A monotonically increasing sequence bounded above converges

From Maths
Jump to: navigation, search
Stub grade: C
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Good enough for now, routine first year work anyway


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]


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]


  • 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
      • 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>\ell-a_n[/ilmath] and as [ilmath]\ell\ge a_n[/ilmath] always, we see [ilmath]\epsilon>\ell-a_n\ge 0[/ilmath]
              • So [ilmath]\vert a_n-\ell\vert\eq\vert \ell-a_n\vert\eq \ell-a_n < \epsilon[/ilmath]
              • Or just [ilmath]\vert a_n-\ell\vert < \epsilon[/ilmath] - as required.


Grade: D
This page requires references, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:
Routine and not important