Register to reply

Problem demonstration

by JasonPhysicist
Tags: demonstration
Share this thread:
JasonPhysicist
#1
Jan22-07, 09:09 PM
P: 12
Here goes a theorem and its demonstration(attachment) .Sorry,I couldn't find it in english,so it's in portuguese).

We have that the intervals In=[An,Bn] wich are closed and limited.


What I want to know is: what consideration(s) is/are NOT valid on the demonstration,when we consider an open intervals?

Thank you in advance.
Attached Files
File Type: doc 1.doc (42.0 KB, 3 views)
Phys.Org News Partner Science news on Phys.org
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
HallsofIvy
#2
Jan23-07, 07:26 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,345
This is the "nested interval property", used to prove that any closed and bounded set of real numbers is compact.

"Let [itex][a_n,b_n][/itex] be a nested set of intervals (nested: each interval is inside the previous interval [itex]a_n\le a_{n+1}\le b_{n+1}\le b_n[/itex]). Then there exist a real number [itex]\zeta[/itex] contained in the intersection of all the intervals. Further, if [itex]lim_{n\rightarrow \infty}b_n-a_n= 0[/itex], that intersection consists of the single number [itex]\zeta[/itex].

It can be shown that, for all n, [itex]a_n\le b_1[/itex] so that [itex]b_1[/itex] is an upper bound on the set [itex]{a_n}[/itex] and so, by the least upper bound property that set has a least upper bound (sup). Let [itex]\zeta[/itex] be [itex]sup{a_n}[/itex]. Then it can be shown that [itex]\zeta[/itex] is a lower bound on the set [itex]{b_n}[/itex] and, so lies in all intervals [itex][a_n, b_n][/itex]

If the intervals are not closed, then it might happen that that [itex]\zeta[/itex] is NOT in some or all of the intervals.

For example, suppose [itex](a_n, b_n)= (0, \frac{1}{n})[/itex]. The set [itex]{a_n}[/itex] is the set {0} which has 0 as its "least upper bound. But 0 is not in any of those intervals. The same example with closed sets [itex][0, \frac{1}{n}][/itex] would give the same set of "left endpoints" {0} having the same sup, 0, but now 0 is contained in each interval. [itex][0, \frac{1}{n}][/itex] has intersection {0} while [itex](0, \frac{1}{n})[/itex] has empty intersection.


Register to reply

Related Discussions
I need to create a demonstration.. General Physics 2
Demonstration Introductory Physics Homework 1
Help with physics demonstration Calculus & Beyond Homework 4