Monotonically normal space

The space ${\displaystyle X}$ is T₁ and there is a function ${\displaystyle G}$ that assigns to each ordered pair ${\displaystyle (A,B)}$ of disjoint closed sets in ${\displaystyle X}$ an open set ${\displaystyle G(A,B)}$ such that:

(i) ${\displaystyle A\subseteq G(A,B)\subseteq {\overline {G(A,B)}}\subseteq X\setminus B}$;

(ii) ${\displaystyle G(A,B)\subseteq G(A',B')}$ whenever ${\displaystyle A\subseteq A'}$ and ${\displaystyle B'\subseteq B}$.

Condition (i) says ${\displaystyle X}$ is a normal space, as witnessed by the function ${\displaystyle G}$. Condition (ii) says that ${\displaystyle G(A,B)}$ varies in a monotone fashion, hence the terminology monotonically normal. The operator ${\displaystyle G}$ is called a monotone normality operator.

One can always choose ${\displaystyle G}$ to satisfy the property

${\displaystyle G(A,B)\cap G(B,A)=\emptyset }$,

by replacing each ${\displaystyle G(A,B)}$ by ${\displaystyle G(A,B)\setminus {\overline {G(B,A)}}}$.

The space ${\displaystyle X}$ is T₁ and there is a function ${\displaystyle G}$ that assigns to each ordered pair ${\displaystyle (A,B)}$ of separated sets in ${\displaystyle X}$ (that is, such that ${\displaystyle A\cap {\overline {B}}=B\cap {\overline {A}}=\emptyset }$) an open set ${\displaystyle G(A,B)}$ satisfying the same conditions (i) and (ii) of Definition 1.

The space ${\displaystyle X}$ is T₁ and there is a function ${\displaystyle \mu }$ that assigns to each pair ${\displaystyle (x,U)}$ with ${\displaystyle U}$ open in ${\displaystyle X}$ and ${\displaystyle x\in U}$ an open set ${\displaystyle \mu (x,U)}$ such that:

(i) ${\displaystyle x\in \mu (x,U)}$;

(ii) if ${\displaystyle \mu (x,U)\cap \mu (y,V)\neq \emptyset }$, then ${\displaystyle x\in V}$ or ${\displaystyle y\in U}$.

Such a function ${\displaystyle \mu }$ automatically satisfies

${\displaystyle x\in \mu (x,U)\subseteq {\overline {\mu (x,U)}}\subseteq U}$.

(Reason: Suppose ${\displaystyle y\in X\setminus U}$. Since ${\displaystyle X}$ is T₁, there is an open neighborhood ${\displaystyle V}$ of ${\displaystyle y}$ such that ${\displaystyle x\notin V}$. By condition (ii), ${\displaystyle \mu (x,U)\cap \mu (y,V)=\emptyset }$, that is, ${\displaystyle \mu (y,V)}$ is a neighborhood of ${\displaystyle y}$ disjoint from ${\displaystyle \mu (x,U)}$. So ${\displaystyle y\notin {\overline {\mu (x,U)}}}$.)[5]

Let ${\displaystyle {\mathcal {B}}}$ be a base for the topology of ${\displaystyle X}$. The space ${\displaystyle X}$ is T₁ and there is a function ${\displaystyle \mu }$ that assigns to each pair ${\displaystyle (x,U)}$ with ${\displaystyle U\in {\mathcal {B}}}$ and ${\displaystyle x\in U}$ an open set ${\displaystyle \mu (x,U)}$ satisfying the same conditions (i) and (ii) of Definition 3.

The space ${\displaystyle X}$ is T₁ and there is a function ${\displaystyle \mu }$ that assigns to each pair ${\displaystyle (x,U)}$ with ${\displaystyle U}$ open in ${\displaystyle X}$ and ${\displaystyle x\in U}$ an open set ${\displaystyle \mu (x,U)}$ such that:

(i) ${\displaystyle x\in \mu (x,U)}$;

(ii) if ${\displaystyle U}$ and ${\displaystyle V}$ are open and ${\displaystyle x\in U\subseteq V}$, then ${\displaystyle \mu (x,U)\subseteq \mu (x,V)}$;

(iii) if ${\displaystyle x}$ and ${\displaystyle y}$ are distinct points, then ${\displaystyle \mu (x,X\setminus \{y\})\cap \mu (y,X\setminus \{x\})=\emptyset }$.

Such a function ${\displaystyle \mu }$ automatically satisfies all conditions of Definition 3.