## 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 $$I_{n}=\left [ a_{n},b_{n} \right ]$$ and$$I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq...$$ then $$\bigcap_{n=1}^{\infty}I_{n}\neq \varnothing$$ In addition if $$b_{n}-a_{n}\rightarrow 0$$ as $$n \to \infty$$ then $$\bigcap_{n=1}^{\infty}I_{n}$$ consists of a single point.

Monotone Convergence Theorem(MCN): If $$a_{n}$$ is a monotone and bounded sequence of real numbers then $$a_{n}$$ converges.
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 Blog Entries: 1 Here's an approach you could try. Let an be a bounded increasing sequence, which means that the sequence has an upper bound b. Then ([an,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.

 Quote by lugita15 Here's an approach you could try. Let an be a bounded increasing sequence, which means that the sequence has an upper bound b. Then ([an,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.
If by property 1 you mean the least upper bound property the point here is not to use it!
I want a proof 2-3 without using 1,3,4,5

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## Equivalence of Completeness Properties

 Quote by 3.1415926535 If by property 1 you mean the least upper bound property the point here is not to use it! I want a proof 2-3 without using 1,3,4,5
Yes, sorry. I think you may still be able use my suggestion to prove 2 implies 3 without using 1,4, or 5.

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.

 Quote by lugita15 Yes, sorry. I think you may still be able use my suggestion to prove 2 implies 3 without using 1,4, or 5. 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.
Even though I would like a more direct approach 2-5 will suffice.
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?

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 Quote by 3.1415926535 Even though I would like a more direct approach 2-5 will suffice. 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?
It's really quite simple. For convenience, I'll refer to half the length of an interval as it's "radius". Since (x_n) is Cauchy, there exists an x_n1 such that all subsequent elements of the sequence are within an interval I1 of radius r1=1/2 centered at x_n1. And there exists an n2>n1 such that all subsequent elements of the sequence are within an interval I2 centered at x_n2, which is within I1 and has radius r2<1/4. And there exists an n3>n2 such that all subsequent elements are within an interval I3 centered at x_n3, which is within I2 and has radius r3<1/8. I think you get the picture: we have a nested sequence (In) of intervals, with radii rn→0 as n→∞.