Let $\tau $ be a topology: $\tau=\{A \subset \mathbb{R}^2 | S^1 \subset(\mathbb{R}^2 \setminus A ) \} \cup\{\mathbb{R}^2\}$ where $S^1=\{ (x,y) \in \mathbb{R}^2 | x^2+y^2=1 \}$
Prove $\tau$ is a Hausdorff space and that it is compact
The solution I was given for the fist part is: if $p \in S^1$, the only neighborhood of $p$ is $\mathbb{R}^2$, then it is not a Hausdorff's space.
I don't get it. the open sets of this topology are disjoint set with the circle $S^1$, so I should not be taking a neighborhood(an open set) on the circle, should I? Can someone elaborate on it and prove compactness as well?
It is not Hausdorff. Take $x \in S^1$, then the only open set that contains $x$ is $ \mathbb{R}^2$, so for the points in $S^1$, there aren't disjoint open sets containing them. Now that being said, if there is a cover $\mathcal{A}$ for $ \mathbb{R}^2$, then $ \mathbb{R}^2 \in \mathcal{A}$ ( since a point of $S^1$ is in the cover), so it has a finite subcover.