Quotient of $R^2$

Question

Consider $X$=$R^2$ and let A $\in X$ be A={$(-1,0),(1,0)$}. Consider now the quotient space ${X}/{A}$.

Is it Hausdorff? Connected? Compact?

My answer: the projection from $X$ to $A$ is surjective and continuous, $X$ is connected and so is ${X}/{A}$.

The projection of the open cover of $X$ made up of ball centered at $(0,0)$ is an open cover of $X/A$ as they are all both open and saturated. We can not extract any finite subcover and because of that ${X}/{A}$ is not compact.

Honestly I'm not sure about the proof about compactness and I'm left stuck with the Hausdorff question. If I could demonstrate that the projection is open I could work from there. Working with the intervals left me with a proof that I'm not 100% sure about.

Thanks in advance for any help.

Answer 1

Your proof of noncompactness is fine. For the Hausdorff condition, there's really only one unusual point in this space. For a direct proof, you just have to separate this point from points outside.

Answer 2

Hint: Prove that, in general, if $X$ is Hausdorff and $K \subset X$ is compact then the contraction $X/K$ is Hausdorff