My question was Prove, using the ε-characterisation of a limit, that a sequence $(a_n)_{n≥1}$ of real numbers which is bounded below and decreasing is convergent. You may assume that a set of real numbers which is bounded below has an infimum.
I wrote if a sequence of real numbers is decreasing and bounded below, then the infimum is the limit.
We prove if $A=\left\{a_n,n\in\mathbb N^\ast\right\}$ is bounded below then it is convergent and the limit is $c=\inf_{n\ge 1}\left\{a_n\right\}$.
Since $A$ is a non-empty and bounded below subset of $\mathbb R$, $c$ is the greatest lower bound of $A$ if it satisfies
- $c$ is a lower bound of $A$,
- for all $k\in\mathbb R$, if $k$ is a lower bound for $A$, then $k\le c$.
Now for every $\epsilon>0$ there exists $N\in\mathbb N^\ast$ such that $a_N<c+\epsilon$ since otherwise $c+\epsilon$ is an lower bound which contradicts c being the infimum of $A$.
Then since $(a_n)$ is decreasing, for $n\ge N$ we have $a_N\ge a_n$, so: $$c-\epsilon < c\le a_n\le a_N < c+\epsilon$$
We see for $n\ge N$, $c-\epsilon < a_n< c+\epsilon$ and so $\left|a_n−c\right|<\epsilon$. Since $\epsilon>0$ is arbitrary, we have $\lim_{n\to\infty} a_n=c$, that is $(a_n)$ is convergent to $c$.
I modified the proof from https://en.wikipedia.org/wiki/Monotone_convergence_theorem#Proof can someone check if I did it right please
We want to prove that if $(A_n)_{n=1}^\infty$ is a decreasing and bounded sequence of real numbers then $(A_n)$ is converging:
Theorem: $A_n$ will be called converging to limit $L\in\mathbb R$ if for every $\varepsilon>0$ there exists $N\in \mathbb N$ such that [complete it...]
Given $A_n$ is bounded and decreasing there exists $c\in\mathbb R$ it infimum [largest lower bound]
Claim: c satisfies the Limit Theorem
proof: c is a lower limit so for all $n$ we have $A_n>c$, c is also the infimum then for every $\varepsilon>0$ we have $N\in\mathbb N$ such that: $a_m<c+\varepsilon$ for all $N<m$
Can you complete the proof now?