View source for Subsequence

==[[Subsequence/Definition|Definition]]==
{{:Subsequence/Definition}}
==Immediate properties==
{{Begin Inline Theorem}}
{{M|1=\forall n\in\mathbb{N}[k_n\ge n]}} - the {{M|k_i}}<sup>th</sup> term of a subsequence cannot correspond to any element before the (but not including) {{M|i}}<sup>th</sup> term of the main sequence 
{{Begin Inline Proof}}
This is a standard [[proof by induction]].
# Suppose {{M|n\eq 1}}, show {{M|k_1\ge 1}}
#* Well {{M|k_1}} is the first term of the subsequence and this may be the 1st term of the sequence, in which case we have {{M|k_1\eq 1}} thus {{M|k_1\ge 1}} holds
#* If the first term of the subsequence is any term other than the first we see {{M|k_1>1}} and so {{M|k_1\ge 1}} holds
# Assuming we have {{M|k_n\ge n}} show that this [[implies]] {{M|k_{n+1}\ge n+1}}
#* We have {{M|k_n\ge n}} by assumption, and we wish to show {{M|k_n\ge n\implies k_{n+1}\ge n+1}}
#** By definition of subsequence we see {{M|k_n < k_{n+1} }} so {{M|n\le k_n < k_{n+1} }} means {{M|n<k_{n+1} }}
#** As {{M|k_{n+1} }} and {{M|n}} are integer valued, we can use {{M|(n<m)\iff(n+1\le m)}} for integers {{M|n}} and {{M|m}}<ref group="Note">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!</ref>
#** Thus we see {{M|(n<k_{n+1})\iff(n+1\le k_{n+1})}}, we have the LHS, so now we have the RHS: {{M|n+1\le k_{n+1} }}
# Draw conclusions
#* Since {{M|k_n\ge n}} is true for {{M|n\eq 1}} and if it is true for {{M|n\eq m}} then it is true for {{M|n\eq m+1}} we have proved by induction that for all {{M|n\in\mathbb{N} }} we have {{M|k_n\ge n}}, as required
{{End Proof}}{{End Theorem}}
==See also==
* [[Sequence]]
==Notes==
<references group="Note"/>
==References==
<references/>
{{Definition|Real Analysis|Set Theory|Functional Analysis}}