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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!

Community
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yung_Pabs
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    In your intuitions about part (a), be careful: If $S_i = [i, \infty) \subseteq \Bbb{R}$, the $S_i$ comprise a nest of closed sets with empty intersection. The hypothesis that $X$ is compact is important. – Rob Arthan Sep 12 '16 at 20:48
  • More generally a family $F$ of sets has the Finite Intersection Property (F.I.P.) iff $F\ne \emptyset$ and $\emptyset\ne \cap G$ whenever $G$ is a finite non-empty subset of $F.$... Theorem: A space is compact iff every family $F$ of closed sets, that has the F.I,P., satisfies $\cap F \ne \emptyset.$ – DanielWainfleet Sep 13 '16 at 04:01

2 Answers2

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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 $\;S_1\supset S_2\supset\ldots\;$

DonAntonio
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  • When you use the compactness of $X$ in the second part, are we able to claim this because of the first line where you showed that the infinite union of the collection of $X\setminus S_{i}$ covers the entire set $X$? So then finitely many of the collection $X\setminus S_{i}$ must also cover $X$? – yung_Pabs Sep 12 '16 at 23:04
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    @yung_Pabs. Yes. If ${X$ \ $S_i:i\in \mathbb N}$ is an open cover of $X$ then it has a finite sub-cover $T$, and $T\subset U={X$ \ $S_i:i\leq N}$ for some $N\in \mathbb N$, so $U$ is also a cover of $X$. This implies $S_N=\cap_{i=1}^NS_i=\emptyset.$ – DanielWainfleet Sep 13 '16 at 04:08
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I propose a different solution:

Once again we let $U_i := X \setminus S_i$.

Let's assume to the contrary that $\bigcap S_i = \emptyset$. Which means there exists a set $S_j$ such that no point of $S_j$ belong to every $S_i$. This means that the set of complements $U_i$ form an open cover of $S_j$. But observe that $S_j$ is a closed subset of a compact space $X$, so $S_j$ is compact. This means that there exists a finite amount of sets $U_{i_1}, \dots, U_{i_n}$ which still covers $S_j$. But this means that the finite intersection $S_j \cap S_{i_1} \cap \dots \cap S_{i_n} = \emptyset$, in contradiction with our hypothesis because the sets $S_i$s are nested.

kjQtte
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