Proximity induced by a topology Let $\operatorname{cl}$ is the closure operator for some topology.
I will call induced proximity the proximity defined by the formula:
$$A\delta B\Leftrightarrow \operatorname{cl}(A)\cap\operatorname{cl}(B)\ne\varnothing.$$
Is induced proximity really a proximity for every given topological space?
Also: What I call here induced proximity is the weakest proximity generating our topology, right?
 A: From Wikipedia, I learn about proximities:

The resulting topology is always completely regular. 

Thus either induced proximity fails to be a proximity if the given topological space is not regular.
Or the generated topology may differ from the given topology.
The latter kind of failure occurs in the space $\{1,2\}$ with open sets $\emptyset$, $\{1\}$, $\{1,2\}$. Here, $\{2\}$ is closed in the given topology, but in the generated topology, the closure of $\{2\}$ is $\{x\mid \{x\}\delta\{2\}\}=\{1,2\}$  (because $\operatorname{cl}(\{1\})=\{1,2\}$).
A: $\newcommand{\cl}{\operatorname{cl}}$Check the axioms from this answer. It’s immediate that $P_0$ and $P_2$ are satisfied. Suppose that $A\delta(B\cup C)$; then $\varnothing\ne\cl A\cap\cl(B\cup C)=\cl A\cap\big(\cl B\cup\cl C\big)=(\cl A\cap\cl B)\cup(\cl A\cap\cl C)$, so at least one of $\cl A\cap\cl B$ and $\cl A\cap\cl C$ is non-empty, and $A\delta B$ or $A\delta C$; thus, $P_1$ is satisfied.
$P_3$, however, becomes $\cl\{x\}\cap\cl\{y\}=\varnothing$ iff $x\ne y$ in this setting; for this you want $X$ to be $T_1$.
$P_4$ is also problematic. If $A\bar\delta B$, then $\cl A\cap\cl B=\varnothing$. We want to find $C,D\subseteq X$ such that $A\bar\delta C$, $B\bar\delta D$, and $C\cup D=X$, i.e., such that $\cl A\cap\cl C=\cl B\cap\cl D=\varnothing$ and $C\cup D=X$. Setting $U=X\setminus\cl C$ and $V=X\setminus\cl D$, we see that this requires finding open sets $U,V\subseteq X$ such that $\cl A\subseteq U$, $\cl B\subseteq V$, and $U\cap V=\varnothing$, so for this we want $X$ to be normal.
You do get a proximity if $X$ is $T_4$.
