# Quotient to make $X$ a $T_1$ space

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

• Note that you can just use $[0,\infty)$ instead of $\mathbb{R}$, which maybe makes the example a bit easier to think about. – Eric Wofsey Apr 18 '19 at 15:34
• @freakish Ok, I misunderstood you. My bad. I thought you take $[x,\infty)$ as open sets – YuiTo Cheng Apr 18 '19 at 15:50
• @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 $\overline{\{x\}}$. – principal-ideal-domain Apr 18 '19 at 19:11
• @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 $0\not\sim x$. – freakish Apr 18 '19 at 19:33
• @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. – principal-ideal-domain Apr 18 '19 at 19:38