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?


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) $\{\emptyset, \{[1]\}, X/\sim\}$.

  • $\begingroup$ Note that you can just use $[0,\infty)$ instead of $\mathbb{R}$, which maybe makes the example a bit easier to think about. $\endgroup$ – Eric Wofsey Apr 18 '19 at 15:34
  • $\begingroup$ @freakish Ok, I misunderstood you. My bad. I thought you take $[x,\infty)$ as open sets $\endgroup$ – YuiTo Cheng Apr 18 '19 at 15:50
  • $\begingroup$ @freakish Thank you very much for your answer. I don't think that you need transfinite induction to prove $0\not\sim x$ for $x>0$. It follows directly from your characterization of $\overline{\{x\}}$. $\endgroup$ – principal-ideal-domain Apr 18 '19 at 19:11
  • $\begingroup$ @principal-ideal-domain Eee, it seems that to obtain $\sim$ you need the transitive closure of $\approx$ first (is $\approx$ transitive?) and then you apply the symmetric closure. These have to applied to show that $0\not\sim x$. $\endgroup$ – freakish Apr 18 '19 at 19:33
  • $\begingroup$ @freakish The relation $\approx$ is already transitive. When you have $x \in \overline{\{y\}}$ and $y \in \overline{\{z\}}$, then it follows $\overline{\{y\}} \subseteq \overline{\{z\}}$, so you have $x \in \overline{\{z\}}$ as well. $\endgroup$ – principal-ideal-domain Apr 18 '19 at 19:38

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