Recall that a nonempty topological space $X$ is said to be totally separated iff, for every distinct points $x,y \in X$, there is a separation $U,V$ of $X$ such that $x \in U$ and $y \in V$. It can be readily seen that every such space is completely Hausdorff. As such, I tried to find stronger conditions, while associating with separation axioms.
A stronger condition involving regularity:
Definition 1. A nonempty topological space $X$ is said to be regularly separated iff, it is $T_1$ and for every closed subset $A \subset X$ and a point $x \in X \setminus A$, there is a separation $U,V$ of $X$ such that $x \in U$ and $A \subset V$.
Obviously, this implies total separatedness. The implication is strict. Deleted Tychonoff Corkscrew (Counterexamples In Topology by Steen & Seebach, Part II, Section 91.) is one counterexample. Note that this implies also complete regularity.
An even stronger condition involving normality:
Definition 2. A nonempty topological space $X$ is said to be normally separated iff, it is $T_1$ and for every disjoint closed subsets $A,B \subset X$, there is a separation $U,V$ of $X$ such that $A \subset U$ and $B \subset V$.
Obviously, this implies regular separatedness. But is this implication strict? I have a strong intuition that Sorgenfrey Plane is one counterexample. Since Sorgenfrey Plane is totally separated, completely regular, but not normal, it would suffice to prove that Sorgenfrey Plane is regularly separated. How?
Best Answer
The separations as described above are clopen. Hence a space is "totally separated", iff all quasi components are singletons. Such spaces are sometimes called totally disconnected (eg. in Engelking's book). A space is "regularly separated", iff it is zero dimensional and T1. Finally, a space is "normally separated", iff it is ultranormal.
The Sorgenfrey line is zero-dimensional (each half-open interval $[x, y[$ is clopen), products of zero-dimensional spaces are zero-dimensional, hence the Sorgenfrey Plane is "regularly separated".