In Willard's General Topology, section 13.2.c, for any topological space $X$ is defined a quotient space $X/\sim$ such that $x \sim y$ iff $cl(\{x\})=cl(\{y\})$ where $cl(.)$ is the topological closure.
Then it is stated that the quotient space $X/\sim\, = \widetilde X$ is a $T_0$ space.
The proof of this statement has been given in my instructor's lecture notes and it follows the following line of reasoning:
Let $[x]\neq [y]$ be two elements of $\widetilde X$. Then either $x \notin cl\{y\}$ or vice versa.
Suppose $x \notin cl(\{y\})$. Now we know that $X \backslash cl(\{y\})$ is open in X. Also, it is a saturated subset of $X$ and $x \in X \backslash cl(\{y\})$.
Therefore, $\varphi(X \backslash cl\left(\{y\})\right)$ is open in $\widetilde X,\,$ where $\varphi :X \rightarrow \widetilde X$ is the canonical mapping. This is due to the fact that open sets in quotient space, carrying the final topology w.r.t. the canonical mapping, are in one-one correspondence with open saturated subsets of the original space.
Further, $[x] \in \varphi(X \backslash cl\left(\{y\})\right)$ but $[y] \notin \varphi(X \backslash cl\left(\{y\})\right)$. Hence, $\widetilde X$ is a $T_0$ space.
My doubts are:
1) How do we know that $X \backslash cl(\{y\})$ is a saturated subset?
2) What is the actual form of $[x]$? I have managed to prove that $[x] \subset cl(\{x\})$ (quite trivial) but it seems to be a dead end.
Best Answer
$O=X\setminus \overline{\{y\}}$ is saturated:
If $z \in O$ and $z' \sim z$, we have to show that $z' \in O$ as well. (This is the definition of being saturated in this context).
So $z \in O$ impies $z \notin \overline{\{y\}}$ and $z \sim z'$ means $\overline{\{z\}} = \overline{\{z'\}}$. Now, $z' \notin \overline{\{y\}}$, or else $\{z'\} \subseteq \overline{\{y\}}$ and hence $\overline{\{z'\}} \subseteq \overline{\{y\}}$, but then also $z \in \overline{\{z\}} = \overline{\{z'\}} \subseteq \overline{\{y\}}$ which is a contradiction. So $z' \in O$, as required.
So when $x \notin \overline{\{y\}}$, $x \in O$ and $O$ is saturated so $\phi[O]$ is open in $X{/}\sim$ and contains $[x]=\phi(x)$ but not $[y]=\phi(y)$( because if $[y] \in \phi[O]$, this would mean that some $z$ with $z \sim y$ would be in $O$, but this cannot be as $z \in \overline{\{z\}} = \overline{\{y\}}$ which is disjoint from $O$ by definition). This thus shows the $T_0$-ness of the quotient.
The actual form of $[x]$ doesn't really matter: it's just the set of all $y$ that have the same closure as $\{x\}$, by definition. The idea is to identify all points that are in exactly the same open sets: if $x \sim y$ and $x \in O$ then $y \in O$, and if $y \in O$ then $x \in O$, for any open set. So we can never distinguish $x$ and $y$ by using open sets.
A simple example: $X=\{1,2,3,4\}$, $\mathcal{T}=\{\emptyset,\{1,2\},\{3,4\},X\}$. The quotient relation identifies $1$ and $2$, and $3$ and $4$, and the resulting quotient space is a two point discrete space (indeed $T_0$). Leaving out $\{3,4\}$ as an open set, gives the same quotient two-point set but with a Sierpiński topology instead. Try it out.