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.