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Equivalence of Completeness Properties 
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#1
Feb1512, 11:41 AM

P: 80

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. 


#2
Feb1512, 01:49 PM

P: 1,583

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.



#3
Feb1612, 10:12 AM

P: 80

I want a proof 23 without using 1,3,4,5 


#4
Feb1612, 10:19 AM

P: 1,583

Equivalence of Completeness Properties
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. 


#5
Feb1612, 01:34 PM

P: 80

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? 


#6
Feb1612, 05:41 PM

P: 1,583




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