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Property of topological spaces stronger than normality
In mathematics, specifically in the field of topology, a monotonically normal space is a particular kind of normal space, defined in terms of a monotone normality operator. It satisfies some interesting properties; for example metric spaces and linearly ordered spaces are monotonically normal, and every monotonically normal space is hereditarily normal.
A topological space X {\displaystyle X} is called monotonically normal if it satisfies any of the following equivalent definitions:[1][2][3][4]
The space X {\displaystyle X} is T1 and there is a function G {\displaystyle G} that assigns to each ordered pair ( A , B ) {\displaystyle (A,B)} of disjoint closed sets in X {\displaystyle X} an open set G ( A , B ) {\displaystyle G(A,B)} such that:
Condition (i) says X {\displaystyle X} is a normal space, as witnessed by the function G {\displaystyle G} . Condition (ii) says that G ( A , B ) {\displaystyle G(A,B)} varies in a monotone fashion, hence the terminology monotonically normal. The operator G {\displaystyle G} is called a monotone normality operator.
One can always choose G {\displaystyle G} to satisfy the property
by replacing each G ( A , B ) {\displaystyle G(A,B)} by G ( A , B ) ∖ G ( B , A ) ¯ {\displaystyle G(A,B)\setminus {\overline {G(B,A)}}} .
The space X {\displaystyle X} is T1 and there is a function G {\displaystyle G} that assigns to each ordered pair ( A , B ) {\displaystyle (A,B)} of separated sets in X {\displaystyle X} (that is, such that A ∩ B ¯ = B ∩ A ¯ = ∅ {\displaystyle A\cap {\overline {B}}=B\cap {\overline {A}}=\emptyset } ) an open set G ( A , B ) {\displaystyle G(A,B)} satisfying the same conditions (i) and (ii) of Definition 1.
The space X {\displaystyle X} is T1 and there is a function μ {\displaystyle \mu } that assigns to each pair ( x , U ) {\displaystyle (x,U)} with U {\displaystyle U} open in X {\displaystyle X} and x ∈ U {\displaystyle x\in U} an open set μ ( x , U ) {\displaystyle \mu (x,U)} such that:
Such a function μ {\displaystyle \mu } automatically satisfies
(Reason: Suppose y ∈ X ∖ U {\displaystyle y\in X\setminus U} . Since X {\displaystyle X} is T1, there is an open neighborhood V {\displaystyle V} of y {\displaystyle y} such that x ∉ V {\displaystyle x\notin V} . By condition (ii), μ ( x , U ) ∩ μ ( y , V ) = ∅ {\displaystyle \mu (x,U)\cap \mu (y,V)=\emptyset } , that is, μ ( y , V ) {\displaystyle \mu (y,V)} is a neighborhood of y {\displaystyle y} disjoint from μ ( x , U ) {\displaystyle \mu (x,U)} . So y ∉ μ ( x , U ) ¯ {\displaystyle y\notin {\overline {\mu (x,U)}}} .)[5]
Let B {\displaystyle {\mathcal {B}}} be a base for the topology of X {\displaystyle X} . The space X {\displaystyle X} is T1 and there is a function μ {\displaystyle \mu } that assigns to each pair ( x , U ) {\displaystyle (x,U)} with U ∈ B {\displaystyle U\in {\mathcal {B}}} and x ∈ U {\displaystyle x\in U} an open set μ ( x , U ) {\displaystyle \mu (x,U)} satisfying the same conditions (i) and (ii) of Definition 3.
The space X {\displaystyle X} is T1 and there is a function μ {\displaystyle \mu } that assigns to each pair ( x , U ) {\displaystyle (x,U)} with U {\displaystyle U} open in X {\displaystyle X} and x ∈ U {\displaystyle x\in U} an open set μ ( x , U ) {\displaystyle \mu (x,U)} such that:
Such a function μ {\displaystyle \mu } automatically satisfies all conditions of Definition 3.
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