# Kolmogorov topological spaces ( T0-spaces)

I think it should be easy but: how can it be shown that this two definitions of T0-spaces are equivalent:

1- Given $x,y\in X$ with $x\neq y$ exists an open set $U$ that $x\in U$ and $y\notin U$ or $y\in U$ and $x\notin U$.

2- Given $x,y \in X$ with $x\neq y$ then $\overline{x}\neq \overline{y}$ (Different points has different closures

I have seen this equivalence many times mentionated in the literature but I don´t know how to prove it.

• Do you know this result: For any subset $A$ of the topological space $X$, $x\in \overline{A} \iff \forall U$ open in $X$, $U\cap A \neq \emptyset$? – Indrayudh Roy Feb 6 '17 at 17:48
Assume Statement $1$. Take $x,y\in X$ with $x\neq y$. Let $U$ be an open set containing $x$ such that $y\notin U$. Since $U \cap \{y\} = \emptyset$, then $x\notin \overline{\{y\}}$, and hence $\overline{\{x\}}\neq \overline{\{y\}}$. Therefore, Statement $2$ holds.
Conversely, suppose Statement $2$ holds. Let $x,y\in X$ with $x\neq y$. Then $\overline{\{x\}} \neq \overline{\{y\}}$, so either $x\notin \overline{\{y\}}$ or $y\notin \overline{\{x\}}$; in the former case there is an open set $U$ such that $x\in U$ and $y\notin U$, and in the latter case there is an open set $V$ such that $y\in V$ and $x\notin V$. Thus, Statement $1$ holds.
• Why $x \notin \overline{\{y\}}$ or $y\notin \overline{\{x\}}$? – matemagreek Feb 6 '17 at 20:07
• @matemagreek Having $x\in \overline{\{y\}}$ is equivalent to having $\overline{\{x\}}\subset \overline{\{y\}}$. Similarly for the condition $y\in \overline{\{x\}}$. – kobe Feb 6 '17 at 21:18