I was trying to do this problem this way:
Let $\mathcal{B}=\{B_i\}$ be a collection of compact sets. By Heine-Borel, each of the $B_i$'s are closed and bounded. We already know that the intersection of a collection of closed sets is once again closed, so $\cap \mathcal{B}$ is closed.
If I can prove $\cap\mathcal{B}$ is bounded as well, I can use Heine-Borel to say it is compact. So I was not sure how to show it was bounded, but a friend showed me his way of doing it. He did it the following way:
Let $b,a\in\mathbb{R}$ be an upper and lower bound for $B_1$, respectively. Since $\cap\mathcal{B}\subset B_1, a$ is a lower bound for $\cap \mathcal{B}$ and $b$ is an upper bound for $\cap\mathcal{B}$. Therefore, $\cap\mathcal{B}$ is bounded and then by Heine-Borel, it is compact.
My question is on the paragraph right above. I don't understand why what he did means that $a$ and $b$ are lower and upper bounds for $\cap\mathcal{B}.$ Can anyone explain why this is? Thanks.
Best Answer
$I=\cap B_i$ is closed because the arbitrary intersection of closed sets is closed hence $\cap B_i$ is closed. As you mentioned, any compact subset of $\mathbb{R}^n$ is closed and bounded
Now $I$ is a closed subset of a compact set (any $B_i$ will suffice). Hence it is also compact (Why? Proof below). To answer your question about the boundedness of $I$, it is a subset of a bounded set (any $B_{i}$ suffices) hence it is bounded.
We need to prove that a closed subset of a compact space is closed.
Let $X$ be any hausdorff space. ($\mathbb{R}^n$ certainly is.)
Let $A\subset C$ be closed with $C$ compact. Take an open covering of $M=\{A_{\alpha}\}$. Since $A$ is closed, $U=X-A$ is open. Hence $N= M\cup \{U\}$ is an open covering of $C$. Since $C$ is compact, $M$ has a finite subcover for $C$. This obviously covers $A$ as well thus $A$ has a finite subcover. If the finite subcover of $C$ involves $U$, we can always eliminate it.
I assumed $X$ is hausdorff because it can be shown that any compact subset of a hausdorff space is closed.