🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

Let $S\subseteq \mathbb{R}$ and $S\ne \varnothing$, because of that let $s\in S$ and set $a_0=s$, quickly note that if it so happens that $a_0$ is an upper bound then we can take $a_0$ to be our least upper bound. We also know that $S$ is bounded above, so let $b_0$ be any upper bound of $S$, consider $d:=b_0-a_0$, and note that we must have $d>0$, for if not then $b_0<a_0$ which would mean that $b_0$ is not an upper bound. If $d=0$ then $a_0=b_0$, and so as before $S$ has the supremum $a_0$.

Therefore we can assume that $a_0<b_0$ and also assume that $a_0$ is not an upper bound. Now consider the value $M_n:=\frac{a_n+b_n}{2}$. Let's define two sequences simultaneously, if $M_n$ is an upper bound of $S$ then define $a_{n+1}=a_n$ and $b_{n+1}=M_n$, if $M_n$ is not an upper bound of $S$ then we define $a_{n+1}=M_n$ and $b_{n+1}=b_n$. The goal of this is to consider $M_n$ as an approximation of our supremum, if its too big we adjust our upper bound to force the next iteration to get closer, if it's inside the set we push up our lower bound to force it outside the set.

We claim that every $n\in {\mathbb{N}}_0,b_n$ is an upper bound of $S$ and that every $a_n$ is not an upper bound of $S$. We prove both at once through induction, we assumed $a_0$ was not an upper bound, and $b_0$ by definition was so the base case holds. Assume that the statement holds true for $k\in {\mathbb{N}}_0$ and we'll show that it holds true for $k+1$. Suppose that $M_k$ is an upper bound to $S$, in that case $b_{k+1}=M_k$ and so $b_{k+1}$ is an upper bound, in this case $a_{k+1}=a_k$ and since we knew that $a_k$ was not an upper bound $a_{k+1}$ is also not an upper bound, in the case that $M_k$ is not an upper bound, then $b_{k+1}=b_k$ and so $b_{k+1}$ is still an upper bound, but since $a_{k+1}=M_k$ we see that $a_{k+1}$ still remains not an upper bound as needed.

Consider the sequence of intervals $I_n:=[a_n,b_n]$ and define $d_j=\ell \left(I_j\right)$ then we claim that $${\displaystyle d_j=\frac{d_0}{2^j}}$$ For $j=0$ it holds trivially, then suppose for $k\in {\mathbb{N}}_0$ it holds true, and we'll prove it holds true for $j+1$, so we know that $\ell \left([a_j,b_j]\right)=\frac{d_0}{2_j}$. If $M_j$ is an upper bound then $a_{j+1}=a_j$ and $b_{j+1}=M_j$ therefore $\ell \left([a_{j+1},b_{j+1}]\right)=\frac{d_0}{2_j}·\left(\frac{1}{2}\right)=\frac{d_0}{2^{j+1}}$ as needed, the case when $M_j$ is not an upper bound follows symmetrically, thus the statement holds true for all $j\in {\mathbb{N}}_0$.

By the definition of $a_n,b_n$ it follows that for every $n,m\in {\mathbb{N}}_0$ we have that $a_n<b_m$ for if there was a $k,j\in {\mathbb{N}}_0$ such that $a_k\ge b_j$ since $b_j$ is an upper bound, so is $a_k$, but that's a contradiction because we proved that $a_k$ is not an upper bound for any $k$. Also consider through the definition of $a_n,b_n$ that we must have that $a_n\le a_{n+1}$ and that $b_n\le b_{n+1}$. This shows that for any $J\in {\mathbb{N}}_0$ and any $j\ge J,a_j,b_j\in [a_J,b_J]$ because we know that $a_J\le a_j\le b_j\le b_J$

Now consider the sequence $\left(c_n\right)=(a_0,b_0,a_1,b_1,a_2,b_2,\dots )$, defined so that

• $c_n=a_{\frac{n}{2}}$ if $n$ even
• $c_n=b_{\frac{n}{2}-1}$ if $n$ odd

We will show that this sequence is cauchy. So let $ϵ\in {\mathbb{R}}^{+}$ and let $K\in {\mathbb{N}}_0$ such that $d{2}^{-K}<ϵ$, note that we know that $\ell \left(I_K\right)=\frac{d_0}{2^K}$, take $N=2K+2$ and let $n,m\ge N$ we must prove that $|c_n-c_m|\le ϵ$ , since we know that $\frac{d_0}{2^K}\le ϵ$ it's enough to show that $c_n,c_m\in I_K$, recall that $c_n$ is either equal to $c_{\frac{n}{2}}$ or $b_{\frac{n}{2}-1}$ but since $N\ge 2K+2$ we see that the indices are greater than $K$ that is $${\displaystyle \frac{n}{2}>\frac{n}{2}-1\ge \frac{2K+2}{2}-1=K}$$ therefore no matter what $c_n,c_m\in I_K$ as hinted at earlier, in other words $|c_n-c_m|\le \frac{d_0}{2^K}$ as needed, so that $\left(c_n\right)$ is cauchy.

Since it is cauchy it converges to some limit $L$, we will prove that $L$ is the supremum of $S$, that is, it is the greatest upper bound, first of all it must be an upper bound, for if it wasn't then there would be some $s\in S$ such that $s>L$, recall that since $\left(c_n\right)\to L$ then $\left(b_n\right)\to L$ as it's a subsequence, in that case consider $ϵ=\frac{s-L}{2}$, then there exists some $N\in {\mathbb{N}}_0$ such that $|b_N-L|<ϵ$ so that $b_N<s$ but this implies that $b_N$ is not an upper bound which is a contradiction, therefore $L$ must be an upper bound.

But it also is the least upper bound, so suppose for the sake of contradiction that there was a smaller upper bound $B$ so that $B<L$ and set $ϵ=\frac{L-B}{2}$ similarly to our previous paragraph we can obtain some $N$ such that $|a_N-L|<ϵ$ which implies that $B<a_N$ but that would imply that $a_N$ is an upper bound, which is not possible, and therefore we have a contradiction so $L$ is the least upper bound.