Why is the condition that the intervals be closed necessary? Could someone give me an example of a sequence of nonempty, bounded, nested intervals whose intersection is empty? I can't think of one, so why does the theorem require it?
Here is the theorem:
If $I_1 \supset I_2 \supset I_3 \supset \dots$ is a sequence of nested, closed, bounded, nonempty intervals, then $\bigcap_{n=1}^{\infty}I_n$ is nonempty. If, in addition, the length of $I_n$ approaches zero, then $\bigcap_{n=1}^{\infty}I_n$ consists of a single point.
My apologies if this has been asked before but I couldn't find it.
Best Answer
Consider the family $A_n = (0, \frac{1}{n})$. We have $A_{n+1} \subset A_n$ for every $n \in \mathbb{N}$, and the length of $A_n$ approaches zero as $n$ approaches infinity, but $\bigcap_{n=1}^{\infty} A_n$ is empty. To see this, note that any element of the intersection would be greater than zero, yet less than $\frac{1}{n}$ for every $n \in \mathbb{N}$; by the Archimedian property of the real numbers, there is no such element.