Heine-Borel Theorem states that if a set has an open cover and if we can find a finite subcover from that open cover that covers the set, the set would be compact.
I got this question while I was trying to prove that Heine-Borel property will imply that the set is closed.
Is there any restriction on open sets that are the contained in the open cover? If not, can't I just make one of the open sets to be a set of real numbers? Then, the set would always have a finite subcover, that is, the set of all real numbers that covers the set. Hence, it can cover $(0,1)$ or any other open intervals but still $(0,1)$ is not closed.
Best Answer
The Heine Borel Theorem states that in $\mathbb{R}^n$, a set is compact iff it is closed and bounded. Perhaps in your class you initially defined compactness as being closed and bounded, but in general a set $K$ is compact if every open cover has a finite sub-cover.
An open cover $\mathcal{U}$ for as set $K$ is a collected of sets that are open in $\mathbb{R}^n$ such that $$K\subset\bigcup_{U\in\mathcal{U}}U.$$ So, the sets simply must be open. I think the key point you are missing here is that all open covers have to have a finite subcover, not just some open cover. Otherwise, every subset of $\mathbb{R}^n$ would be compact, because $\mathcal{U}=\{\mathbb{R}^n\}$ is an open cover for every subset of $\mathbb{R}^n$.
To see that every compact set is closed, suppose $K$ is compact and $p\not\in K$. For each $x\in K$, let $U_x$ be an open set containing $x$ and $V_x$ be an open set containing $p$ such that $U_x\cap V_x=\emptyset$. Then $\mathcal{U}=\{U_x:x\in K\}$ is an open cover of $K$. Thus, we can extract a finite subcover $\{U_{x_1},\dots, U_{x_n}\}$. Note that $$\left(\bigcap_{i=1}^n V_{x_i}\right)\cap\left(\bigcup_{i=1}^n U_{x_i}\right)=\emptyset$$ by our construction. Since $K\subset \bigcup_{i=1}^n U_{x_i}$, it follows that $V=\bigcap_{i=1}^n V_{x_i}$ is an open set containing $p$ that contains no element of $K$. Thus, $p$ cannot be a limit point of $K$. It follows that $K$ contains all of its limit points, and is closed.
Note that the point where we used a finite subcover is so that we could take a finite intersection of open sets to ensure $V$ was open.