2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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???

$\endgroup$
  • $\begingroup$ Yes, my definition of $\xi$ implies that it is $\mathrm {sup} \{a_n\}$. $\endgroup$ – Lucas Alanis Feb 18 '14 at 4:28
  • $\begingroup$ @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. $\endgroup$ – Ishfaaq Feb 18 '14 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.