Equivalence of Completeness Properties
The completeness properties are 1)The least upper bound property, 2)The Nested Intervals Theorem, 3)The Monotone Convergence Theorem, 4)The Bolzano Weierstrass, 5) The convergence of every Cauchy sequence.
I can show 1→2 and 1→3→4→5→1 All I need to prove is 2→3 I therefore need the proof of the Monotone Convergence Theorem using Nested intervals Theorem The theorems: Nested Interval Theorem(NIT): If [tex]I_{n}=\left [ a_{n},b_{n} \right ][/tex] and[tex]I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq...[/tex] then [tex]\bigcap_{n=1}^{\infty}I_{n}\neq \varnothing[/tex] In addition if [tex]b_{n}a_{n}\rightarrow 0[/tex] as [tex]n \to \infty[/tex] then [tex]\bigcap_{n=1}^{\infty}I_{n}[/tex] consists of a single point. Monotone Convergence Theorem(MCN): If [tex]a_{n}[/tex] is a monotone and bounded sequence of real numbers then [tex]a_{n}[/tex] converges. 
Re: Equivalence of Completeness Properties
Here's an approach you could try. Let a_{n} be a bounded increasing sequence, which means that the sequence has an upper bound b. Then ([a_{n},b]) is a nested sequence of ntervals. Can you take it from here, using properties 1 and 2 to prove 3? And then you can do the analogous thing for bounded decreasing sequences.

Re: Equivalence of Completeness Properties
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I want a proof 23 without using 1,3,4,5 
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On a seperate note, you can try proving 2 implies 5 instead (because you've already proven that 1,3,4, and 5 are equivalent, so the fact that 1 implies 2 and 2 implies 5 means that 2 is equivalent to the rest). One simple strategy is to try constructing a nested sequence of intervals whose lengths go to zero using the elements of a Cauchy sequence. 
Re: Equivalence of Completeness Properties
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Suppose that I want to prove that a Cauchy sequence x_n converges How can I create a sequence of nested intervals whose lengths go to 0 when x_n is not necessarily monotonous? 
Re: Equivalence of Completeness Properties
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