Cantor's Intersection Theorem

Question

If the subsets of the compact space are already non-empty, isn't it obvious that the even the smallest subset is non-empty, and so the intersection is also non-empty because it would be the smallest set? What point am I missing here? I understand the proof on wikipedia but I'm missing something crucial here.

Answer 1

Why is there a smallest set?

Consider the family of intervals $[-1,\frac1n]$ for $n\in\Bbb N\setminus\{0\}$. For every interval in our family, there is a strictly smaller interval in our family of intervals. You might want to argue that $[-1,0]$, their intersection, is in the family but it is not.

Answer 2

Another example might be $A_n=\left[\sqrt{2}-\frac{1}{n}, \sqrt{2}+\frac{1}{n}\right]\cap\mathbb{Q}$. The intersection is empty, because there is no fixed rational number arbitrarily close to $\sqrt{2}$, but each $A_n$ is clearly non-empty. $$\bigcap_{n\geq1}A_n=\varnothing$$ How does this fail the hypothesis of Cantor's intersection theorem?