Before we look at the Nested Intervals Theorem let's first look at the following important lemma that will be used to prove the Nested Intervals Theorem. Maybe it is true in general that an in nite intersection of nested closed intervals is non-empty. Consider the nested sequence of open intervals ⋯ …

Then the intersection of all of these intervalsn+1n+1 nn© is either a closed interval or a single point. The situation is different for closed intervals. The analogous theorem for open intervals is false. We have just looked at what exactly a Nested Interval is, and we are about to look at a critically important theorem in Real Analysis. If, in addition, the length of I n approaches zero, then ⋂ n = 1 ∞ I n consists of a single point. The proof by contradiction allows us to assume that the infinite nested intersection ⋂ n = 1 ∞ I n is empty. We may apply the nested intervals theorem to conclude that the intersection of this family of nested intervals, [an,bn], consists of a single pointc, and moreover limn!1an=c= limn!1bnso that by … The nested interval theorem for the real numbers says that if you have a sequence of nested, bounded, nonempty, closed intervals, then their intersection is nonempty. That attempt to prove "by contradiction" something like the Nested Interval Theorem for open intervals (rather than closed ones) is invalidated by introducing an extra assumption other than the one for which a contradiction is sought. 1 Nested Intervals: a sequence of intervals I 1;I 2; is called nested if The fact that an in nite intersection of nested open intervals can be empty while (this partic-ular) in nite intersection of nested closed intervals is not empty is suggestive. We start with a bounded sequence (x n) {\displaystyle (x_{n})} : Because (x n) n ∈ N {\displaystyle (x_{n})_{n\in \mathbb {N} }} is bounded, this sequence has a lower bound s {\displaystyle s} and an upper bound S {\displaystyle S}. Nested Intervals Theorem Math 4200 Nested Intervals Theorem: Suppose { [a, b ] } is a sequence of closed intervals suchnn that for each n, [a, b ] [a, b ]. Choose one of these intervals, call it [an+1,bn+1], such thatf(an+1)f(bn+1≤0. There is also an alternative proof of the Bolzano–Weierstrass theorem using nested intervals. My apologies if this has been asked before but I couldn't find it. The Nested Intervals Theorem. The nested intervals theorem states that if each I n is a closed and bounded interval, say I n = [ a n, b n ] Here is the theorem: If I 1 ⊃ I 2 ⊃ I 3 ⊃ … is a sequence of nested, closed, bounded, nonempty intervals, then ⋂ n = 1 ∞ I n is nonempty.