In Willard's General Topology, section 13.2.c,
for any topological space X is defined a quotient space X/~ such that x ~ y iff $cl({\{x\}}) = cl(\{y\})$ where $cl(.)$ is the topological closure.
Then it is stated that if in the new space X/~ there are x and y such that their closure is the same, then they are actually equal.
Somewhere on the web (Jianfei Shen's "A Solution Manual for Willard (2004)" p38), I found this "proof" but this seems too much incomplete and does not satisfy me:
take any $cl(\{x\}) \neq cl(\{y\})$ in X/~. Then $cl(cl(\{x\})) = cl(\{x\}) \neq cl(\{y\}) = cl(cl(\{y\}))$
This is not clear at all why we should have $cl_{X/\sim}(cl_{X}(\{x\})) = cl_X(\{x\})$ as, inside $X/\sim$, the $cl_X(\{x\})$ are just standard elements (not sets, let alone closed sets) defined as equivalence classes from the relation $\sim$.
I've also looked into the quotient topology of $X/\sim$, where closed sets are the sets $C$ such that $\cup\{f^{-1}(y), y \in C\}$ is closed in $X$, where $f : x \in X \mapsto cl(\{x\}) \in X/\sim$
but I did not manage to sort out the path to the proof.
It should be obvious, from the fact that the author did not give any detailed explanation, but I am not getting it…
Best Answer
Let $[x]$ denote the equivalence class of $x \in X$ with respect to $\sim$. We have $[x] \subset cl_{X}(\{x\})$ because if $y \sim x$, the $y \in cl_{X}(\{y\}) = cl_{X}(\{x\})$. However, in general $[x] \subsetneqq cl_{X}(\{x\})$. As an eaxmple take the Sierpinski space $X = \{0,1\}$ with open sets $\emptyset, \{0\}, X$. Then $cl_{X}(\{0\}) = X, cl_{X}(\{1\}) = \{1\}$. Hence $[0] = \{0\} \subsetneqq cl_{X}(\{0\})$.
This shows that Jianfei Shen's solution manual erroneously assumes that the points of $X/\sim$ are the sets $cl_{X}(\{x\})$ with $x \in X$.
The space $X/\sim$ is nothing else than the Kolmogoroff quotient of $X$. In fact, we have $cl_{X}(\{x\}) = cl_{X}(\{y\})$ iff each open $U \subset X$ contains either both $x,y$ or none of $x,y$.
Now see Prove that the Kolmogorov Quotient is $T_0$ for a correct proof.