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$_{n}$|$\rightarrow$0. Then $\cap^{∞}_{n=1}$A$_{n}$={p} for some p$\in$R.

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$\in$N is monotonically nondecreasing.

Since a$_{m}$≤b$_{m}$ and a$_{n}$≤b$_{n}$, then a$_{i}$≤b$_{j}$ for i,j$\in$N. And {b$_{j}$} is monotonically nonincreasing and bounded below $\forall$a$_{i}$,i$\in$N.

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$\in$N, 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$\in$R.
∴ $\cap^{∞}_{n=1}$A$_{n}$={p} for some p$\in$R.

 

[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$_{n}$|$\rightarrow$0. Then $\cap^{∞}_{n=1}$A$_{n}$={p} for some p$\in$R.

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$\in$N is monotonically nondecreasing.

Since a$_{m}$≤b$_{m}$ and a$_{n}$≤b$_{n}$, then a$_{i}$≤b$_{j}$ for i,j$\in$N. And {b$_{j}$} is monotonically nonincreasing and bounded below $\forall$a$_{i}$,i$\in$N.

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$\in$N, 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$\in$R.
∴ $\cap^{∞}_{n=1}$A$_{n}$={p} for some p$\in$R.

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.