# How to show that $[\xi, \eta ]$ is equal to the intersection of the following closed, nested, and bounded intervals?

Let $\eta = \mathrm {inf} \{b_n : n \in \mathbb N \}$ and $\xi$ be the unique number contained in the closed, nested, and bounded intervals $I_n = [a_n,b_n]$ for all $n \in \mathbb N$ if $\mathrm {inf} \{b_n - a_n: n \in \mathbb N \} =0$. How to show that $$[\xi, \eta ] = \bigcap_{n=1}^\infty I_n$$.

I attempted to show this by the definition of set equality,i.e, to show that $$[\xi, \eta ] \subseteq \bigcap_{n=1}^\infty I_n$$ and $$\bigcap_{n=1}^\infty I_n \subseteq [\xi, \eta ]$$ but I do not know how to proceed from here or if this approach is correct.

Approach is right.

Hint:

$$x \in [\xi, \eta ] \implies a_n \le \xi \le x \le \eta \le b_n \forall n \in \Bbb N$$

and

$$x \in \bigcap_{n=1}^\infty I_n \implies a_n \le x \le b_n \forall n \in \Bbb N \implies \text{ x is a lower bound for {b_n}}$$

Ususally $\xi$ is defined to be $\sup \{a_n\}$ but I think your definition should also work since you claim that it is a unique number contained in all the nested intervals $I_n$. The $x$ in the latter case is in all the intervals and hence must be equal to a certain singular value. What???

• Yes, my definition of $\xi$ implies that it is $\mathrm {sup} \{a_n\}$. – Lucas Alanis Feb 18 '14 at 4:28
• @Lance Ferd: The fact that $\xi = \sup \{a_n\}$ is not very clear to me. But if you're sure about it then a proof should now be obvious to you. – Ishfaaq Feb 18 '14 at 4:31