Monotonically normal implies normal

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., monotonically normal space) must also satisfy the second topological space property (i.e., normal space)
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Statement

Any monotonically normal space is a normal space.

Definitions used

Monotonically normal space

Further information: monotonically normal space

A topological space $X$ is termed monotonically normal if it is a $T_1$ space and there exists an operator (called a monotone normality operator) $G$ from pairs of disjoint closed subsets $(A,B)$ to open subsets such that:

• $G(A,B)$ is an open subset containing $A$ and whose closure is disjoint from $B$.
• If $A\subseteq A^{\prime }$, $B^{\prime }\subseteq B$, with $A,B$ disjoint and $A^{\prime },B^{\prime }$ disjoint, we have $G(A,B)\subseteq G(A^{\prime },B^{\prime })$.

Normal space

Further information: normal space

A topological space $X$ is termed normal if it is a $T_1$ space and, for any two disjoint closed subsets $A$ and $B$, there exist disjoint open subsets $U$ and $V$ such that $U$ contains $A$ and $V$ contains $B$.

Proof

Given: A monotonically normal space $X$ with a monotone normality operator $G$.

To prove: $X$ is a normal space.

Proof: The $T_1$-space part follows by definition. For the other part, note that $U=G(A,B)$ and $V=X\setminus \overline{G(A,B)}$ are disjoint open subsets containing $A$ and $B$ respectively.