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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.

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  • $\begingroup$ Note that $|u_{n+1}-u_n|\leq\frac12(l_n-u_n)=\frac1{2^n}(l_0-u_0)$ and use $|u_{n+m}-u_n|\leq\sum_{l=0}^{m-1}|u_{n+l+1}-u_{n+l}|$. $\endgroup$ – Jens Schwaiger Mar 30 '20 at 5:22
  • $\begingroup$ I don't see why $|𝑒_𝑛+1βˆ’π‘’_𝑛|≀\frac{1}{2}(𝑙_π‘›βˆ’π‘’_𝑛)=\frac{1}{2^𝑛}(𝑙_0βˆ’π‘’_0)$? $\endgroup$ – Yeet Mar 30 '20 at 9:41
  • $\begingroup$ By construction $|u_{n+1}-u_n|$ equals either $0$ or $|\frac12(u_n+l_n)-u_n|=|\frac12(l_n-u_n)|$. $\endgroup$ – Jens Schwaiger Mar 31 '20 at 3:36
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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.

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