# A space is $T_{0}$ if and only if the closures of singletons are distinct

This is exercise $1.5$A from Engelking's book, page $47$.

Verify that $X$ is $T_{0}$ if and only if $\overline{\{x\}} \neq \overline{\{y\}}$ for every pair of distinct points $x,y$.

My try:

Suppose first that $X$ is $T_{0}$ and let $x,y$ be two distinct points of $X$. Then since $X$ is $T_{0}$ wlog we can find an open set $U$ such that $x \in U$ and $y \not \in U$. Let us show that $\overline{\{x\}} \neq \overline{\{y\}}$. Suppose otherwise, then $x \in \overline{\{x\}} \subseteq \overline{\{y\}}$. This in turn implies that $U \cap \{y\} \neq \emptyset$ so that $y \in U$, a contradiction.

For the other direction, let $x,y$ be distinct points of $X$. Then either $\overline{\{x\}}$ is not contained in $\overline{\{y\}}$ or $\overline{\{y\}}$ is not contained in $\overline{\{x\}}$. Assume the former, then $x \not \in \overline{\{y\}}$ so we can find an open set $U$ such that $U$ contains $x$ and $U$ does not intersects $\{y\}$. If the other case happens we can find an open set $V$ such that $V$ contains $y$ and $x \not \in V$. Therefore $X$ is $T_{0}$.

Is this OK?

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1. How do you justify ‘this in turn implies ...’? 2. How do justify ‘then $x \notin \overline{\{ y \}}$’? – Zhen Lin Aug 7 '11 at 10:48
@Zhen Lin: $x \in \overline{\{y\}}$ and $U$ is an open set containing $x$ therefore $U$ intersects $\{y\}$. For the other one: note that if $x \in \overline{\{y\}}$ then $\{x\} \subset \overline{\{y\}}$, taking closures on both sides yields $\overline{\{x\}} \subset \overline{\{y\}}$, a contrdiction. – user10 Aug 7 '11 at 10:52
The argument is detailed, clear, and correct. Maybe, for added clarity, in the second half of the proof, one could insert "Then $\overline{\{x\}} \ne \overline{\{y\}}."$ between the first two sentences, and begin the next sentence with "So". Or maybe that's overkill.