A monotonically increasing sequence bounded above converges
From Maths
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Good enough for now, routine first year work anyway
Contents
Statement
Let [ilmath](a_n)_{n\in\mathbb{N} }\subset\mathbb{R} [/ilmath] be a real sequence. Suppose [ilmath]\forall n\in\mathbb{N}[a_n\le a_{n+1}][/ilmath] (the sequence is monotonically increasing) and is bounded above (i.e. [ilmath]\exists b\in\mathbb{R}\exists K\in\mathbb{N}\forall n\in\mathbb{N}[n>K\implies a_n\le b][/ilmath]  [ilmath]b[/ilmath] is the bound) then:
 [math]\lim_{n\rightarrow\infty}(a_n)\eq\sup_{n\in\mathbb{N} }(a_n)[/math]
Proof
By the axiom of completeness any set of real numbers with an upper bound has a supremum. Define: [ilmath]\ell:\eq\sup_{n\in\mathbb{N} }(a_n)[/ilmath]
We can do this as we know there's an upper bound (denoted [ilmath]b[/ilmath] on the diagram)
We must now show [ilmath]\lim_{n\rightarrow\infty}(a_n)\eq\ell[/ilmath], which is of course equivalent to [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies \vert a_n\ell\vert<\epsilon][/ilmath]
Proof:
 Let [ilmath]\epsilon>0[/ilmath] be given.
 Let [ilmath]N\in\mathbb{N} [/ilmath] be such that [ilmath]\forall n\in\mathbb{N}[n>N\implies a_n>\ell\epsilon][/ilmath]
 We must first prove such an [ilmath]N[/ilmath] exists, we shall do so by contradiction.
 Suppose there is no such [ilmath]N[/ilmath] such that [ilmath]\forall n\in\mathbb{N}[n>N\implies a_n>\ell\epsilon][/ilmath]
 That means for [ilmath]n>N[/ilmath] we always have [ilmath]a_n\le \ell\epsilon[/ilmath]
 TODO: Be more formal with the proof work

 But this means [ilmath]\ell\epsilon[/ilmath] is an upper bound of [ilmath]a_n[/ilmath] lower than [ilmath]\ell[/ilmath] which is the supremum!
 That contradicts that [ilmath]\ell[/ilmath] is the surpremum in the first place
 That means for [ilmath]n>N[/ilmath] we always have [ilmath]a_n\le \ell\epsilon[/ilmath]
 Now we know such an [ilmath]N[/ilmath] exists
 Let [ilmath]n\in\mathbb{N} [/ilmath] be given
 Suppose [ilmath]n\le N[/ilmath]  by the nature of logical implication we do not care about the truth or falsity of [ilmath]\vert a_n\ell\vert<\epsilon[/ilmath], either way the implication holds. We are done in this case
 Suppose [ilmath]n>N[/ilmath]  we must show that in this case we have [ilmath]\vert a_n\ell\vert<\epsilon[/ilmath]
 By definition of [ilmath]N[/ilmath] we have for all [ilmath]n>N[/ilmath] that [ilmath]\ell\epsilon<a_n[/ilmath] so
 we see [ilmath]\epsilon<a_n\ell[/ilmath], so [ilmath]\epsilon>\ella_n[/ilmath] and as [ilmath]\ell\ge a_n[/ilmath] always, we see [ilmath]\epsilon>\ella_n\ge 0[/ilmath]
 So [ilmath]\vert a_n\ell\vert\eq\vert \ella_n\vert\eq \ella_n < \epsilon[/ilmath]
 Or just [ilmath]\vert a_n\ell\vert < \epsilon[/ilmath]  as required.
 By definition of [ilmath]N[/ilmath] we have for all [ilmath]n>N[/ilmath] that [ilmath]\ell\epsilon<a_n[/ilmath] so
 Let [ilmath]n\in\mathbb{N} [/ilmath] be given
 Let [ilmath]N\in\mathbb{N} [/ilmath] be such that [ilmath]\forall n\in\mathbb{N}[n>N\implies a_n>\ell\epsilon][/ilmath]
References
Grade: D
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