Problem Statement: Let $X$ be a compact topological space.
(a) Let $S_{i}$,$i\in \mathbb{N}$ be a closed, nonempty subset of $X$, such that $S_{1}\supset S_{2}\supset \cdots$. Show that $$\overset{\infty}{\underset{i=1}{\bigcap}}S_{i}\neq \emptyset.$$ [Hint: Consider the collection $\mathcal{U}$ of open sets $X\setminus S_{i}$, $i\in \mathbb{N}$.]
For part (a) of this problem, I think I understand why this is true intuitively, because since each $S_{i}$ is closed, and non-empty, then as $i\rightarrow \infty$, $S_{i}$ is at least one point in $X$, and because of the containment of each successive $S_{i}$, the infinite intersection would be at the very least a point.
But I am not quite understanding how I should use compactness to show that the intersection is non-empty. From the hint, $X\setminus S_{i}=X\cap S_{i}^{^c}=S_{i}^{^c}$, where $S_{i}^{^c}$ is the complement of $S_{i}$ in $X$, which is open in $X$, with $S_{1}^{^c}\subset S_{2}^{^c}\subset\cdots$
I know how to show that $S_{i}$ has a finite subcovering (i.e. compact), but can I use this to show that $S_{i}\neq \emptyset$ as $i\rightarrow \infty$?
Part (b) of this problem uses part (a) to prove the Finite Intersection Property, but I am pretty sure I know how to approach this once I figure out part (a).
I am still in the early stages of grasping analysis, so any suggestions are appreciated!
Best Answer
An idea:
Supose the claim is false, and as proposed take $\;U_i:= X\setminus S_i\;$ . This is an open set in $\;X\;$ , and (observe we use de Morgan's Laws)
$$\bigcup_{i=1}^\infty U_i=X\setminus\bigcap_{i=1}^\infty S_i=X\setminus\emptyset=X$$
Since $\;X\;$ is compact, there exists $\;N\in\Bbb N\;$ such that
$$X=\bigcup_{i=1}^N U_i=X\setminus\bigcap_{i=1}^NS_i\implies \bigcap_{i=1}^NS_i=\emptyset$$
which of course is absurd as $\;S_1\supset S_2\supset\ldots\;$