Subsequence

Definition

Given a sequence $(x_n)_{n=1}^\infty$ we define a subsequence of $(x_n)^\infty_{n=1}$ ^[1]^[2] as follows:

Given any strictly increasing monotonic sequence^{[Note 1]}, (kn)∞n=1⊆N
- That means that $\forall n\in\mathbb{N}[k_n<k_{n+1}]$ ^{[Note 2]}

Then the subsequence of $(x_n)$ given by $(k_n)$ is:

(xkn)∞n=1, the sequence whose terms are: xk1,xk2,…,xkn,…
- That is to say the $i$ ^th element of $(x_{k_n})$ is the $k_i$ ^th element of $(x_n)$

As a mapping

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

$\forall a,b\in\mathbb{N}[a<b\implies k(a)<k(b)]$

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

Now $(x_{k_n})_{n=1}^\infty$ is a subsequence

Immediate properties

[Expand]

$\forall n\in\mathbb{N}[k_n\ge n]$ - the $k_i$ ^th term of a subsequence cannot correspond to any element before the (but not including) $i$ ^th term of the main sequence

This is a standard proof by induction.

Suppose n=1, show k1≥1
- Well $k_1$ is the first term of the subsequence and this may be the 1st term of the sequence, in which case we have $k_1\eq 1$ thus $k_1\ge 1$ holds
- If the first term of the subsequence is any term other than the first we see $k_1>1$ and so $k_1\ge 1$ holds
Assuming we have kn≥n show that this implies kn+1≥n+1
- We have kn≥n by assumption, and we wish to show kn≥n⟹kn+1≥n+1
  - By definition of subsequence we see $k_n < k_{n+1}$ so $n\le k_n < k_{n+1}$ means $n<k_{n+1}$
  - As $k_{n+1}$ and $n$ are integer valued, we can use $(n<m)\iff(n+1\le m)$ for integers $n$ and $m$ ^{[Note 3]}
  - Thus we see $(n<k_{n+1})\iff(n+1\le k_{n+1})$ , we have the LHS, so now we have the RHS: $n+1\le k_{n+1}$
Draw conclusions
- Since $k_n\ge n$ is true for $n\eq 1$ and if it is true for $n\eq m$ then it is true for $n\eq m+1$ we have proved by induction that for all $n\in\mathbb{N}$ we have $k_n\ge n$ , as required

Notes

Jump up ↑
Note that strictly increasing cannot be replaced by non-decreasing as the sequence could stay the same (ie a term where mi=mi+1 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: kn:=5 for all n.
- Then $(x_{k_n})_{n\in\mathbb{N} }$ is a constant sequence where every term is $x_5$ - the 5^th term of $(x_n)$ .
Jump up ↑
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
Jump up ↑ The proof of this is elementary and omitted here. Usually I avoid "exercise for the reader" but they really ought to be able to see it themselves, kids in year 3 can!

References

[3] Jump up ↑ Note that strictly increasing cannot be replaced by non-decreasing as the sequence could stay the same (ie a term where $m_i\eq m_{i+1}$ 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: $k_n:\eq 5$ for all $n$ .
Then $(x_{k_n})_{n\in\mathbb{N} }$ is a constant sequence where every term is $x_5$ - the 5^th term of $(x_n)$ .

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

[4] Jump up ↑ 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 $\iff$ every sequence contains a convergent subequence. If we only require that:
$k_n\le k_{n+1}$
Then we can define the sequence: $k_n:=1$ . This defines the subsequence $x_1,x_1,x_1,\ldots x_1,\ldots$ of $(x_n)_{n=1}^\infty$ 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.
The mapping definition directly supports this, as the mapping can be thought of as choosing terms

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

[5] Jump up ↑ The proof of this is elementary and omitted here. Usually I avoid "exercise for the reader" but they really ought to be able to see it themselves, kids in year 3 can!

[APIKM-1] Jump up ↑ Analysis - Part 1: Elements - Krzysztof Maurin

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

[1]

[2]

[Note 1]

[Note 2]

[Note 3]

Subsequence

Contents

Definition

As a mapping

Immediate properties

See also

Notes

References

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