} as a sequence: * {{M|k:...")
 
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<noinclude>
 
<noinclude>
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{{Requires cleanup|grade=A|msg=The notes are a bit messy}}
 
==Definition==
 
==Definition==
</noinclude>Given a sequence {{M|1=(x_n)_{n=1}^\infty}} we define a ''subsequence of {{M|1=(x_n)^\infty_{n=1} }}''{{rAPIKM}} as a sequence:
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</noinclude>Given a [[sequence]] {{M|1=(x_n)_{n=1}^\infty}} we define a ''subsequence of {{M|1=(x_n)^\infty_{n=1} }}''{{rAPIKM}}{{rFAVIDMH}} as follows:
* {{M|k:\mathbb{N}\rightarrow\mathbb{N} }} which operates on an {{M|n\in\mathbb{N} }} with {{M|1=n\mapsto k_n:=k(n)}} where:
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* Given any ''strictly'' increasing [[monotonic sequence]]<!--
** {{M|k_n}} is increasing, that means {{M|k_n\le k_{n+1} }}
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We denote this:
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START OF FIRST NOTE
* {{M|1=(x_{k_n})_{n=1}^\infty}}<noinclude>
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--><ref group="Note">Note that ''strictly increasing'' cannot be replaced by ''non-decreasing'' as the sequence could stay the same (ie a term where {{M|m_i\eq m_{i+1} }} for example), it didn't decrease, but it didn't increase either. It must be STRICTLY increasing.<br/><br/>If it was simply "non-decreasing" or just "increasing" then we could define: {{M|k_n:\eq 5}} for all {{M|n}}.
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* Then {{M|(x_{k_n})_{n\in\mathbb{N} } }} is a constant sequence where every term is {{M|x_5}} - the 5<sup>th</sup> term of {{M|(x_n)}}.</ref><!--
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END OF FIRST NOTE
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-->, {{M|1=(k_n)_{n=1}^\infty\subseteq\mathbb{N} }}
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** That means that {{M|\forall n\in\mathbb{N}[k_n<k_{n+1}]}}<!--
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START OF LONG NOTE
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--><ref group="Note">Some books may simply require ''increasing'', this is wrong. Take the theorem from [[Equivalent statements to compactness of a metric space]] which states that a [[metric space]] is [[compact]] {{M|\iff}} every [[sequence]] contains a [[convergent (sequence)|convergent]] subequence. If we only require that:
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* {{M|k_n\le k_{n+1} }}
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Then we can define the sequence: {{M|1=k_n:=1}}. This defines the subsequence {{M|x_1,x_1,x_1,\ldots x_1,\ldots}} of {{M|1=(x_n)_{n=1}^\infty}} which obviously converges. This defeats the purpose of subsequences.
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A subsequence should preserve the "forwardness" of a sequence, that is for a sub-sequence the terms are seen in the same order they would be seen in the parent sequence, and also the "sub" part means building a sequence from it, we want to built a sequence by choosing terms, suggesting we ought not use terms twice. <br/>
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The mapping definition directly supports this, as the mapping can be thought of as choosing terms</ref><!--
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END OF LONG NOTE
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-->
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Then the subsequence of {{M|(x_n)}} given by {{M|(k_n)}} is:
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* {{M|1=(x_{k_n})_{n=1}^\infty}}, the sequence whose terms are: {{M|x_{k_1},x_{k_2},\ldots,x_{k_n},\ldots}}
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** That is to say the {{M|i}}<sup>th</sup> element of {{M|(x_{k_n})}} is the {{M|k_i}}<sup>th</sup> element of {{M|(x_n)}}
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===As a mapping===
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Consider an ([[injection|injective]]) [[mapping]]: {{M|k:\mathbb{N}\rightarrow\mathbb{N} }} with the property that:
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* {{M|1=\forall a,b\in\mathbb{N}[a<b\implies k(a)<k(b)]}}
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This defines a sequence, {{M|1=(k_n)_{n=1}^\infty}} given by {{M|1=k_n:= k(n)}}
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* Now {{M|1=(x_{k_n})_{n=1}^\infty}} is a subsequence
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<noinclude>
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==Notes==
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<references group="Note"/>
 
==References==
 
==References==
 
<references/>
 
<references/>
 
{{Definition|Set Theory|Real Analysis|Functional Analysis}}
 
{{Definition|Set Theory|Real Analysis|Functional Analysis}}
 
</noinclude>
 
</noinclude>

Latest revision as of 21:54, 16 November 2016

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Definition

Given a sequence [ilmath](x_n)_{n=1}^\infty[/ilmath] we define a subsequence of [ilmath](x_n)^\infty_{n=1}[/ilmath][1][2] as follows:

  • Given any strictly increasing monotonic sequence[Note 1], [ilmath](k_n)_{n=1}^\infty\subseteq\mathbb{N}[/ilmath]
    • That means that [ilmath]\forall n\in\mathbb{N}[k_n<k_{n+1}][/ilmath][Note 2]

Then the subsequence of [ilmath](x_n)[/ilmath] given by [ilmath](k_n)[/ilmath] is:

  • [ilmath](x_{k_n})_{n=1}^\infty[/ilmath], the sequence whose terms are: [ilmath]x_{k_1},x_{k_2},\ldots,x_{k_n},\ldots[/ilmath]
    • That is to say the [ilmath]i[/ilmath]th element of [ilmath](x_{k_n})[/ilmath] is the [ilmath]k_i[/ilmath]th element of [ilmath](x_n)[/ilmath]

As a mapping

Consider an (injective) mapping: [ilmath]k:\mathbb{N}\rightarrow\mathbb{N} [/ilmath] with the property that:

  • [ilmath]\forall a,b\in\mathbb{N}[a<b\implies k(a)<k(b)][/ilmath]

This defines a sequence, [ilmath](k_n)_{n=1}^\infty[/ilmath] given by [ilmath]k_n:= k(n)[/ilmath]

  • Now [ilmath](x_{k_n})_{n=1}^\infty[/ilmath] is a subsequence

Notes

  1. Note that strictly increasing cannot be replaced by non-decreasing as the sequence could stay the same (ie a term where [ilmath]m_i\eq m_{i+1} [/ilmath] for example), it didn't decrease, but it didn't increase either. It must be STRICTLY increasing.

    If it was simply "non-decreasing" or just "increasing" then we could define: [ilmath]k_n:\eq 5[/ilmath] for all [ilmath]n[/ilmath].
    • Then [ilmath](x_{k_n})_{n\in\mathbb{N} } [/ilmath] is a constant sequence where every term is [ilmath]x_5[/ilmath] - the 5th term of [ilmath](x_n)[/ilmath].
  2. Some books may simply require increasing, this is wrong. Take the theorem from Equivalent statements to compactness of a metric space which states that a metric space is compact [ilmath]\iff[/ilmath] every sequence contains a convergent subequence. If we only require that:
    • [ilmath]k_n\le k_{n+1} [/ilmath]
    Then we can define the sequence: [ilmath]k_n:=1[/ilmath]. This defines the subsequence [ilmath]x_1,x_1,x_1,\ldots x_1,\ldots[/ilmath] of [ilmath](x_n)_{n=1}^\infty[/ilmath] which obviously converges. This defeats the purpose of subsequences. A subsequence should preserve the "forwardness" of a sequence, that is for a sub-sequence the terms are seen in the same order they would be seen in the parent sequence, and also the "sub" part means building a sequence from it, we want to built a sequence by choosing terms, suggesting we ought not use terms twice.
    The mapping definition directly supports this, as the mapping can be thought of as choosing terms

References

  1. Analysis - Part 1: Elements - Krzysztof Maurin
  2. Functional Analysis - Volume 1: A gentle introduction - Dzung Minh Ha