If a sequence ${a_n}$ is monotonically increasing. then $\lim_{n \to \infty} a_n = \sup{(a_n)}$

Question

Can you please tell me if my proof is correct:

If a sequence ${a_n}$ is monotonically increasing. Then $$\lim_{n \to \infty} a_n = \sup{(a_n)}$$

Proof: $$a_n\leq a_{n+1}\leq \sup(a_n)$$ Assume for the sake of contradiction that $$|a_n-sup(a_n)|>\epsilon$$ $$a_n-\sup(a_n)>\epsilon$$ $$a_n>\epsilon+\sup(a_n)$$ !!contradiction with definition of supremum Hence, $$|a_n-\sup(a_n)| \leq \epsilon$$ which implies $$\lim_{n \to \infty} a_n = \sup{(a_n)}$$

Answer 1

Your proof is incomplete. When you assume by contradiction of $|a_n−\sup(a_n)|\le\epsilon$, i.e., $-\epsilon\le a_n-\sup(a_n)\;\;\text{and}\;\;a_n-\sup(a_n)\le\epsilon$, the negation is $-\epsilon> a_n-\sup(a_n)\;\;\text{or}\;\;a_n-\sup(a_n)>\epsilon$. So you need to prove two cases.

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Proof. Let $\epsilon>0$ be a positive real number and let $x:=\sup(a_n)_{n=m}^\infty$. Since $x-\epsilon<x$, by Lemma 1, there exists a integer $m'$ such that $x-\epsilon<a_{m'}\le x$, i.e., $x-\epsilon<a_{m'}<x+\epsilon$. Also, since $a_n \le a_{n+1}$ for every $n$; we can show by induction (using base case $m'$) that $x-\epsilon<a_n<x+\epsilon$ for every $n\ge m'$, i.e., $-\epsilon<a_n-x<\epsilon$ for every $n\ge m'$. Thus we have $|a_n-x|<\epsilon$ for ervery $n\ge m'$. By Lemma 2, $(a_n)_{n=m}^\infty$ converges to $x$.

NOTE. The Lemma 2 is subtle. But I do not know with what premises you are working.

Lemma 1. Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers, and let $x:=\sup(a_n)_{n=m}^\infty$. Then for every real number $y$ for which $y<x$, there exists at least one $n\ge m$ for which $y<a_n\le x$.

Lemma 2. Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers, let $L$ be areal number, and let $m'\ge m$ be an integer. Then $(a_n)_{n=m}^\infty$ converges to $L$ if and only if $(a_n)_{n=m'}^\infty$ converges to $L$.