nested intervals theorem?

Question

Why open intervals will make the conclusion false?
Anyone can explain to me in details?
If possible, show me an example...
Thanks:)

Answer 1

Let's see an example. For each n =2, 3,4... let I_n = (0, 1/n). Then, {I_n} is a nested sequence of open and bounded intervals.

If x is in I_n for some n, then 0 < x < 1/n. But if we take m > 1/x (the Archimedean property of the real field), then 0 < 1/m < x, which means x does not belong to I_m if m > 1/x. Therefore, no real x belongs to all of the intervals I_n, which implies their intersection is empty.

This shows a collection of nested open intervals doesn't need to satisfy the nested interval theorem which holds for nested collections of closed and bounded (compact)intervals.

Oserve that if, in our example, we had [0, 1/n] instead of (0, 1/n), then the number 1 would belong to all intervals I_n, and their intersection would be {1}. But with open intervals, the intersection is empty.

Of course there are infinite collections of nested intervals that have non empty intesection, like {(-1/n, 1/n), n= 2,3,4...} (it's intersection is {0}). But this is not always the case.

Answer 2

Nested Interval Theorem

Answer 3

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RE:
nested intervals theorem?
Why open intervals will make the conclusion false?
Anyone can explain to me in details?
If possible, show me an example...
Thanks:)

Answer 4

Take S_n = (0, 1/n), n=1, 2, ...

We have S_(n+1) contained in S_n for all n, yet
S = intersection(S_n, n=1..infinity) is empty.

For any positive number A, by the Archimedian property we can find a positive integer N such that A>1/N. Thus A is not in S_N (or S_n for n>N for that matter), and cannot be in the intersection above. Thus the intersection is empty, as non-positive numbers aren't in any of the sets above.

A reason it doesn't work is that (a,a) is empty, but [a,a] is not.