$A_{\alpha,\beta} = \{(x,y) \in \mathbb{R}^2 \vert x = r \cos(\theta), y = r\sin(\theta), r > 0, \alpha < \theta <\beta \}$ Let $A_{\alpha,\beta} = \{(x,y) \in \mathbb{R}^2 \vert x = r \cos(\theta), y = r\sin(\theta), r > 0, \alpha < \theta <\beta \}$. Then define:
$$ \mathcal{B} = \{ A_{\alpha,\beta} \vert \alpha < \beta \}$$
I want to know if the topology induced by $\mathcal{B}$ on $\mathbb{R}^2 \setminus \{(0,0)\}$ is compact. 
To me it seems it is not compact, because the open sets are part of disks, and I would need $r$ to go to infinity to cover all $\mathbb{R}^2 \setminus 
\{(0,0)\}$, so I cannot do that with a finite number of open sets. Is that right? Would it work as a formal proof?
 A: Let $X = ℝ^2 \setminus \{(0,0)\}$.
(1) It $A_{α,β}$ are cones extending infintely from the origin, then $X$ is compact with respect to the topology induced by $\mathcal B$. The space would then be basically an artificially enlarged circle.
(2) If $A_{α,β}$ are arcs of circles with radius $r$ (and really should be named $A_{r,α,β}$ in this case), then the space is not compact.
Here are the reasons:
(1) Notice that all sets $A_{α,β}$ are open in the Euclidean topology of $ℝ^2$. Also note that $\mathcal B$ is stable under intersections, so it yields a base for a topology (if you add the empy set).
Hence, every open covering of $X$ may be assumed to be a covering consisting of sets in $\mathcal B$. Such covering can be viewed as a covering of the circle in the Euclidean topology of $ℝ^2$. As the circle is compact, finitely many of these base sets suffice to cover it. But a covering of the circle with elements of $\mathcal B$ already has to cover all of $X$ because all angles are covered.
(2) Again, $A_{r,α,β}$ are stable under intersections, so they form a base (if you, again, add the empty set). Note that $A_{r,α,β}$ are bounded in the usual metric of $ℝ^2$.
Hence, every finite open covering of base sets must be bounded, so you never can cover $X$ with finitely many them. Therefore $X$ can’t be compact.
