Finite Covering of Closed Unit Disk? I'm in an analysis class, and we just learned about the Heine-Borel theorem... But it seems so counterintuitive to me. Even with one of the simplest examples of a closed and bounded set, the closed unit disk, I can't seem to come up with an open cover for it, let alone a finite cover. Could anyone give an example of a finite covering for the closed unit disk in $\mathbb{R}^2$?
Also, on a separate note, $\mathbb{C}$ is isomorphic to $\mathbb{R}^2$, right? So does the Heine-Borel theorm imply that every closed and bounded subset of the complex numbers is also compact?
 A: Let me answer your questions and make a few points.

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*There are many finite covers of the closed unit disc! Here's one: $\{B_2(0,0)\}$, where $B_2(0,0)$ is the open ball of radius $2$ about the origin $(0,0)$. Here is another: $\{B_2(0,0), B_1(5,0)\}$. Here is another: $\{B_1(1,0), B_1(0,1), B_1(-1,0), B_1(0,-1), B_{1/2}(0,0)\}$. Do you see why these sets of open sets cover the closed unit disc?


*However, the Heine-Borel theorem does not just say there is some finite open cover of a closed, bounded set. That would be not very interesting: any set in any topological space has some finite open cover; we can just take the cover consisting of only the entire space, $\{X\}$.
Heine-Borel says that closed and bounded sets are compact, which is to say that for any open cover there is a finite subcover. It doesn't just say there is some finite cover by open sets that we can make. It says: if you give me any cover by open sets, I can find a finite subcover (subcover meaning that I use only the sets you gave me, but I am allowed to use only finitely many of them.)


*Yes, because $\mathbb{C}$ is isomorphic to $\mathbb{R}^2$ (isomorphic really meaning they are the same except for complex number multiplication, which we don't care about when we are talking about topology), Heine-Borel tells us that any closed and bounded subset of the complex numbers is compact.
