Is my proof that Cauchy sequences converge valid During study for my math course, I decided to practice my proof skills by trying to prove the following theorem:
Theorem: If $ (a_n)_{n=1}^\infty $ is a Cauchy sequence in $ \mathbb{R} $, then it converges.
I have finished the proof, and afterwards googled how other people proved it. As I expected (because I'm an amateur), my proof was quite convoluted compared to others. Despite this, I am wondering if my proof is valid, and if it isn't, what I would need to change. The outline of my proof is as follows;


*

*Prove that the sequence is bounded

*Choose a bunch of arbitrary epsilons, and from that create smaller and smaller intervals, which are all subsets of each other.

*Going forever, as the intervals get smaller, eventually we can "hone in" onto the limit.

*Use the definition of all the intervals (sets in $\mathbb{R}$) to prove that a number $L$ is the limit.


Note: the second triangle inequality is as follows $||x|-|y||\leq|x-y|$ for all $x,y\in\mathbb{R}$.
Proof Fix $ \epsilon_0 = 1 $. There exists $N_1\in \mathbb{N}$ such that when $n,k \geq N_1$, then $|a_n-a_k|<1$. By the second triangle inequality, $|a_n|-|a_k|<1$ which implies that $|a_n|<1+|a_k|$.
Thus, there exists $M=\max\{a_1,a_2,...,a_{N_1-1},1+|a_k|\}>0$ such that $|a_n|<M$ for all $n \in \mathbb{N}$. By definition, the sequence $ (a_n)_{n=1}^\infty $ is bounded. This means we can let 
$$x_1=\min\{a_N,a_{N+1},a_{N+2},...\}$$
$$y_1=\max\{a_N,a_{N+1},a_{N+2},...\}$$
$$z_1=\frac {x_1+y_1} {2}$$
Now define the set $\Omega_1=[z_1-\epsilon_0,z_1+\epsilon_0]=[z_1-1,z_1+1]$. Clearly $a_n\in\Omega_1$ for all $n\geq N_1$.
Now fix $\epsilon_0=\frac12$. There exists $N_2>N_1\in\mathbb{N}$ such that when $n,k \geq N_2$, then $|a_n-a_k|<\frac12$. Define the set $\Omega_2=[z_1-\epsilon_0,z_1+\epsilon_0]=[z_1-\frac12,z_1+\frac12]$ where $z_2$ is defined analogously to $z_1$. Because $N_2>N_1$ and for all $n\geq N_1$, $a_n\in\Omega_1$, we conclude that $\Omega_2\subseteq\Omega_1$. Furthermore, $\frac12<1$ which means that $\Omega_2\subset\Omega_1$.
Now fix $\epsilon_0=\frac13$. There exists $N_3>N_2$ such that when $n,k\geq N_3$, then $|a_n-a_k|<\frac13$. We define the set $\Omega_3$ analogously to $\Omega_2$ and by the same reasoning conclude that $\Omega_3\subset\Omega_2$
Continue like this for $\epsilon_0=\frac14,\frac15,...,\frac1m,...$ and so on. We are left with infinitely many sets $\Omega_m\subset\mathbb{R}$ and numbers $N_m\in\mathbb{N}$ such that $a_n\in\Omega_m$ for all $n\geq N_m$. Now we define the set $X$ as follows:
$$X=\bigcap\limits_{m\in\mathbb{N}} \Omega_m=\Omega_1 \cap \Omega_2 \cap \Omega_3 \cap \Omega_4 \cap ...$$
The interval that $\Omega_m$ is defined on is $[z_m-\frac1m,z_m+\frac1m]$. This interval has a width of $\frac2m$ which can be made smaller than any positive number, for large enough $m$. Also, there are infinitely many such sets $\Omega_m$ and $\Omega_m\subset\Omega_{m+1}$ for all $m\in\mathbb{N}$. Therefore, the set $X$ must contain only one element, which we will call $L$.
Claim $\lim_{n\to\infty}a_n=L$
Proof of claim Fix $\epsilon>0$. Consider the set $\Omega_m$ defined on the interval $[z_m-\frac1m,z_m+\frac1m]$. Choose $m\in\mathbb{N}$ so that $\Omega_m\subseteq(L-\epsilon,L+\epsilon)$. Such an $m$ always exists because $L\in\Omega_m$ and we can make the width of the interval, $\frac2m$, smaller than any positive number, including $\epsilon$.
Now consider the number $N_m$ defined along with the set $\Omega_m$. If $n\geq N_m$, then
$$a_n\in\Omega_m=\bigg[z_m-\frac1m,z_m+\frac1m\bigg]\subseteq(L-\epsilon,L+\epsilon)\implies|a_n-L|<\epsilon\quad\blacksquare$$
How did I do? If it is valid, is it "rigorous" enough by modern standards? If it isn't valid, what can be changed, if anything?
Update: This proof assumes the completeness of the real numbers (in assuming that the narrowing intervals intersect to one point), so in a way, it is a circular proof. :) I still like it though.
 A: Firstly, let me say this is a cool proof idea! I've never seen it before but I think the overall structure works! Nice job. I can see a lot of thinking and effort must have gone into trying to formalise this. Hopefully what I say will be helpful. Everything will be in "chronological order" rather than order of importance, but then again I think all of it's pretty important.
Secondly, if a set is bounded, it doesn't mean there's a minimum or maximum. For example, what is $\min\{1/n:n\in\mathbb{N}\}$ equal to? The infimum is clearly $0$, but there's no minimum. However, the way you've used $z_1$, you can just redefine $x_1$ and $y_1$ in terms of the supremum and infimum. (Naturally this needs to be changed for each of your $z_n$ terms.)
Hence Thing to be careful of number 1: There are often edge cases or slightly pathological cases that you need to deal with. Think about the worst case scenario.
Thirdly, you should also be careful, because it's not necessarily true that $\Omega_{m+1} \subset \Omega_m$. For example, consider the tail of a sequence $\{a_{N_3}, a_{N_3+1}, \cdots\}$ = $\{0.3, 0, 0, 0, \cdots \}$. Then $\Omega_3 = [0.15-1/3, 0.15+1/3] \approx [-0.183, 0.483]$ However, when you let $\epsilon_3 = 1/4$, $\Omega_4 = [0-1/4, 0+1/4] = [-0.25, 0.25]$. Clearly $\Omega_4 \not\subset \Omega_3$. However, it is clear that they intersect. In fact for your proof, the fact that $\Omega_{m+1} \subset \Omega_m$ isn't even important. The result that matters (which also happens to be true) is that $\bigcap_{i=1}^m\Omega_i$ is nonempty for all $i$. I understand that you effectively "proved this" with your subset argument, but because the subset argument isn't actually true, this is a significant flaw. (Fun fact: If your epsilons were of the form $1/2^n$ instead of $1/n$, then that part of your proof would be fine. Can you prove that fact?)
Hence Thing to be careful of number 2: Often people claim or prove stronger results than required in order to prove the final result. This means you're doing more work than required, so it increases the chances of making a mistake. You want to be concise.
The last thing I'll talk about is your paragraph before the "proof" section. You used a lot of words without any mathematical evidence to prove a very important result. When I say "important", I mean it's equivalent to the epsilon proof you did at the end, without any epsilons.
Well, technically, it wasn't important. If you showed that $X$ is non-empty, all you had to do was say "Let $L \in X$" and commence in exactly the same way. Uniqueness will follow if you can prove that any $L$ in $X$ is the "limit" first.
The first thing to note here is just a lack of rigour, but the more important fact is you once again stated something stronger than it had to be and it lead to potential problems. (Refer to thing to be careful about number 2). If you're more comfortable proving that $X$ is a singleton anyway, to be consistent with the rest of the proof, I would suggest you use epsilons.
Actually I lied that wasn't the last thing - one more thing!! In a proof it's not very good style to redefine the same symbol. You continued to redefine $\epsilon_0$, but what I would do is start at $\epsilon_1$, then number them the same way as $N$ and $z$.
But yeah nice proof, I loved the concept!
