Two points $x,y$ of a topological space are said to be distinguishable if at least one has a neighborhood not containing the other, and separated if each has a neighborhood not containing the other.
It is readily seen that in an $R_0$ space (one in which distinguishable points are separated), separated points have disjoint closures. I'm curious, however, if the $R_0$ condition is necessary for separated points to have disjoint closures. Does anyone know of an example of a non-$R_0$ space in which separated points have disjoint closures? Alternately, does anyone know of a proof that non-$R_0$ spaces will necessarily have separated points with non-disjoint closures?
Edit: coffeemath cleverly pointed out a trivial counterexample in the comments below. Does anyone know of a non-trivial counterexample? In particular, a space such that:
- there exist separated points in the space,
- there exist distinguishable, non-separated points in the space, and
- separated points in the space have disjoint closures.
$\newcommand{\cl}{\operatorname{cl}}$For $n\in\Bbb Z^+$ let $U_n=\{1,\dots,n\}$, and let let $X=\Bbb N$ with the topology generated by the base
$$\{\{0\}\}\cup\{U_n:n\in\Bbb Z^+\}\;.$$
Then all pairs of points are distinguishable: if $m<n$, then $m$ has an open nbhd not containing $n$. $0$ and $1$ are isolated points and hence separated. Finally, $\cl\{0\}=\{0\}$ and $\cl\{1\}=\Bbb Z^+$, so $\{0\}$ and $\{1\}$ have disjoint closures.