Why a circle with a hole in it is not compact?

Question

Take the real plane $\mathbb R^2$ with the standard topology, and take $A\subset \mathbb R^2$ which is a closed circle minus a smaller closed circle:

I am trying to understand why this set is not compact, the hint is to use the Heine-Borel theorem. For me, intuitively, the set is bounded because it can be enclosed by an open ball, so the property it lacks is being closed. One definition for a closed set states that the set must contain all its frontier points, in this case it doesn't contain the "interior" frontier points, but i don't think is a good explanation. Another way to prove it is not closed is $\mathbb R^2-A$ is open, but the resultant set has a closed ball at its center, so i think that this doesn't explain it either. Any insights will be greatly appreciated.

Answer 1

You are on the right track! Indeed, as the hint suggests, you should show that $A$ is not closed, i.e. it does not contain all of its limit points.

To do this, all you need to do is find a sequence in $A$ whose limit is not contained in $A$. Can you think of one?

Answer 2

If we have the standard unit circle with hole at $(0,0)$, consider the sequence $(\frac1n, 0)$ which lies in the puntured circle and converges to the hole, in the plane at least. This shows your set is not closed hence not compact. In the case of an annulus, as you describe, the set would be something like $\{(x,y):r<\|x\| \le R\}$ ($R$ being the lareg radius, $r$ the small one). Its boundary is two circles, the inner one of radius $r$ not being a subset of the annulus, also showing the non-closedness that way.

Answer 3

Using the Heine-Borel theorem just means saying, $A$ is not closed, hence not compact.

If you want, however, an explanation as to what fails here, let's use this special case: let $A$ be the closed unit disc, centered at the origin, with the center removed. (So, again, $A$ fails to be closed.) Consider the open sets $$ U_{n} = \left\{ x \in \mathbb{R}^2 : ||x|| > {1 \over n} \right\}, \quad n = 1, 2, \ldots. $$ These are the open exteriors of the circles with radii $1/n$ and center at the origin. The sets $U_{n}$ constitute an open cover of $A$. But there is no finite subcover.

Answer 4

I suggest that you ignore the imprecision of the "interior" language.

I also suggest you look at an actual mathematical formula for your set $A$. Let $C$ be the center of the two circles, let $r$ be the radius of the smaller circle, and let $R$ be the radius of the larger circle. With this notation we have $$A = \{P \in \mathbb R^2 \mid r < d(C,P) \le R\} $$ Note the strict inequality on the left versus the nonstrict inequality on the right.

Now take any point $Q$ on the smaller circle, i.e. any point $Q$ such that $d(C,Q)=r$. The point $Q$ is a limit point of $A$, and the point $Q$ is not an element of the set $A$, and therefore $A$ is not closed.