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There are many equivalent versions of completeness in the real number system:
i) LUB/supremum property
ii) Monotone Convergence property
iii) Nested Interval property
iv) Bolzano Weierstrass property
v) Cauchy Criterion property

I've been able to prove: (i)$\implies$(ii)$\implies$(iii)$\implies$(iv)$\implies$(v)
I need help with
a) (v)$\implies$(i)
b) (iii)$\implies$(i)

P.S. In proving (v)$\implies$(i), we use the construction of 2 sequences by using mid-points. I am having problem with showing that they are Cauchy sequences

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Can you explain the cauchy criterion? – Henrique Tyrrell Oct 25 '11 at 20:02
Are you looking to explicitly prove (iii) $\implies$ (i) (perhaps as an exercise), without going through (iv) and (v)? – Srivatsan Oct 25 '11 at 20:09
It is economical to do a "round trip" of implications but it is very instructive to consider direct implications. Some work out nicely in a natural way; others seem to need to go through an auxiliary property. (I'm not sure this is the case here, but I've seen this happen often.) – lhf Oct 25 '11 at 20:18
Why does (iii) imply (i)? How do you know the naturals are not bounded? – scineram Oct 25 '11 at 22:05
As scineram has suggested, some of these equivalences may need the Archimedean property, which is actually a consequence of (i). – lhf Oct 27 '11 at 10:08

Given a non-empty set $T\subset \mathbb{R}$ with an upper bound, $y_0$. Let $x_0$ be any elment of $T$.

Given $x_n$ and $y_n$, define $z_n=\frac{x_n+y_n}{2}$.

If $z_n$ is an upper bound for $T$, then let $x_{n+1}=x_n$ and $y_{n+1}=z_n$.

If $z_n$ is not an upper bound for $T$, let $x_{n+1}>z_n$ be a member of $T$ greater than $z_n$, and let $y_{n+1}=y_n$.

Lemma: For every $n$, $|x_n-y_n|\leq \frac{|x_0-y_0|}{2^n}$

Lemma: If $m>n$, then $|y_m-y_n|\leq \frac{|x_0-y_0|}{2^n}$

So $\{y_n\}$ is Cauchy. Now you just have to prove the limit of $\{y_n\}$ is the LUB of $T$. (Hint: By the same reasoning, $\{x_n\}$ is Cauchy, and the two limits must be equal.)

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You don't need to prove that (iii) $\implies$ (i) if you prove that (v) $\implies$ (i). But here is a sketch.

Let $S \subset \mathbb R$ be non-empty and bounded above. Let $a \in S$ and $b$ be an upper bound for $S$. Start with the interval $I_0=[a,b]$. Consider the midpoint $m=(a+b)/2$. If $m \in S$, let $I_1=[m,b]$. If $m$ is an upper bound for $S$, let $I_1=[a,m]$. Otherwise, there is $a'\in S$ such that $m<a'$. Let $I_1=[a',b]$. Repeat. You get a nested sequence of intervals that converge to $\sup S$.

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Actually, I don't think you need the first case ($m\in S$), but it helps intuition a bit. – lhf Oct 25 '11 at 20:15
Any ideas on how to prove (ii)$\Rightarrow$(i)?… – user87274 Sep 30 '13 at 4:33

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