nested interval theorem

**particlejohn** · July 5th 2008, 09:12 PM

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

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

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

Now what?

**kalagota** · July 6th 2008, 05:46 AM

Originally Posted by particlejohn

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

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

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

Now what?

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

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

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

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

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

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

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

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

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

**particlejohn** · July 6th 2008, 04:25 PM

Originally Posted by kalagota

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

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

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

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

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

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

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

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

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

So $A_1 = \{a_1 \}, \ A_2 = \{a_2 \} \ \ldots$ ? Each one is a singleton set? Does this only show the existence of $x$ in the interval (e.g. it does not show that the intersection of the intervals is $x$ )?

**Plato** · July 6th 2008, 05:09 PM

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

**particlejohn** · July 6th 2008, 05:14 PM

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 $x,y \in \bigcap_{n \in \mathbb{N}} I_N$ , then $x =y$ .

Originally Posted by Plato

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

**kalagota** · July 6th 2008, 06:18 PM

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

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