Intersection of a sequence of intervals equals a point

[The Captain]

Intersection of a sequence of intervals equals a point (Analysis)

Homework Statement

Let A_{n} = [a_{n}, b_{n}] be a sequence of intervals s.t. A_{n}>A_{n+1} and |b_{n}-a_{n}|\rightarrow0. Then \cap^{∞}_{n=1}A_{n}={p} for some p\inR.

Homework Equations

Monotonic Convergent Theorem
If {a_{n}} is a sequence of real numbers, then this sequence has a finite limit if and only if the sequence is bounded.

R = Real Numbers
N = Natural Numbers

The Attempt at a Solution

Since [a_{m}, b_{m}]≤[a_{n},b_{n} for n≤m, then a_{n}≤a_{m} for n≤m and implies {a_{n}}, n\inN is monotonically nondecreasing.

Since a_{m}≤b_{m} and a_{n}≤b_{n}, then a_{i}≤b_{j} for i,j\inN. And {b_{j}} is monotonically nonincreasing and bounded below \foralla_{i},i\inN.

Then by Monotonic Convergent Theorem, lim_{n\rightarrow∞}{a_{n}} and lim_{n\rightarrow∞}{b_{n}} exist. Also, lim_{n\rightarrow∞}{a_{n}}=sup(a_{n}}) = α and lim_{n\rightarrow∞}{b_{n}}=inf(b_{n}}) = β.

Then since α=sup(a_{n}), α≤b_{n}, n\inN, β=inf(b_{n}), α≤β, this implies that [α,β]≠0. Since a_{n}≤α≤β≤b_{n}, then [α,β]\subset[a_{n}, b_{n}]. This implies \cap^{∞}_{n=1}=[α,β].

Then lim_{n\rightarrow∞}(b_{n}-a_{n})=0. This implies lim_{n\rightarrow∞}(b_{n})-lim_{n\rightarrow∞}(a_{n})=0. This implies β-α=0, implies β=α.

Then [α,β] = {α} = {β}, implies [α,β] = {p} for some p\inR.
∴ \cap^{∞}_{n=1}A_{n}={p} for some p\inR.

 

[jbunniii]

Homework Statement

Let A_{n} = [a_{n}, b_{n}] be a sequence of intervals s.t. A_{n}>A_{n+1} and |b_{n}-a_{n}|\rightarrow0. Then \cap^{∞}_{n=1}A_{n}={p} for some p\inR.

Homework Equations

Monotonic Convergent Theorem
If {a_{n}} is a sequence of real numbers, then this sequence has a finite limit if and only if the sequence is bounded.

R = Real Numbers
N = Natural Numbers

The Attempt at a Solution

Since [a_{m}, b_{m}]≤[a_{n},b_{n} for n≤m, then a_{n}≤a_{m} for n≤m and implies {a_{n}}, n\inN is monotonically nondecreasing.

Since a_{m}≤b_{m} and a_{n}≤b_{n}, then a_{i}≤b_{j} for i,j\inN. And {b_{j}} is monotonically nonincreasing and bounded below \foralla_{i},i\inN.

Then by Monotonic Convergent Theorem, lim_{n\rightarrow∞}{a_{n}} and lim_{n\rightarrow∞}{b_{n}} exist. Also, lim_{n\rightarrow∞}{a_{n}}=sup(a_{n}}) = α and lim_{n\rightarrow∞}{b_{n}}=inf(b_{n}}) = β.

OK up to here, I think.

Then since α=sup(a_{n}), α≤b_{n}, n\inN, β=inf(b_{n}), α≤β, this implies that [α,β]≠0. Since a_{n}≤α≤β≤b_{n}, then [α,β]\subset[a_{n}, b_{n}]. This implies \cap^{∞}_{n=1}=[α,β].

I'm not sure how you concluded that last bit. All this demonstrates is that [\alpha,\beta] \subset \cap_{n=1}^{\infty}[a_n,b_n].

Then lim_{n\rightarrow∞}(b_{n}-a_{n})=0. This implies lim_{n\rightarrow∞}(b_{n})-lim_{n\rightarrow∞}(a_{n})=0. This implies β-α=0, implies β=α.

Then [α,β] = {α} = {β}, implies [α,β] = {p} for some p\inR.
∴ \cap^{∞}_{n=1}A_{n}={p} for some p\inR.

I think you have all the necessary ingredients for the proof, but it is more cluttered than it needs to be. It suffices to prove the following two statements:

1. The intersection is non-empty. For example, it contains \sup a_n. You have already proved this.

2. The intersection cannot contain more than one point. For suppose it contained two distinct points x and y. Then there is some nonzero distance between them: |x - y| = d > 0. Use the fact that b_n - a_n \rightarrow 0 to show that this cannot happen.