Demonstração de Problema: Considerações Inválidas em Intervalos Abertos

[JasonPhysicist]

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] which 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.

 

[HallsofIvy]

This is the "nested interval property", used to prove that any closed and bounded set of real numbers is compact.

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

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

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

For example, suppose (a_n, b_n)= (0, \frac{1}{n}). The set {a_n} 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 [0, \frac{1}{n}] would give the same set of "left endpoints" {0} having the same sup, 0, but now 0 is contained in each interval. [0, \frac{1}{n}] has intersection {0} while (0, \frac{1}{n}) has empty intersection.