Example of Hausdorff Space Whose Quotient Space is Not Hausdorff

Question

I'm self-studying Croom's Principles of Topology. Exercise 7.4.13 says:

Give an example of a Hausdorff space which has a quotient space that is not Hausdorff.

I came up with the example below, and I'd like your feedback on whether it is correct. Thank you in advance.

Let $A = \mathbb{R} \backslash \{0\}$. Consider $\mathbb{R}$ in the standard topology, which is of course Hausdorff, and $\mathbb{R}/A$, where we identify the members of $A$ to a point. We see $\mathbb{R}/A = \{ \{0\}, A \}$. The only subsets of $\mathbb{R}/A$ containing $\{0\}$ are $\mathbb{R}/A$ and $\{\{ 0 \}\}$. Since $\bigcup \{\{0\}\} = \{0\}$ is not open in $\mathbb{R}$, the set $\{\{0\}\}$ is not open in $\mathbb{R}/A$. As a result, the only open subset of $\mathbb{R}/A$ containing $\{0\}$ is $\mathbb{R}/A$. Consequently, any open set in $\mathbb{R}/A$ containing $A$ belongs necessarily intersects every open set in $\mathbb{R}/A$ containing $\{0\}$. It follows that $\mathbb{R}/A$ is not Hausdorff.

Answer 1

The example is fine, but your argument could be simplified a bit: consequently, any open set in the quotient space containing $\{0\}$ necessarily contains $A$, so the quotient not only isn't Hausdorff, it's not $T_1$ (although it is $T_0$, since $\{A\}$ is open).

Answer 2

Even simpler: $X=\Bbb R$ (usual top.). Let the two classes (for the equivalence relation to induce the quotient) be $\Bbb Q$ and $\Bbb P$ (the irrationals).

Then the only non-empty open set of the quotient space $Y=\{q,p\}$ is $Y$ as all non-empty open subsets of $X$ contain both rationals and irrationals. So $Y$ has the trivial (indiscrete) topology and is not Hausdorff.