# Lower separation axioms

Let $(X,\mathcal{T})$ be a topological space and $A$ be a subset of $X$. $A$ is said to be $g$-closed if $A$ is a subset of $c(O)$ ; whenever $A$ is a subset of $O$, where $O$ is an open set and $c(O)$ is the closure of $O$ in X.

$X$ is said to be $T_{1/2}$ if every $g$-closed set is closed.

My doubt is how can we prove that every $T_{1/2}$ space is $T_0$.

• We always have $M \subseteq c(M)$. Hence the assumption $A\subseteq O$ trivially implies $A\subseteq c(O)$, so every set would be $g$-closed. Something went wrong with the definition of $g$-closedness. Dec 18 '16 at 13:44
• Some search indicates you probably mean that a set $A$ is $g$-closed if for all open $O$ with $A\subseteq O$, we also have $c(A) \subseteq O$. Dec 18 '16 at 13:55

HINT: I will assume that you meant the usual definition of $g$-closed sets: a set $A$ in a space $\langle X,\tau\rangle$ is $g$-closed if and only if $\operatorname{cl}A\subseteq U$ whenever $A\subseteq U\in\tau$. For $x\in X$ let $\tau(x)=\{U\in\tau:x\in U\}$.
Suppose that $X$ is not $T_0$. Then there are distinct $x,y\in X$ such that $\tau(x)=\tau(y)$. Let $A=\operatorname{cl}\{x\}$, and show that $A\setminus\{y\}$ is $g$-closed but not closed.