1. ## nested interval theorem

Prove that if $\displaystyle I_n = [a_{n}, b_{n}]$ is a nested collection of closed intervals such that $\displaystyle \lim_{n \to \infty} b_{n} - a_{n} = 0$, then there is an $\displaystyle x \in \mathbb{R}$ such that $\displaystyle \bigcap_{n \in \mathbb{N}} I_N = \{x \}$.

So $\displaystyle a_n$ is an increasing sequence and $\displaystyle b_n$ is a decreasing sequence. So choose an $\displaystyle N \in \mathbb{N}$ so that $\displaystyle |b_{N}- a_{N}| < \varepsilon$. Then $\displaystyle a_{N} \leq a_{n} < b_{n} \leq b_{N}$ for all $\displaystyle n \geq N$. So $\displaystyle a_{n} \to \alpha$ and $\displaystyle b_n \to \beta$ for $\displaystyle \alpha, \beta \in \mathbb{R}$. We need to show that $\displaystyle x = \alpha = \beta$.

Then $\displaystyle a_{N} \leq \sup \{a_{n}: n \geq N \}$ and $\displaystyle b_{N} \geq \inf \{b_{n}: n \geq N \}$.

Now what?

2. Originally Posted by particlejohn
Prove that if $\displaystyle I_n = [a_{n}, b_{n}]$ is a nested collection of closed intervals such that $\displaystyle \lim_{n \to \infty} b_{n} - a_{n} = 0$, then there is an $\displaystyle x \in \mathbb{R}$ such that $\displaystyle \bigcap_{n \in \mathbb{N}} I_N = \{x \}$.

So $\displaystyle a_n$ is an increasing sequence and $\displaystyle b_n$ is a decreasing sequence. So choose an $\displaystyle N \in \mathbb{N}$ so that $\displaystyle |b_{N}- a_{N}| < \varepsilon$. Then $\displaystyle a_{N} \leq a_{n} < b_{n} \leq b_{N}$ for all $\displaystyle n \geq N$. So $\displaystyle a_{n} \to \alpha$ and $\displaystyle b_n \to \beta$ for $\displaystyle \alpha, \beta \in \mathbb{R}$. We need to show that $\displaystyle x = \alpha = \beta$.

Then $\displaystyle a_{N} \leq \sup \{a_{n}: n \geq N \}$ and $\displaystyle b_{N} \geq \inf \{b_{n}: n \geq N \}$.

Now what?
consider the set $\displaystyle \{a_{n}: n \geq N \}$. $\displaystyle a_1 \leq a_n \leq b_n \leq b_1$ for all n.. this means that the set is bounded above by $\displaystyle b_1$ and hence supremum exists and $\displaystyle a_n \leq \sup \{a_{n}: n \geq N \}$ for all $\displaystyle n$.

show that $\displaystyle \sup \{a_{n}: n \geq N \} \leq b_n$ for all $\displaystyle n$.

let $\displaystyle x=\sup \{a_{n}: n \geq N \}$ and fix $\displaystyle n \in \mathbb{N}$

consider $\displaystyle A_k = \{a_k | k \in \mathbb{N}\}$

If $\displaystyle n \leq k$, we have $\displaystyle a_k \leq b_k \leq b_n$ for all $\displaystyle k \geq n$

If $\displaystyle n > k$, we get $\displaystyle a_k \leq a_n \leq b_n$ for all $\displaystyle k < n$

This means that $\displaystyle a_k \leq b_n$ for all $\displaystyle k$ and for all $\displaystyle n$..

Thus, $\displaystyle b_n$ is an upper bound for the set $\displaystyle \{a_{n}: n \geq N \}$

Therefore $\displaystyle \sup \{a_{n}: n \geq N \} = x \leq b_n$ for all $\displaystyle n$ and hence $\displaystyle a_n \leq x \leq b_n$ for all $\displaystyle n$..

3. Originally Posted by kalagota
consider the set $\displaystyle \{a_{n}: n \geq N \}$. $\displaystyle a_1 \leq a_n \leq b_n \leq b_1$ for all n.. this means that the set is bounded above by $\displaystyle b_1$ and hence supremum exists and $\displaystyle a_n \leq \sup \{a_{n}: n \geq N \}$ for all $\displaystyle n$.

show that $\displaystyle \sup \{a_{n}: n \geq N \} \leq b_n$ for all $\displaystyle n$.

let $\displaystyle x=\sup \{a_{n}: n \geq N \}$ and fix $\displaystyle n \in \mathbb{N}$

consider $\displaystyle A_k = \{a_k | k \in \mathbb{N}\}$

If $\displaystyle n \leq k$, we have $\displaystyle a_k \leq b_k \leq b_n$ for all $\displaystyle k \geq n$

If $\displaystyle n > k$, we get $\displaystyle a_k \leq a_n \leq b_n$ for all $\displaystyle k < n$

This means that $\displaystyle a_k \leq b_n$ for all $\displaystyle k$ and for all $\displaystyle n$..

Thus, $\displaystyle b_n$ is an upper bound for the set $\displaystyle \{a_{n}: n \geq N \}$

Therefore $\displaystyle \sup \{a_{n}: n \geq N \} = x \leq b_n$ for all $\displaystyle n$ and hence $\displaystyle a_n \leq x \leq b_n$ for all $\displaystyle n$..
So $\displaystyle A_1 = \{a_1 \}, \ A_2 = \{a_2 \} \ \ldots$? Each one is a singleton set? Does this only show the existence of $\displaystyle x$ in the interval (e.g. it does not show that the intersection of the intervals is $\displaystyle x$)?

4. I find this very strange! You do not connect this with previous problems that we have helped you with.
Let $\displaystyle A = \left\{ {a_n :n \in Z^ + } \right\}\,\& \,B = \left\{ {b_n :n \in Z^ + } \right\}$.
Now any element of $\displaystyle B$ is an upper bound for $\displaystyle A$.
Now apply one of the first problems we did.
Do you see that $\displaystyle \sup \left\{ A \right\} = \inf \left\{ B \right\} = x$ is in the intersection?

5. yes I see that (Rudin does this to prove greatest lower bound property). I was just thinking that we would have to prove uniqueness also.

But now I see that if $\displaystyle x,y \in \bigcap_{n \in \mathbb{N}} I_N$, then $\displaystyle x =y$.

Originally Posted by Plato
I find this very strange! You do not connect this with previous problems that we have helped you with.
Let $\displaystyle A = \left\{ {a_n :n \in Z^ + } \right\}\,\& \,B = \left\{ {b_n :n \in Z^ + } \right\}$.
Now any element of $\displaystyle B$ is an upper bound for $\displaystyle A$.
Now apply one of the first problems we did.
Do you see that $\displaystyle \sup \left\{ A \right\} = \inf \left\{ B \right\} = x$ is in the intersection?

6. the proof i have given shows that there will always an element in $\displaystyle [a_n,b_n]$ for all $\displaystyle n$. and what happens when $\displaystyle \lim b_b - a_n = 0$? it must be the element.