For all $n \in \mathbb N$, let $D_n = [a_n, b_n]$ be a closed interval in $ \mathbb R$ with $b_n - a_n > 0$. Suppose that $$D_1 \supset D_2 \supset ....$$ Moreover, suppose that $\lim_{n\rightarrow\infty}(b_n - a_n) = 0$. Prove that there exists precisely one $p \in \mathbb R$ such that $p \in \bigcap_{n=1}^\infty D_n$. (Then I will continue to check if this is also valid everywhere in $\mathbb Q$.
Intuitively, it's very simple to draw a picture and see what is going on, and I have concluded that I have to work with the supremum of {$(a_n)_{n\in\mathbb N}$} but I have no idea how to get started.
This is general preparation for a test, so hints would be preferred, thanks a lot.
(EDIT) My progress:
Define $p = sup${$(a_n)_{n \in \mathbb N}$}.
It goes without saying that $p\in D_n = [a_n, b_n]$ for all $n\in\mathbb N$ given by the properties of a supremum and how $b_n - a_n > 0$. Moreover by hypothesis; $e>0$, $ |b_n-p| \le |b_n-a_n| \le e$ (since $p \ge a_n$) hence $b_n$ converges to p. I guess I need to show that the infimum of $a_n = p$ or something and thus by properties of infimum and supremum, there lies nothing inbetween.
Left to prove: p is the only thing in $\bigcap_{n=1}^\infty D_n$