Separated sets: $\bar{A}\cap B=A\cap \bar{B}=\mathrm{\varnothing }$

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $A,B\subseteq X$, with closures taken in $X$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space). Then $A$ and $B$ are separated when

$\bar{A}\cap B=\mathrm{\varnothing }\text{and}A\cap \bar{B}=\mathrm{\varnothing }\mathrm{.}$

Equivalently, neither set meets the closure of the other. The condition is symmetric in $A$ and $B$ by construction, and it is inherited downwards: if $A$ and $B$ are separated and $A^{\mathrm{\prime }}\subseteq A$, $B^{\mathrm{\prime }}\subseteq B$, then $A^{\mathrm{\prime }}$ and $B^{\mathrm{\prime }}$ are separated, because $A^{\mathrm{\prime }}\subseteq A$ forces $\overline{A^{\mathrm{\prime }}}\subseteq \bar{A}$, the closure $\bar{A}$ being a closed superset of $A^{\mathrm{\prime }}$ and $\overline{A^{\mathrm{\prime }}}$ the smallest such (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$; the closure is the smallest closed superset and equals $A$ together with its derived set, claim 2).

Separated sets are disjoint, and being disjoint is not enough. From $A\subseteq \bar{A}$ one gets $A\cap B\subseteq \bar{A}\cap B=\mathrm{\varnothing }$. The converse fails: in $\mathbb{R}$ with its usual topology the sets $A=(0,1)$ and $B=[1,2)$ are disjoint, yet $1\in \bar{A}\cap B$, so they are not separated.

Two sufficient conditions, both used constantly below.

1. Disjoint closed sets are separated. If $A$ and $B$ are closed and disjoint then $\bar{A}=A$ and $\bar{B}=B$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$; the closure is the smallest closed superset and equals $A$ together with its derived set, claim 2), so both displayed intersections are $A\cap B=\mathrm{\varnothing }$.
2. Disjoint open sets are separated. Let $U,V$ be open and disjoint. If $y\in V$ then $V$ is an open set containing $y$ and missing $U$, so $y\mathrm{\notin }\bar{U}$ by clause (c) of A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$; the closure is the smallest closed superset and equals $A$ together with its derived set; hence $\bar{U}\cap V=\mathrm{\varnothing }$, and symmetrically $U\cap \bar{V}=\mathrm{\varnothing }$.

Separation is absolute rather than relative to a subspace. Let $A,B\subseteq S\subseteq X$ with $S$ carrying the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then $A$ and $B$ are separated in the space $S$ if and only if they are separated in $X$. Indeed ${\mathrm{cl}}_S(A)=\bar{A}\cap S$ (For $A\subseteq S\subseteq X$ the closure of $A$ in $S$ is ${\bar{A}}^X\cap S$, while the interior only contains ${\mathrm{int}}^X(A)\cap S$, with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$, claim 1), so

${\mathrm{cl}}_S(A)\cap B=\bar{A}\cap S\cap B=\bar{A}\cap B$

because $B\subseteq S$, and symmetrically for the other intersection. So the phrase "$A$ and $B$ are separated" needs no ambient space named once both sets are fixed, and this is exactly what makes the notion the right hypothesis for complete normality later on this page.

• Why the notion is not "disjoint closures". Requiring $\bar{A}\cap \bar{B}=\mathrm{\varnothing }$ is strictly stronger, and it is too strong to be useful: in $\mathbb{R}$ the sets $(0,1)$ and $(1,2)$ are separated in the sense above, while their closures $[0,1]$ and $[1,2]$ meet. The definition asks only that each set avoid the other's closure.

• The vocabulary collides with two others and neither is meant here. "$A$ and $B$ are separated by disjoint open sets" is a different, stronger condition, and it is the conclusion of the normality and complete-normality axioms below, not the hypothesis. "Separable", meaning "has an at most countable dense subset", is unrelated and is defined later in Separability: the existence of an at most countable dense subset ↗.

• Nothing here needs a separation axiom. The definition and all four observations above hold in an arbitrary topological space, points closed or not.