Elements of a $T_{0}$ quotient space are closed in $X$.

Define an equivalence relation on a topological space $$(X,\tau)$$ as $$x\sim y$$ iff $$\overline{\{x\}}=\overline{\{y\}}$$.

I want to show that each equivalence class $$[x]$$ is closed in $$X$$.

My attempt: Let $$y\in X\backslash[x]$$. Then $$y\notin [x]$$ which means that $$[x]\neq[y]$$. Because $$(X\backslash \sim)$$ is $$T_{0}$$, we have an open set $$O\subseteq (X\backslash \sim)$$ containing $$[y]$$ and not containing $$[x]$$. Because $$f:X\longrightarrow (X\backslash \sim)$$ is continuous, we have $$y\in f^{-1}(O)\in \tau$$ and $$x\notin f^{-1}(O)$$. It's easy to see that $$f^{-1}(O)\subseteq X\backslash[x]$$ which shows that $$X\backslash[x]$$ is open. Thus $$[x]$$ is closed.

What if $$O$$ contains $$[x]$$ and not $$[y]$$? How would I show that $$X\backslash[x]$$ is open under that case? That is my problem.

Or is $$[x]$$ not necessarily closed in $$X$$?.

Note that the classes of $$\sim$$ are exactly the sets $$\overline{\{x\}}$$, where $$x$$ ranges over $$X$$: these closures are disjoint or they are the same: if $$z \in \overline{\{x\}} \cap \overline{\{y\}}$$, then $$\overline{\{x\}} = \overline{\{y\}} = \overline{\{z\}}$$. So the classes are closed in $$X$$ (which means the standard quotient space is $$T_1$$).
• @Henno_Brandsma isn't $\overline{\{x\}}=\bigcup_{x\in \overline{\{x\}}}[x]$? – Percy Nov 29 '18 at 8:34
• Closures of singletons don't necessarily have to be either disjoint or the same: in the Sierpinski space, $\overline{\{0\}} = \{0,1\}$ whereas $\overline{\{1\}} = \{1\}$. – Daniel Schepler Dec 13 '18 at 17:51
Consider $$\mathbb{N}$$ with the topology that specifies a subset $$U \subseteq \mathbb{N}$$ is open if and only if $$U = \emptyset$$, or $$0 \in U$$ and $$\mathbb{N} \setminus U$$ is finite. (This is homeomorphic to $$\operatorname{Spec} \mathbb{Z}$$ with the Zariski topology, which was the inspiration for me to come up with this example.) Then $$\overline{\{ 0 \}} = \mathbb{N}$$ whereas for $$n > 0$$, we have $$\overline{\{ n \}} = \{ n \}$$. Therefore, all points of $$\mathbb{N}$$ are in distinct equivalence classes of the defined relation, so $$[x] = \{ x \}$$ for all $$x$$; however, $$ = \{ 0 \}$$ is not closed.
• I think $\{0\}$ is closed since $\mathbb{N}\backslash \{0\}$ is open. – Percy Dec 13 '18 at 16:05
• @Percy You're right, I got the definition of the topology that was intended to be homeomorphic to $\operatorname{Spec} \mathbb{Z}$ wrong. Fixed now. – Daniel Schepler Dec 13 '18 at 17:41
• Come to think of it, just the Sierpinski space $( \{ 0, 1 \}, \{ \emptyset, \{ 0 \}, \{ 0, 1 \} )$ is also sufficient to form a counterexample: $\overline{\{0\}} = \{0,1\}$ whereas $\overline{\{1\}}=\{1\}$. – Daniel Schepler Dec 13 '18 at 17:48