I am currently trying to improve my skills doing epsilon-delta proves and I just attempted the following one. Since I'm such a newbie in calculus I would like to improve learning form my mistakes (lack of rigour...). Any help will be appreciated, thanks in advance.
PS: it looks long but it's actually really easy to follow
Theorem
$\{a_n\}_{n\in\mathbb{N}}$ converges $\iff$ $\{a_n\}_{n\in\mathbb{N}}$ is Cauchy
- My attempt to proof:
$\Rightarrow$:
Converges, thus, $\forall\epsilon>0,\ \exists n_0\in\mathbb{N}: \forall n\ge n_o \Rightarrow\lvert a_n-l\rvert \lt \epsilon/2$
Take $n,m \ge n_0$, then, $\lvert a_n-a_m\rvert \le \lvert a_n-l\rvert +\lvert a_m-l\rvert \lt 2\epsilon/2 = \epsilon$
$\Leftarrow$:
Cauchy: $\forall\epsilon>0,\ \exists n_0\in\mathbb{N}: \forall n,m\ge n_o \Rightarrow\lvert a_n-a_m\rvert \lt \epsilon/3$
First we have to prove that $a_n$ is bounded for every $m, n \ge n_0$: $\lvert a_n\rvert \le \lvert a_n - a_{n_0}\rvert + \lvert a_{n_0}\rvert \lt \epsilon + \lvert a_{n_0}\rvert = M$. Thus, $\lvert a_n\rvert \le M$.
Since its bounded we can define...
- $ \sup(n) = \sup\{a_m: m\ge n\} $
- $ \inf(n) = \inf\{a_m: m\ge n\} $
From the definition, it follows $ 0 \le \sup(n) - \inf(n)$
From the characterization of "supremum" and "infimum" we get that $ \forall \delta\gt 0,\ \forall t_0,\ \exists s, t\ge t_0$ such that...
$\sup(t_0) - \delta/3 \lt a_s$
$\inf(t_0) + \delta/3 \gt a_t$
Let $\delta = \epsilon$, $t_0 = n_0$ and $m, n \ge n_0$. Subtracting the second equation to the first one: $\sup(n_0) - \inf(n_0) - 2\epsilon/3\lt a_n - a_m \le \lvert a_n-a_m\rvert \lt \epsilon/3 \Rightarrow \sup(n_0) - \inf(n_0) - 2\epsilon/3 \lt \epsilon/3 \Rightarrow 0 \stackrel{\text{seen above}}{\le} \sup(n_0) - \inf(n_0) \lt \epsilon$ (note that this is not true in general but our hypothesis guarantees that this happens)
This implies that $\lim_{n\to \infty } \sup(n) - \inf(n) = 0$
Now, since $\{\sup(n)\}_{n\in \mathbb{N}}$ is decreasing and bounded below, it converges (and its limit exists).
Decreasing: since $\{a_m: m\ge n+1\} \subseteq \{a_m: m\ge n\} \Rightarrow \sup(n) = \sup\{a_m: m\ge n\} \ge \sup(n+1) = \sup\{a_m: m\ge n+1\}$
Bounded below: As we saw before, $\lvert a_n\rvert \lt M \iff -M \lt a_n \lt M \Rightarrow - M \lt a_n \le \sup(n)$
Similarly we can prove that $\{\inf(n)\}_{n\in \mathbb{N}}$ is increasing and bounded above, so it converges too (and its limit exists).
So, we can conclude that $0 = \lim_{n\to \infty } \sup(n) - \inf(n) \stackrel{\text{both converge}}{=} \lim_{n\to \infty } \sup(n) - \lim_{n\to \infty } \inf(n)$. Let's define $L = \lim_{n\to \infty } \sup(n) = \lim_{n\to \infty } \inf(n)$
From the definition of "sup" and "inf" we get that: $\inf(n) \le a_n \le \sup(n)$
And we also know that: $L = \lim_{n\to \infty } \sup(n) = \lim_{n\to \infty } \inf(n)$
By squeeze theorem:
$\exists \lim_{n\to \infty } a_n \iff \{a_n\}_{n\in\mathbb{N}}$ converges (QED)
$\lim_{n\to \infty } a_n = L$
$\quad\square$
Please, let me know if there's something wrong or it is not rigorous enough. As I said before, any help will be highly appreciated since it's my first "hard" proof and I want to keep improving