Register to reply

Equivalence of Completeness Properties

Share this thread:
3.1415926535
#1
Feb15-12, 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.
Phys.Org News Partner Science news on Phys.org
Flapping baby birds give clues to origin of flight
Prions can trigger 'stuck' wine fermentations, researchers find
Socially-assistive robots help kids with autism learn by providing personalized prompts
lugita15
#2
Feb15-12, 01:49 PM
P: 1,583
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.
3.1415926535
#3
Feb16-12, 10:12 AM
P: 80
Quote Quote by lugita15 View Post
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

lugita15
#4
Feb16-12, 10:19 AM
P: 1,583
Equivalence of Completeness Properties

Quote Quote by 3.1415926535 View Post
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.
3.1415926535
#5
Feb16-12, 01:34 PM
P: 80
Quote Quote by lugita15 View Post
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?
lugita15
#6
Feb16-12, 05:41 PM
P: 1,583
Quote Quote by 3.1415926535 View Post
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→∞.


Register to reply

Related Discussions
Equivalence relation and equivalence class Precalculus Mathematics Homework 2
Prove Relationship between Equivalence Relations and Equivalence Classes Calculus & Beyond Homework 1
Equivalence of 8 properties in Real Analysis Calculus & Beyond Homework 9
Multiplication Properties of Equivalence Classes Calculus & Beyond Homework 2
Equivalence relations and equivalence classes Differential Geometry 4