Show that the infinite intersection of nested non-empty closed subsets of a compact space is not empty

Question

I'm given the following problem:

Suppose that for every $n\in \mathbb{N}$ $V_n$ is a non-empty, closed subset of a compact space $X$, with $V_n \supseteq V_{n+1}$.

Now I have to show that $V_{\infty}= \bigcap_{n=1}^{\infty} V_n \neq \emptyset$.

How can I do that? I know the nested interval property from real analysis...

The 'answer' should be that the family $\{V_n: n\in \mathbb{N} \}$ has the finite intersection property - the intersection of any finite subfamily $\{V_{n_1}, V_{n_2}, ..., V_{n_r} \}$ is $V_N$, where $N$ is $\max \{n_1,n_2, ..., n_r \}$ and $V_N \neq \emptyset$. Hence since $X$ is compact another exercise $(*)$ says that $\bigcap_{n=1}^{\infty} V_n$ is non-empty.

Exercise $(*)$ was about to prove that for a space $X$ to be compact, it is necessary and sufficient condition that if $\{V_i: i\in I \}$ is any indexed family of closed subsets of $X$ such that $\bigcap_{j\in J}V_j$ is non-empty for any finite subset $J \subseteq I$, then $\bigcap_{i\in I}V_i$ is non-empty.

So I don't understand the proof now.... Can somebody clarify this stuff? :-)

Thanks for your trouble !

Definition of open cover:

Let $A$ be a subset in $X$.

A family $\mathcal{U}=\{U_i: i\in I\}$ of subsets of $X$ is called a cover for $A$ is $A\subseteq \bigcup U_i$.

If each $\{U_k\}$ is open in $X$, then $\mathcal{U}$ is an open cover for $A$

Answer 1

Claim: A topological space $\,X\,$ is compact iff it has the Finite Intersection Property (=FIP):

Proof: (1) Suppose $\,X\,$ is compact and let $\,\{V_i\}\,$ be a family of closed subsets s.t. $\,\displaystyle{\bigcap_{i}V_i=\emptyset}\,$. Putting now $\,A_i:=X-V_i\,$ , we get that $\,\{A_i\}\,$ is a family of open subsets , and

$$\bigcap_{i}V_i=\emptyset\Longrightarrow \;X=X-\emptyset=X-\left(\bigcap_iV_i\right)=\bigcup_i\left(X-V_i\right)=\bigcup_iA_i\Longrightarrow\;\{A_i\}$$

is an open cover of $\,X\,$ and thus there exists a finite subcover of it:

$$X=\bigcup_{i\in I\,,\,|I|<\aleph_0}A_i=\bigcup_{i\in I\,,\,|I|<\aleph_0}(X-V_i)=X-\left(\bigcap_{i\in I\,,\,|I|<\aleph_0}V_i\right)\Longrightarrow \bigcap_{i\in I\,,\,|I|<\aleph_0}V_i=\emptyset\Longrightarrow$$

The family $\,\{V_i\}\,$ has the FIP.

(2) Suppose now that every family of closed subsets of $\,X\,$ hast the FIP, and let $\,\{A_i\}\,$ be an open cover of it. Put $\,U_i:=X-A_i\,$ , so $\,U_i\, $ is closed for every $\,i\,$:

$$\bigcap_iU_i=\bigcap_i(X-A_i)=X-\bigcup_i A_i=X-X=\emptyset$$

By assumption, there exists a finite set $\,J\,$ s.t. $\,\displaystyle{\bigcap_{i\in J}U_i=\emptyset}\,$ , but then

$$X=X-\emptyset=X-\bigcap_{i\in J}U_i=\bigcup_{i\in J}X-U_i)=\bigcup_{i\in J}A_i\Longrightarrow$$

$\,\{A_i\}_{i\in J}$ is a finite subcover for $\,X\,$ and thus it is compact....QED.

Please be sure you can follow the above and justify all steps. Check where we used Morgan Rules, for example, and note that we used the contrapositive of the FIP's definition...

Answer 2

If you are working in a metric space then this can be proved using the sequential definition of compactness.

For each $V_n$ choose $x_n \in V_n$. Then all the $x_n's$ live in $V_1$, since this is compact there is a subsequence $x_{n_k}$ converging to some $x \in V_1$. Now we claim that in fact $x \in V_i$ for all $i$. Suppose not; so that there exists $ i> 0$ such that $x \notin V_i$.

Then $V_i$ being closed implies that we can find an open set $U$ about $x$ such that $x \in U$ and $U \cap V_i = \emptyset$. Then $U$ can only meet finitely many terms of the sequence $x_{n_k}$, contradicting $x_{n_k} \to x$.

Answer 3

Since $V_{n+1}\subset V_n$ and $V_n\neq\varnothing$ for each $n\in\mathbb{N}$, we have that the family of closed subsets of $X$, $\{V_n\}_{n\in\mathbb{N}}$, has the finite intersection property. Since $X$ is compact and $\{V_n\}_{n\in\mathbb{N}}$ has the finite intersection property, $\bigcap\limits_{n\in\mathbb{N}} V_n\neq\varnothing$.