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If we state the Least Upper Bound property as an axiom for the real numbers, we can use this along with the other axioms for $\mathbb{R}$ to show that $\mathbb{R}$ is a complete metric space; so I would like to ask about the reverse argument:

How can one show, using the assumption that $\mathbb{R}$ is a complete metric space (with the usual metric) and the ordered field axioms for $\mathbb{R}$, that $\mathbb{R}$ satisfies the Least Upper Bound property?

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    $\begingroup$ I think you need the Archimedean property in addition to it being a complete ordered field. $\endgroup$ – Daniel Fischer Sep 16 '13 at 20:48
  • $\begingroup$ Nonstandard real numbers form a counter example to this claim. But, as Daniel said, AP + completeness implies the least upper bound axiom. However, if you define completeness using filters rather than sequences, you will be fine. $\endgroup$ – Moishe Kohan Sep 16 '13 at 21:10
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You can do this as follows.

Let $A\subset\mathbb R$ be nonempty and bounded above. Choose $a_0\in A$ and an upper bound $b_0$ for $A$. Then, look at the middle point $m$ of the interval $[a_0,b_0]$. If $m$ is an upper bound for $A$, put $b_1:=m$ and $a_1:=a_0$. Otherwise, put $b_1:=b_0$ and choose a point $a_1\in A$ such that $a_1>m$. In either case, you have $a_0\leq a_1\leq b_1\leq b_0$, and $b_1-a_1\leq \frac12 \,( b_0-a_0)$. Moreover, $a_1\in A$ and $b_1$ is an upper bound for $A$.

Repeating this procedure, you can construct by induction a non-decreasing sequence $(a_n)\subset A$ and a non-increasing sequence $(b_n)$ of upper bounds for $A$, with $a_n\leq b_n$ for all $n$ and $b_{n+1}-a_{n+1}\leq \frac12 (b_n-a_n)$.

By the Archimedean property (which you do need as pointed out by Daniel), the diameter of the interval $[a_n,b_n]$ goes to $0$. It follows that both sequences $(a_n)$ and $(b_n)$ are Cauchy. So they are both convergent, to the same limit since $b_n-a_n\to 0$. If you call this limit $l$, then $l$ is an upper bound for $A$ because the set of all upper bounds for $A$ is closed and $b_n\to l$, and no upper bound for $A$ can be smaller than $l$ because $l$ is in the closure of $A$.

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  • $\begingroup$ Etienne, how you have showed ${a_n}$ and ${b_n}$ are both Cauchy? And how you concluded $l$ is least upper bound? $\endgroup$ – Biswarup Saha Aug 1 '18 at 17:14

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