# Sequence

A sequence is one of the earliest and easiest definitions encountered, but I will restate it.

I was taught to denote the sequence [math]\{a_1,a_2,...\}[/math] by [math]\{a_n\}_{n=1}^\infty[/math] however I don't like this, as it looks like a set. I have seen the notation [math](a_n)_{n=1}^\infty[/math] and I must say I prefer it. This notation is inline with that of a tuple which is a generalisation of an ordered pair.

## Definition

Formally a sequence [ilmath](A_i)_{i=1}^\infty[/ilmath] is a function^{[1]}^{[2]}, [math]f:\mathbb{N}\rightarrow S[/math] where [ilmath]S[/ilmath] is some set. For a finite sequence it is simply [math]f:\{1,...,n\}\rightarrow S[/math]. Now we can write:

- [ilmath]f(i):=A_i[/ilmath]

This naturally then generalises to indexing sets

## Notation

To specify that the points of a sequence, the [ilmath]x_i[/ilmath] are from a space, [ilmath]X[/ilmath] we may write:

- [ilmath](x_n)^\infty_{n=1}\subseteq X[/ilmath]

This is an abuse of notation, as [ilmath](x_n)^\infty_{n=1}[/ilmath] is not a subset of [ilmath]X[/ilmath]. It plays on:

- [ilmath][(x_n)^\infty_{n=1}\subseteq X]\iff[x\in(x_n)_{n=1}^\infty\implies x\in X][/ilmath]

Note that the elements of [ilmath](x_n)_{n=1}^\infty[/ilmath] are ether:

- Elements of a relation (if we consider the sequence as a mapping) or
- So using this, [ilmath]x\in(x_n)_{n=1}^\infty[/ilmath] may look like [ilmath]x=(a,b)[/ilmath] (indicating [ilmath]f(a)=b[/ilmath]) which is an Ordered pair, not in [ilmath]X[/ilmath]

- Elements of a tuple (which is a generalisation of ordered pairs where (usually) [ilmath](a,b)=\{\{a\},\{a,b\}\}[/ilmath]
- So using this, [ilmath]x\in(x_n)_{n=1}^\infty[/ilmath] may indeed look like [ilmath]x=\{\{a\},\{a,b\}\}\notin X[/ilmath]

**As such the notation [ilmath](x_n)^\infty_{n=1}\subseteq X[/ilmath] having no other sensible meaning** is a notation to say that [ilmath]\forall i[x_i\in X][/ilmath]

## Subsequence

Given a **sequence** [ilmath](x_n)_{n=1}^\infty[/ilmath] we define a *subsequence of [ilmath](x_n)^\infty_{n=1}[/ilmath]*^{[3]}^{[4]} 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]}

- That means that [ilmath]\forall n\in\mathbb{N}[k_n<k_{n+1}][/ilmath]

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]

- That is to say the [ilmath]i[/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

## See also

- Subsequence
- Monotonic sequence
- Bolzano-Weierstrass theorem
- Cauchy sequence (Alternatively: Cauchy criterion for convergence)
- Convergence of a sequence (Or Limit (sequence) - the page
*Convergence of a sequence*is being refactored into it)

## 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].

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

## References

- ↑ p46 - Introduction To Set Theory, third edition, Jech and Hrbacek
- ↑ p11 - Analysis - Part 1: Elements - Krzysztof Maurin
- ↑ Analysis - Part 1: Elements - Krzysztof Maurin
- ↑ Functional Analysis - Volume 1: A gentle introduction - Dzung Minh Ha