D-{0} is not compact as a subset of $\mathbb{R}^2$.

Question

So one of the problems I did this week involved proving that $D-\{(0,0)\}$ is not compact in $\mathbb{R}^2$, a counter example of compactness is quite easy and I have already done it. You can just take a sequence of complements of closed balls converging on $(0,0)$

But I thought of an example that I think is more elegant...

For all $\epsilon \in \mathbb{S}^{(1)}$ we take a open ball of radius 1.

Now this is clearly a open cover so know we come to the problem... proving that there is no finite sub-cover

So if there is a finite collection of open balls we order them form 1 to $n$ so between$B(\epsilon,1)$ and $B(\epsilon_1,1)$ and $B(\epsilon_2,1)$ and since $\mathbb{R}$ is a metric space then $d(B(\epsilon_1,1),B(\epsilon_2,1))=\delta$ but I can't fined the exact point that is missing from theses balls.

Thank you for the help...

Answer 1

I think this argument works. Identify $\epsilon$ as an angle on the plane, about the origin. If there were a finite subcover, there will be two such open sets whose centres are closest, say with angles $\epsilon_1$ and $\epsilon_2$. Prove that there is a point on the punctured disc along the line passing through the origin with angle $\frac{\epsilon_1+\epsilon_2}{2}$ that do not belong to any of the open sets in the finite subcover.