Subsequence/Definition

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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

↑
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 5^th 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]
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

[3] 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 5^th term of [ilmath](x_n)[/ilmath].

[2] Then [ilmath](x_{k_n})_{n\in\mathbb{N} } [/ilmath] is a constant sequence where every term is [ilmath]x_5[/ilmath] - the 5^th term of [ilmath](x_n)[/ilmath].

[4] 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

[4] [ilmath]k_n\le k_{n+1} [/ilmath]

[APIKM-1] Analysis - Part 1: Elements - Krzysztof Maurin

[FAVIDMH-2] Functional Analysis - Volume 1: A gentle introduction - Dzung Minh Ha

[1]

[2]

[Note 1]

[Note 2]

Subsequence/Definition

Contents

Definition

As a mapping

Notes

References

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