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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 $X$?.

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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 $T_1$).

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  • $\begingroup$ @Henno_Brandsma isn't $\overline{\{x\}}=\bigcup_{x\in \overline{\{x\}}}[x]$? $\endgroup$ – Percy Nov 29 '18 at 8:34
  • $\begingroup$ @Percy yes it is, but that doesn’t help you. $\endgroup$ – Henno Brandsma Nov 29 '18 at 17:21
  • $\begingroup$ Closures of singletons don't necessarily have to be either disjoint or the same: in the Sierpinski space, $\overline{\{0\}} = \{0,1\}$ whereas $\overline{\{1\}} = \{1\}$. $\endgroup$ – Daniel Schepler Dec 13 '18 at 17:51
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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, $[0] = \{ 0 \}$ is not closed.

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  • $\begingroup$ I think $\{0\}$ is closed since $\mathbb{N}\backslash \{0\}$ is open. $\endgroup$ – Percy Dec 13 '18 at 16:05
  • $\begingroup$ @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. $\endgroup$ – Daniel Schepler Dec 13 '18 at 17:41
  • $\begingroup$ 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 $\overline{\{1\}}=\{1\}$. $\endgroup$ – Daniel Schepler Dec 13 '18 at 17:48

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