Difference between revisions of "Subsequence/Definition"
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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}} as follows: | </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: | ||
* Given any ''strictly'' increasing sequence, {{M|1=(k_n)_{n=1}^\infty}} | * 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: | + | ** 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 (sequence)|convergent]] subequence. If we only require that: |
* {{M|k_n\le k_{n+1} }} | * {{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. | 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. | ||
Revision as of 18:02, 13 March 2016
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
