Let $\{ C_\lambda \}_{\lambda \in \Lambda}$ be a set of closed and bounded subsets of $\mathbb{R}$. Show that if $\bigcap_{\lambda \in \Lambda}C_\lambda = \varnothing$, then there is a finite subset $F$ of $\Lambda$ such that $\bigcap_{\lambda \in F}C_\lambda = \varnothing$.
Remarks
- For the Heine-Pincherle-Borel's Theorem, every closed and bounded subset of $\mathbb{R}$ is compact.
- A consequence of the above proposition is the
Cantor's Theorem
If $C_0\supseteq C_1\supseteq…\supseteq C_{n+1}\supseteq…$ is a decreasing sequence of closed, bounded and not empty subsets of $\mathbb{R}$, then the intersection $\bigcap_{n \in \mathbb{N}}C_n \neq \varnothing$.
Proof (Reductio ad absurdum)
If $\bigcap_{n \in \mathbb{N}}C_n = \varnothing$, for the proposition above, $\bigcap_{n \in F}C_n =\varnothing$ for some $F\subseteq \mathbb{N}$, $F$ finite; but, called $m=max$ $F$, it results $C_m =\bigcap_{n \in F}C_n =\varnothing$: absurd.
Thanks in advance for any hints or answers.
Best Answer
Proof
$\mathbb{R}=\mathbb{R}-\varnothing=\mathbb{R}-\bigcap_{\lambda\in\Lambda}C_\lambda=\bigcup_{\lambda\in\Lambda}(\mathbb{R}-C_\lambda)$.Therefore, for $\mu\in\Lambda, C_\mu\subseteq\bigcup_{\lambda\in\Lambda}(\mathbb{R}-C_\lambda)$; since $C_\mu$ is compact, there is $F_\mu\subseteq\Lambda, F_\mu$ finite, such that $C_\mu\subseteq\bigcup_{\lambda\in F_\mu}(\mathbb{R}-C_\lambda)=\mathbb{R}-\bigcap_{\lambda\in F_\mu}C_\lambda$, therefore $C_\mu\cap(\bigcap_{\lambda\in F_\mu}C_\lambda)=\varnothing$. Call $F=F_\mu\cup\{\mu\}$, so $\bigcap_{\lambda\in F}C_\lambda=\varnothing$.
Thank you @Jochen