Difference between revisions of "Subsequence/Definition"

Revision as of 18:02, 13 March 2016

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

Given any strictly increasing sequence, (kn)∞n=1
- That means that $\forall n\in\mathbb{N}[k_n<k_{n+1}]$ ^{[Note 1]}

The sequence:

$(x_{k_n})_{n=1}^\infty$ (which is $x_{k_1},x_{k_2},\ldots x_{k_n},\ldots$ ) is a subsequence

Consider an (injective) mapping: $k:\mathbb{N}\rightarrow\mathbb{N}$ with the property that:

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

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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 ⟺ every sequence contains a convergent subequence. If we only require that:
- $k_n\le k_{n+1}$
Then we can define the sequence: kn:=1. This defines the subsequence x1,x1,x1,…x1,… of (xn)∞n=1 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

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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:
 * 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} }}
 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.