I'm just getting started on topology, and having trouble reconciling with the Heine-Borel Theorem. That is:
For a subset S of Euclidean space $\mathbb{R}^n$, the following are equivalent:
- S is closed and bounded
- S is compact; that is, every open cover of S has a finite subcover
I've read the proof, but I've managed to get hooked on a "counterexample" that I can't think my way around. Suppose S is some open set in $\mathbb{R}^n$, and then let C be a collection of sets containing just S. Is C then not an open cover of S with a finite subcover (that subcover being C itself)? This would imply that S is both compact and open.
Any guidance would be greatly appreciated. Thanks
You've exhibited an example of an open cover with a finite subcover. To show a subset is compact, we must show that every open cover admits a finite subcover.
If $S \subseteq \mathbb{R}^n$ is a nonempty open subset, I claim that we can cook up an example of an open cover with no finite subcover. In particular, this will show that $S$ is not compact.
For $n\geq 1$, define $U_n \subseteq S$ to be $$U_n =\{ x \in S : d(x, \mathbb{R}^n \setminus S) > 1/n \}.$$ Here, $d(x, T)$ means "the minimal distance between the point $x$ and the set $T$", and is defined precisely as $$d(x,T) = \inf_{y\in T} d(x,y).$$
In words, $U_n$ is the subset of $S$ consisting of elements that are at least $1/n$ away from the edge of $S$. We have $U_1\subseteq U_2 \subseteq \cdots \subseteq S$. Eventually, every element of $S$ will fall into some $U_n$, so this is an open cover. If we take a finite subcover, then we will miss some points.