# Intersection nested, closed sequence of intervals

1. Nov 21, 2013

### mahler1

The problem statement, all variables and given/known data.
Let $\{I_n\}_{n \in \mathbb N}$ be a sequence of closed nested intervals and for each $n \in \mathbb N$ let $\alpha_n$ be the lenght of $I_n$.
Prove that $lim_{n \to \infty}\alpha_n$ exists and prove that if $L=lim_{n \to \infty}\alpha_n>0$, then $\bigcap_{n \in \mathbb N} I_n$ is a closed interval of lenght $L$.

The attempt at a solution.

I didn't have problems to prove the existence of the limit: if $I_n=[a_n,b_n]$, then $\alpha_n=b_n-a_n$ and $\{a_n\}_n, \{b_n\}_n$ are increasing and decreasing bounded sequences respectively, so both are convergent $\implies \{\alpha_n\}_n$ is convergent and $lim_{n \to \infty} \alpha_n=lim_{n \to \infty}b_n-a_n=lim_{n \to \infty}b_n-lim_{n \to \infty}a_n=b-a$.

Now, I would like to say that if $b-a>0 \implies \bigcap_{n \in \mathbb N} I_n=[a,b]$, but I don't know hot to prove this part.

2. Nov 21, 2013

### LCKurtz

Try showing if x isn't in [a,b] then x isn't in the intersection and if x is in [a,b] then x is in the intersection.

Last edited: Nov 21, 2013
3. Nov 21, 2013

### Dick

I guess I'm not seeing anything difficult about this. Can you prove that if x<a then x is not in $\bigcap_{n \in \mathbb N} I_n$? Same if x>b. And if a<=x<=b then it is? It's just reasoning about what 'intersection' means.

4. Nov 22, 2013

### mahler1

Yes, it was just thinking the definition of intersection and apply it, thanks!