Let $X$ be a topological space. We define a relation on $X$:
$$x \approx y : \quad \Leftrightarrow \quad x \in \overline{\{y\}}.$$
In general $\approx$ is no equivalence relation since it lacks symmetrie. But of course it generates an equivalence relation $\sim$ on $X$ (the smallest equivalence relation containing $\approx$).
My question: Is $X/\sim$ a $T_1$ space?
Or in other words: Are the equivalence classes of $\sim$ closed?
Best Answer
Consider the topology on $X=[0,\infty)$ generated by closed subsets $[a,\infty)$ and with addition of $\{0\}$ as a closed subset. So open subsets are generated by $[0,a)$ and $(0,\infty)$. You can verify that $A\subseteq [0,\infty)$ is closed if and only if $A$ is of the form $\{0\}$ or $[a,\infty)$ or $\{0\}\cup[a,\infty)$. It follows that
$$\overline{\{x\}}=\begin{cases} [x,\infty) &\text{if }x>0 \\ \{0\} &\text{for }x=0 \end{cases}$$
As a consequence $0\not\sim x$ if $x>0$.
On the other hand you can easily check that if $x,y>0$ then $x\sim y$.
Combining these two we get that the equivalence class $[1]_\sim=(0,\infty)$ is not closed. In fact the quotient space $X/\sim$ is equal to $\{[0],[1]\}$ with topology (of open subsets) $\{\emptyset, \{[1]\}, X/\sim\}$.