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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==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 | + | </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 follows: |
| − | * {{M| | + | * Given any ''strictly'' increasing sequence, {{M|1=(k_n)_{n=1}^\infty}} |
| − | + | ** That means that {{M|\forall n\in\mathbb{N}[k_n<k_{n+1}]}}<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]] subequence. If we only require that: | |
| + | * {{M|k_n\le k_{n+1} }} | ||
| + | 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. | ||
| − | + | 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/> | |
| − | * {{M|1=(x_{k_n})_{n=1}^\infty}}<noinclude> | + | The mapping definition directly supports this, as the mapping can be thought of as choosing terms</ref> |
| + | The sequence: | ||
| + | * {{M|1=(x_{k_n})_{n=1}^\infty}} (which is {{M|x_{k_1},x_{k_2},\ldots x_{k_n},\ldots}}) is a ''subsequence'' | ||
| + | ===As a mapping=== | ||
| + | Consider an ([[injection|injective]]) [[mapping]]: {{M|k:\mathbb{N}\rightarrow\mathbb{N} }} with the property that: | ||
| + | * {{M|1=\forall a,b\in\mathbb{N}[a<b\implies k(a)<k(b)]}} | ||
| + | This defines a sequence, {{M|1=(k_n)_{n=1}^\infty}} given by {{M|1=k_n:= k(n)}} | ||
| + | * Now {{M|1=(x_{k_n})_{n=1}^\infty}} is a subsequence | ||
| + | <noinclude> | ||
| + | ==Notes== | ||
| + | <references group="Note"/> | ||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Definition|Set Theory|Real Analysis|Functional Analysis}} | {{Definition|Set Theory|Real Analysis|Functional Analysis}} | ||
</noinclude> | </noinclude> | ||
Revision as of 15:57, 1 December 2015
Contents
Definition
Given a sequence [ilmath](x_n)_{n=1}^\infty[/ilmath] we define a subsequence of [ilmath](x_n)^\infty_{n=1}[/ilmath][1] as follows:
- Given any strictly increasing sequence, [ilmath](k_n)_{n=1}^\infty[/ilmath]
- That means that [ilmath]\forall n\in\mathbb{N}[k_n<k_{n+1}][/ilmath][Note 1]
The sequence:
- [ilmath](x_{k_n})_{n=1}^\infty[/ilmath] (which is [ilmath]x_{k_1},x_{k_2},\ldots x_{k_n},\ldots[/ilmath]) is a subsequence
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
- ↑ 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]
The mapping definition directly supports this, as the mapping can be thought of as choosing terms
