Compact set and a sequence of closed sets, the intersection of all of them is empty

Question

I'm having a hard time to prove this. The problem is:

Let $V \subset \mathbb{R}^d $, be a nonempty compact set and $(A_n)_{n\in\mathbb{N}}$ a sequence of closed nonempty sets in $\mathbb{R}^d$ with the property:

\begin{align} V\cap(\bigcap_{n=1}^{\infty}A_n)=\emptyset \end{align}

Prove there is $N\in\mathbb{N}$ that satisfies: $V\cap(\bigcap_{n=1}^{N}A_n)=\emptyset$

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I tried the following:

Let $(K)_{n\in\mathbb{N}}$ be open cover for $V\cup \mathcal{A}$, where $\mathcal{A}:=\bigcap_{n=1}^{\infty}A_n$ . So, $V\cup\mathcal{A} \subset \bigcup_{n=1}^{\infty}K_n$.

Now we have:

\begin{align}(V\cup\mathcal{A})\setminus\mathcal{A}\subset(\bigcup_{n=1}^{\infty}K_n )\setminus\mathcal{A}\end{align} \begin{align} V \subset \bigcup_{n=1}^{\infty}K_n\setminus\bigcap_{n=1}^{\infty}A_n \end{align}

I thought that would help me, because I can find a finite subset that covers V. I don't know what to do next...

Answer 1

Hint: Consider the set $\mathcal{U}:=\{U_n |\ n \in \Bbb{N}, U_n:=\Bbb{R}^d\setminus A_n\}$. Each member of $\mathcal{U}$ is open since $A_n$ is closed for each $n$.

Furthermore, since $V\cap (\bigcap_\Bbb{N} A_n) = \emptyset$, we have $$V\subset \Bbb{R}^d\setminus \bigcap_\Bbb{N} A_n= \bigcup_\Bbb{N} \Bbb{R}^d\setminus A_n = \bigcup_\Bbb{N} U_n.$$

Then $\mathcal{U}$ is an open cover of $V$ and thus there are finitely many $U_n$ that cover $V$.

Can you finish?

Answer 2

Basically, what you want to show is that any decreasing sequence of nonvoid compact sets has a nonvoid intersection.

Let $\{F_n\}$ be such a sequence. Pick $x_n\in F_n$ for all $n\in\mathbb{N}$. This is a sequence lying in a compact set. It has a subsequential limit $x$. You can now show $x\in\cap_n F_n.$

Answer 3

Hint:
Consider $\{A_n^c\}_{n=1}^{\infty }$ as an open covering of $V$.