# $T_0$-space iff for each pair of $a$ and $b$ distinct members of X, $\overline{\{a\}}\neq\overline{\{b\}}$

Let $$(X,\mathscr T)$$ be topological space. Prove that $$(X,\mathscr T)$$ is $$T_0$$-space iff for each pair of $$a$$ and $$b$$ distinct members of X, $$\overline{\{a\}}\neq \overline{\{b\}}.$$

My attempt:-

Let $$(X,\mathscr T)$$ be topological space. Prove that $$(X,\mathscr T)$$ is $$T_0$$-space. $$a\neq b \implies$$ there is an open set $$U$$ contains $$a$$ but not $$b$$. $$\{b\}\subseteq X\setminus U \implies \overline{\{b\}}\subseteq X\setminus U.$$ $$a\notin X\setminus U$$. Any closed set containing $$a$$ must be completeltely disjoint from $$X\setminus U.$$ So, $$\overline{\{a\}}\neq \overline{\{b\}}.$$

Conversaly, $$\overline{\{a\}}\neq \overline{\{b\}}.$$ Consider $$X\setminus \overline{\{b\}}$$, which is an open set. $$X\setminus \overline{\{b\}} \subseteq X\setminus \{b\}$$. $$a\in X\setminus \{b\}$$ . $$b\notin X\setminus \{b\}\implies b\notin X\setminus \overline{\{b\}}$$. How do I complete the proof?

• The first implication is not true as written: Give $\mathbb{R}$ the topology in which a proper subset $U\subset \mathbb{R}$ is open iff $0\not\in \mathbb{R}$, and put $a = 0$. You can swap $a$ and $b$, though. – anomaly Feb 11 at 15:42
• Which implication? – Unknown x Feb 11 at 15:56
• It is an excercise in the Foundation of topology by C.W Patty. – Unknown x Feb 11 at 15:59
• See @PaulFrost's comment below. – anomaly Feb 11 at 16:21
• You cannot prove both. That would be a $T_1$ space. Example: Sierpinski space : $S=\{c,d\}$ with $c\ne d$, and the only open sets are $S,\{c\}$, and $\emptyset.$ This is $T_0$ but not $T_1$. We have $\overline {\{d\}}=\{d\}$ but $\overline {\{c\}}=\{c,d\}.$ – DanielWainfleet Feb 11 at 17:20

The first part of your attempt starts with a wrong statement.

You say that if $$X$$ is $$T_0$$ and $$a \ne b$$, then there exists an open $$U$$ containing $$a$$ but not $$b$$. However, this is the definition of $$T_1$$. In a $$T_0$$ space you can only say there exists an open $$U$$ containing exactly one of $$a, b$$. Of course you can say that without loss of generality we may assume $$a \in U$$ and $$b \notin U$$. To see the difference, consider the Sierpinski space $$\Sigma = \{ 0, 1 \}$$ with topology $$\mathfrak{T} = \{ \emptyset, \{ 1 \}, \Sigma \}$$. It is $$T_0$$ because the only two distinct points are $$0, 1$$ and $$\{ 1 \}$$ is an open set containing exactly on of these points whereas there does not exist an open set containing $$0$$ but not $$1$$.

Moreover, you cannot conclude that any closed set containing $$a$$ must be completely disjoint from $$X \setminus U$$. But it is irrelevant since $$a \notin \overline{\{b \}}$$, hence $$\overline{\{a \}} \ne \overline{\{b \}}$$. As an example consider $$\Sigma$$ and take $$a = 1, b = 0, U = \{1 \}$$. You have $$\overline{\{ a \}} = \Sigma$$ and $$\overline{\{ b \}} = \{ 0\}$$. In fact $$\overline{\{ b \}} \subset X \setminus U$$ and $$a \notin X \setminus U$$, but $$\overline{\{ a \}}$$ is not completely disjoint from $$X \setminus U$$.

For the converse, let $$a, b$$ distinct points of $$X$$. Then $$\overline{\{a \}} \ne \overline{\{b \}}$$. The set $$S = \overline{\{a \}} \cap \overline{\{b \}}$$ is closed.

It is impossible that both $$a, b \in S$$ because otherwise $$\overline{\{a \}} \subset S \subset \overline{\{b \}}$$ and $$\overline{\{b \}} \subset S \subset \overline{\{a \}}$$, i.e. $$\overline{\{a \}} = \overline{\{b \}}$$.

Case 1. $$S$$ contains none of $$a, b$$. Then $$b \notin \overline{\{a \}}$$ (because otherwise $$b \in \overline{\{a \}} \cap \overline{\{b \}} = S$$), hence $$b \in X \setminus \overline{\{a \}}$$ and we are done since $$a \in \overline{\{a \}}$$, i.e. $$a \notin X \setminus \overline{\{a \}}$$.

Case 2. $$S$$ contains exactly one of $$a, b$$, w.l.o.g. $$a$$. Then $$a \notin X \setminus S$$ and $$b \in X \setminus S$$ and we are done again.

• But my converse is not complete. How do I prove that $a\in X\setminus \overline{ \{b\}}$? – Unknown x Feb 11 at 16:23
• Thank you for correcting me in the first part. – Unknown x Feb 11 at 16:24
• You have an amusing typo "unclean" for "unclear" in the 1st sentence. – DanielWainfleet Feb 11 at 17:07
• @DanielWainfleet It was rather an unclear formulation of a non-native speaker than a typo ;-) I edit my answer. – Paul Frost Feb 12 at 10:53
• My native language is English and I make a lot of typos here. – DanielWainfleet Feb 13 at 4:53