Real Analysis – Heine-Borel Theorem for Any Compact Set

Question

I have recently studied the Heine-Borel theorem i.e. set is compact iff any open cover has a finite subcover. I have proven it only for compact intervals and their cartesian products for higher dimensions. I was told it follows directly and that it applies to any compact set. In particular, the sets $X \subset \mathbb R^n$ (Definition of compact is closed and bounded)

My question is how could one show this? Thank you for any ideas or insights.

Answer 1

I will shorten "any open cover of $A$ has a finite subcover" to "$A$ is cover-compact". Given what you know already, it suffices to prove that cover-compact subsets are compact and closed subsets of cover-compact sets are cover-compact, since any bounded set is a subset of a product of rectangles. Let us check each of these.

1. Cover-compact sets are bounded: If $A$ is cover-compact then its open cover $\{B(0, r) \mid r > 0\}$ has a finite subcover, which implies that $A \subset B(0, r)$ for some $r$.
2. Cover-compact sets are closed: Suppose $A$ is cover-compact and $x \notin A$ is a limit point. Since $x \notin A$, we have an open cover $\{\mathbb{R}^n \setminus \overline{B}(x, r) \mid r > 0\}$ by complements of closed balls around $x$. But the union of a finite subcollection given by radii $r_1 < \dots < r_k$ is just $\mathbb{R}^n \setminus \overline{B}(x, r_1)$. Since $x$ is a limit point of $A$, the ball $B(x, r_1)$ contains some point of $A$ and hence the above finite union cannot cover $A$.
3. Closed subsets of cover-compact sets are cover-compact: Let $A$ be cover-compact and $F \subset A$ be closed. Now for any open cover $\mathcal{U}$ of $F$, $\mathcal{U} \cup \{\mathbb{R}^n \setminus F\}$ is an open cover of $A$, therefore has a finite subcover $\mathcal{U}_A$. Then $\mathcal{U}_A \cap \mathcal{U}$ is a finite cover of $F$ and a subset of $\mathcal{U}$.