Understanding Wiki proof - Least upper bound property

I'm constructing the real numbers as equivalence classes of Cauchy sequences and I'm trying to prove that this construction has the least upper bound property. Wiki has a proof (https://en.wikipedia.org/wiki/Construction_of_the_real_numbers) which goes like this.

Let $$S$$ be a non-empty subset of $$\mathbb{R}$$. Then there exists an $$s\in\mathbb{R}$$. Let $$U$$ be an upper bound of $$S$$, we can assume that $$U$$ is rational. We define two sequences $$(u_n)$$ and $$(\ell_n)$$. Let $$u_0=U$$ and $$\ell_0=L$$ where L is a rational number $$L. Let $$m_n$$ be the arithmetic mean of $$u_n$$ and $$\ell_n$$, $$m_n=\frac{u_n+\ell_n}{2}$$ Now we will construct elements in the two sequences in the following way:

If $$m_n$$ is an upper bound of $$S$$, then $$u_{n+1}=m_n$$ and $$\ell_{n+1}=\ell_n$$

If $$m_n$$ is not an upper bound of $$S$$, then $$u_{n+1}=u_n$$ and $$\ell_{n+1}=m_n$$

Now the proof states that $$(u_n)$$ and $$(\ell_n)$$ are Cauchy sequences of rationals, but I can't seem to prove that these two are Cauchy, any help will be greatly appreciated.

• Note that $|u_{n+1}-u_n|\leq\frac12(l_n-u_n)=\frac1{2^n}(l_0-u_0)$ and use $|u_{n+m}-u_n|\leq\sum_{l=0}^{m-1}|u_{n+l+1}-u_{n+l}|$. – Jens Schwaiger Mar 30 '20 at 5:22
• I don't see why $|𝑢_𝑛+1−𝑢_𝑛|≤\frac{1}{2}(𝑙_𝑛−𝑢_𝑛)=\frac{1}{2^𝑛}(𝑙_0−𝑢_0)$? – Yeet Mar 30 '20 at 9:41
• By construction $|u_{n+1}-u_n|$ equals either $0$ or $|\frac12(u_n+l_n)-u_n|=|\frac12(l_n-u_n)|$. – Jens Schwaiger Mar 31 '20 at 3:36

By construction $$l_0. Assume that $$l_n. Then the construction of $$l_{n+1},u_{n+}$$ shows that either $$l_{n+1}=m_n$$ and $$u_{n+1}=u_n$$ or $$l_{n+1}=l_n$$ and $$u_{n+1}=m_n$$, where $$m_n=\frac12(l_n+u_n)$$. Thus in any case $$l_{n+1} and $$u_{n+1}-l_{n+1}$$ equals either $$m_n-l_n$$ or $$u_n-m_n$$. And these two expression are the same, namely $$\frac12(u_n-l_n)$$. This also implies that $$0\leq u_{n+1}-u_n\leq \frac12(u_n-l_n)$$. Induktion makes clear that $$u_n-l_n=\frac1{2^n}(u_0-l_0)=:\frac1{2^n}c$$ for all $$n$$.
Therefore $$0\leq u_{n+1}-u_n\leq\frac1{2^{n+1}}c$$ and $$0\leq u_{n+k}-u_n=\sum_{j=0}^{k-1} (u_{n+j+1}-u_{n+j})\leq \sum_{j=0}^{k-1}\frac1{2^{n+j+1}}c.$$ But $$\sum_{j=0}^{k-1}\frac1{2^{n+j+1}}c= \frac{c}{2^{n+1}}\sum_{j=0}^{k-1}\frac1{2^j}\leq \frac{c}{2^{n+1}}\sum_{j=0}^{\infty}\frac1{2^j}= \frac{c}{2^{n}}$$ showing that the sequence $$(u_n)$$ is Cauchy.