Equivalence of Completeness Properties

[3.1415926535]

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 ] andI_{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.

 

[lugita15]

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.

 

[3.1415926535]

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.

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

 

[lugita15]

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 separate 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.

 

[3.1415926535]

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 separate 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?

 

[lugita15]

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→∞.