Difference between revisions of "Subsequence/Definition"
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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 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}}{{rFAVIDMH}} as follows: |
| − | * Given any ''strictly'' increasing sequence, {{M|1=(k_n)_{n=1}^\infty}} | + | * Given any ''strictly'' increasing [[monotonic sequence]]<!-- |
| − | ** 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: | + | |
| + | START OF FIRST NOTE | ||
| + | |||
| + | --><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}}. | ||
| + | * 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><!-- | ||
| + | |||
| + | END OF FIRST NOTE | ||
| + | |||
| + | -->, {{M|1=(k_n)_{n=1}^\infty\subseteq\mathbb{N} }} | ||
| + | ** That means that {{M|\forall n\in\mathbb{N}[k_n<k_{n+1}]}}<!-- | ||
| + | |||
| + | START OF LONG NOTE | ||
| + | |||
| + | --><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. | ||
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/> | 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/> | ||
| − | The mapping definition directly supports this, as the mapping can be thought of as choosing terms</ref> | + | The mapping definition directly supports this, as the mapping can be thought of as choosing terms</ref><!-- |
| − | + | ||
| − | * {{M|1=(x_{k_n})_{n=1}^\infty}} | + | END OF LONG NOTE |
| + | |||
| + | --> | ||
| + | Then the subsequence of {{M|(x_n)}} given by {{M|(k_n)}} is: | ||
| + | * {{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}} | ||
| + | ** 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)}} | ||
===As a mapping=== | ===As a mapping=== | ||
Consider an ([[injection|injective]]) [[mapping]]: {{M|k:\mathbb{N}\rightarrow\mathbb{N} }} with the property that: | Consider an ([[injection|injective]]) [[mapping]]: {{M|k:\mathbb{N}\rightarrow\mathbb{N} }} with the property that: | ||
Latest revision as of 21:54, 16 November 2016
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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][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
- ↑ 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].
- ↑ 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
