A monotonically increasing sequence bounded above converges

Stub grade: C

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Good enough for now, routine first year work anyway

Statement

Let $(a_n)_{n\in\mathbb{N} }\subset\mathbb{R}$ be a real sequence. Suppose $\forall n\in\mathbb{N}[a_n\le a_{n+1}]$ (the sequence is monotonically increasing) and is bounded above (i.e. $\exists b\in\mathbb{R}\exists K\in\mathbb{N}\forall n\in\mathbb{N}[n>K\implies a_n\le b]$ - $b$ is the bound) then:

• $$\lim_{n\rightarrow\infty}(a_n)\eq\sup_{n\in\mathbb{N} }(a_n)$$

Proof

caption

By the axiom of completeness any set of real numbers with an upper bound has a supremum. Define:

• $\ell:\eq\sup_{n\in\mathbb{N} }(a_n)$

We can do this as we know there's an upper bound (denoted $b$ on the diagram)

We must now show $\lim_{n\rightarrow\infty}(a_n)\eq\ell$, which is of course equivalent to $\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies \vert a_n-\ell\vert<\epsilon]$

Proof:

• Let $\epsilon>0$ be given.
  • Let $N\in\mathbb{N}$ be such that $\forall n\in\mathbb{N}[n>N\implies a_n>\ell-\epsilon]$
    • We must first prove such an $N$ exists, we shall do so by contradiction.
    • Suppose there is no such $N$ such that $\forall n\in\mathbb{N}[n>N\implies a_n>\ell-\epsilon]$
      • That means for $n>N$ we always have $a_n\le \ell-\epsilon$
        • TODO: Be more formal with the proof work
      • But this means $\ell-\epsilon$ is an upper bound of $a_n$ lower than $\ell$ which is the supremum!
        • That contradicts that $\ell$ is the surpremum in the first place
    • Now we know such an $N$ exists
      • Let $n\in\mathbb{N}$ be given
        • Suppose $n\le N$ - by the nature of logical implication we do not care about the truth or falsity of $\vert a_n-\ell\vert<\epsilon$, either way the implication holds. We are done in this case
        • Suppose $n>N$ - we must show that in this case we have $\vert a_n-\ell\vert<\epsilon$
          • By definition of $N$ we have for all $n>N$ that $\ell-\epsilon<a_n$ so
            • we see $-\epsilon<a_n-\ell$, so $\epsilon>\ell-a_n$ and as $\ell\ge a_n$ always, we see $\epsilon>\ell-a_n\ge 0$
            • So $\vert a_n-\ell\vert\eq\vert \ell-a_n\vert\eq \ell-a_n < \epsilon$
            • Or just $\vert a_n-\ell\vert < \epsilon$ - as required.

References

Grade: D

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Routine and not important