For each $n \in \mathbb{N}$ a closed interval $[x_n, y_n]$ is given.
Assume that $[x_m, y_m]\, \cap \, [x_n, y_n] \neq \emptyset$ for all
$m,n \in \mathbb{N}$. Show that $\bigcap_{n=1}^{\infty} [x_n, y_n] \neq \emptyset$.
What I've done is, assumed for contradiction that $\bigcap_{n=1}^{\infty} [x_n, y_n] = \emptyset$ which implies the existence of some disjoint intervals, i.e: that there exists $m,n \in \mathbb{N}$ such that $[x_m, y_m] \, \cap \, [x_n, y_n] = \emptyset$ which contradicts our hypothesis.
Now, I think this is wrong because it (i) makes the problem far too easy and (ii) infinities are fishy, I'm not convinced that the infinite intersection of intervals being empty implies that there exists at least two disjoint intervals.
Question: Is my "proof" completely off the right train of thought? How do I fix it?
Best Answer
It is not obviously true that $\bigcap_{n=1}^{\infty} [x_n, y_n] = \emptyset$ implies there exist $m$ and $n$ such that $[x_m, y_m] \, \cap \, [x_n, y_n] = \emptyset$. Notice, for instance, that the intersection of the three sets $\{0,1\}$, $\{0,2\}$, and $\{1,2\}$ is empty, even though the intersection of any two of them is nonempty. Or notice that $\bigcap_{n=1}^\infty(0,1/n)$ is empty, even though these intervals not only have nonempty pairwise intersections but actually are nested.
Here's how I would suggest to approach it. First, show that the intersection of any finite collection of your intervals is nonempty (for this, you will need to use the fact that they are intervals, rather than just random sets). Second, show that this implies the infinite intersection is nonempty (for this you will need to use that the sets are not just intervals but closed and bounded intervals; for instance, you could use compactness).