
#1
Jan2207, 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. 



#2
Jan2307, 07:26 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,894

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_na_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. 


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