# Proof review - (lack of rigour?) Convergent sequence iff Cauchy without Bolzano-Weierstrass

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

• Well clearly your idea is the right one (to prove it without Bolzano-Weierstrass) and your proof is almost perfect. Some tiny details : fix $\epsilon$ before doing your calculations using $n_0$ and so on; for instance say "Let $\epsilon >0$". Also, fix $n_0$ when you do your calculations on $sup(n_0)$ and $inf(n_0)$, otherwise it's not really crystal clear (it would seem you proved "$\forall \epsilon >0, \forall n_0, 0\leq sup(n_0) - inf(n_0) \leq \epsilon$", which is false in general). Your problem is essentially that you don't fix your variables – Max Jun 6 '17 at 18:29
• Expanding on Max's comment, I'd caution against using a bunch of symbolic quantifiers in your final proof (it can be really useful shorthand in a draft) as it both degrades readability and makes these sorts of errors really easy to make; it definitely isn't more rigorous. – Christian Sykes Jun 6 '17 at 19:38
• Also, all the mucking about with fractions to get your $\epsilon$ inequality just so is unnecessary (though very common to beginning analysis students). If you get your quantity of interest less than some positive multiple of $\epsilon$, that is good enough. (People do have different opinions on this on aesthetic grounds, but I think it's an unnecessary distraction.) – Christian Sykes Jun 6 '17 at 19:41
• @ChristianSykes I know you know, but just to make it clear to beginner readers: "a positive multiple of $\epsilon$" that does not depend on $n$ (or the variable we're moving around) – Max Jun 6 '17 at 20:04
• @J.Doe if you want a general condition on $f$ you must make sure that: $f>0$; $f(x)\to 0$ as $x\to 0$; $f$ does not depend on $n$ or the variables in question (for instance, $|a_n-l| < n\epsilon$ will not do the trick to show $a_n \to l$ ) . I think this is enough (of course you can mess around a bit and in some examples you will find $f$'s that do not have these properties but you can still make sure that they work; but they will be special cases) – Max Jun 6 '17 at 20:11