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I'm aware how to prove convergence of every monotone and bounded sequence in $\mathbb{R}$ by using the completeness of $\mathbb{R}$ (using least upper bound property). But now I want to prove the equivalency of these two. So, I need to assume that every monotone and bounded sequence in $\mathbb{R}$ is convergent and then deduce the least upper bound property. what should I do?

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  • $\begingroup$ this probably isn't it - assume a MBS in $\mathbb{R}$ doesn't have a LUB, then the bound is a LUB - a contradiction. $\endgroup$ – JonMark Perry Apr 24 '15 at 10:35
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Let $A \subseteq \mathbf R$ be a nonempty set, bounded from above by $x$. We will show that $A$ has a least upper bound. Let $a_0 \in A$ and $b_0 := x$. Inductively define sequences $(a_n)$, $(b_n)$ as follows: If $\frac{a_n + b_n}2$ is an upper bound of $A$, let $a_{n+1} := a_n$, $b_{n+1} := \frac 12(a_n +b_n)$. Otherwise, choose $a_{n+1} \in A$ with $\frac 12(a_n + b_n) \le a_{n+1} \le b_n$, and let $b_{n+1} := b_n$.

Then: $(a_n)$ is increasing, $(b_n)$ is decreasing, both are bounded, $a_n \in A$ for all $n$ and $A \le b_n$ for all $n$. Moreover $(b_n - a_n) \le 2^{-n}(b_0 - a_0)$. Now, by the monotonicity, $a_n$ converges, say $a_n \to s$. As $(b_n - a_n) \le 2^{-n}(b_0 - a_0) \to 0$, $b_n \to s$ also. We will prove that $s$ is the least upper bound of $A$. On one hand, if $a \in A$, then $a \le b_n$ for all $n$, which implies $a \le \lim_n b_n = s$. Hence $s$ is an upper bound. If $t$ is any upper bound, as $a_n \in A$, we have $a_n \le t$ for all $n$, giving $s = \lim_n a_n \le t$. Hence $s$ is the least upper bound.

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