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 $L<s$. 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.
By construction $l_0<u_0$. Assume that $l_n<u_n$. 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 $l_{n+1}<u_{n+1}$ 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.