Monotone normality is hereditary

This article gives the statement, and possibly proof, of a topological space property (i.e., monotonically normal space) satisfying a topological space metaproperty (i.e., subspace-hereditary property of topological spaces)
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Statement

Suppose $X$ is a monotonically normal space with a monotone normality operator $G$. Suppose $Y$ is a subspace of $X$. Then, $Y$ is also a monotonically normal space, with a monotone normality operator defined in terms of $G$.

Related facts

• Monotonically normal implies collectionwise normal
• Monotonically normal implies hereditarily collectionwise normal
• Monotonically normal implies hereditarily normal

Proof

Given: A monotonically normal space $X$ with monotone normality operator $G$. A subspace $Y$ of $X$.

To prove: We can define a monotone normality operator on $Y$ in terms of $G$.

Proof: Note that the part about $X$ being $T_{1}$ implying $Y$ being $T_{1}$ is immediate.

Suppose $A$ and $B$ are disjoint closed subsets of $Y$. Then, by the definition of subspace topology, we have that ${\overline {A}}\cap B$ is empty and $A\cap {\overline {B}}$ is empty.

We define the monotone normality operator on $Y$ as follows:

$G'(A,B)=Y\cap \bigcup _{a\in A}G(\{a\},{\overline {B}})$.

We claim the following:

1. $G'(A,B)$ is open in $Y$: Note first that since $X$ is $T_{1}$, the point $\{a\}$ is closed. Also, ${\overline {B}}$ is closed, and disjoint from $\{a\}$. Thus, $G(\{a\},B)$ is open for each $a\in A$. Thus, the union of all of these is open in $X$, and so, by the definition of subspace topology, the intersection with $Y$ is open in $Y$.
2. The closure of $G'(A,B)$ in $Y$ is disjoint from $B$: Consider $G({\overline {A}},\{b\})$ for $b\in B$. This is an open subset of $X$ by definition, and $G(\{a\},{\overline {B}})\subseteq G({\overline {A}},b)$ for all $a\in A,b\in B$. Thus, we obtain that $G'(A,B)\subseteq \bigcup _{a\in A}G(\{a\},{\overline {B}})\subseteq \bigcap _{b\in B}G({\overline {A}},\{b\})\subseteq \bigcap _{b\in B}{\overline {G({\overline {A}},\{b\})}}$. The last set is closed, and does not contain any of the elements $b\in B$, so is disjoint from $B$. Thus, $G'(A,B)$ is contained in a closed subset of $X$ that is disjoint from $B$. Intersecting with $Y$, we obtain that $G'(A,B)$ is contained in a closed subset of $Y$ that is disjoint from $B$. Thus, the closure of $G'(A,B)$ in $Y$ is disjoint from $B$.
3. If $A\subseteq A'$ and $B'\subseteq B$, with $A,B,A',B'$ all closed in $Y$, $A\cap B=A'\cap B'=\emptyset$, then $G(A,B)\subseteq G(A',B')$: Since $B'\subseteq B$, we have ${\overline {B'}}\subseteq {\overline {B}}$, so $G(\{a\},{\overline {B}})\subseteq G(\{a\},{\overline {B'}})$ for all $a\in A$. Thus, $G(A,B)\subseteq G(A,B')$. In turn, it is clear that $G(A,B')\subseteq G(A',B')$. Thus, $G(A,B)\subseteq G(A',B')$.