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This proof of the Least Upper Bound Property defines sequences $A_1, A_2,... $ and $B_1, B_2,...$ recursively: Let S be a nonempty set of the reals that is bounded above. Let $B_1$ be an upper bound of S, and let $A_1$ be an element of S that is not an upper bound.

1)Compute $(A_n +B_n)/2$.

2) If it is an upper bound of S, then take $A_{n+1}=A_n$ and $B_{n+1}=(A_n +B_n)/2$.

3) Otherwise, there is an $s\in S$ such that $(A_n +B_n)/2<s$. Take $A_{n+1}=s$ and $B_{n+1}=B_n$.

How does one (as the wiki page suggests) show that these sequences are Cauchy? Does one have to provide a direct proof (which I was having a hard time with due to the recursive definition) or use the fact that $A_1 \leq A_2\leq ...\leq B_2 \leq B_1$ and $|A_n -B_n|$ tends to 0?

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up vote 1 down vote accepted

Just use the facts that $A_1\le A_2\le A_3\le\ldots\le B_3\le B_2\le B_1$ and $|A_n-B_n|=B_n-A_n\to 0$ as $n\to\infty$: that $\langle A_n:n\in\Bbb Z^+\rangle$ and $\langle B_n:n\in\Bbb Z^+\rangle$ follows immediately by the following argument.

Fix $\epsilon>0$. There is an $n_0\in\Bbb Z^+$ such that $|A_n-B_n|<\epsilon$ whenever $n\ge n_0$. Thus, for any $m,n\ge n_0$ we have $$A_{n_0}\le A_m,A_n,B_m,B_n\le B_{n_0}$$ and hence $$|A_m-A_n|,|B_m-B_n|\le|A_{n_0}-B_{n_0}|<\epsilon\;,$$ as desired.

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