Consider a sequence of closed (but not necessarily nested) intervals $I_1,I_2, \ldots$ in $\mathbb{R}$ with the property that $\bigcap_{n=1}^N I_n \neq \emptyset$ for all $N \in \mathbb{N}$, but $\bigcap_{n=1}^\infty I_n = \emptyset$.
I must produce an example of this, or show that this sequence cannot exist. My intuition is that this cannot exist but it is very hard for me to articulate why. I know sets like $J_n = (0,1/n)$ have empty intersections at infinity but nonempty ones for all $n \in \mathbb{N}$, but they're open. It seems like if sets are closed and have nonempty intersections in all finite cases, then they'd have a nonempty intersection in the infinite case.
Best Answer
To understand closed sets, it is necessary to understand open sets. An open subset $A$ of $\mathbb R$ is any set such that if $x\in A$, there is a positive number $\epsilon$ such that $(x-\epsilon,x+\epsilon)\subset A$. You are probably already familiar with this, but it is important to note because the definition of a closed set is a set whose complement is open. I propose that any set of the form $[a,\infty)$ is closed. To see this, we examine its complement, $(-\infty,a)$. It is not hard to see that $(-\infty,a)$ is open, so we conclude that $[a,\infty)$ is closed. Now let $I_n=[n,\infty)$. What is $$\bigcap_{n=1}^N I_n?$$ What is $$\bigcap_{n=1}^\infty I_n?$$