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 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:

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

We claim the following:

1. $G^{\prime }(A,B)$ 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 $G^{\prime }(A,B)$ 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 $G^{\prime }(A,B)\subseteq {\bigcup }_{a\in A}G(\{a\},\overline{B})\subseteq {\bigcap }_{b\in B}G(\overline{A},\{b\})\subseteq {\bigcap }_{b\in B}\overline{G(\overline{A},\{b\})}$. The last set is closed, and does not contain any of the elements $b\in B$, so is disjoint from $B$. Thus, $G^{\prime }(A,B)$ is contained in a closed subset of $X$ that is disjoint from $B$. Intersecting with $Y$, we obtain that $G^{\prime }(A,B)$ is contained in a closed subset of $Y$ that is disjoint from $B$. Thus, the closure of $G^{\prime }(A,B)$ in $Y$ is disjoint from $B$.
3. If $A\subseteq A^{\prime }$ and $B^{\prime }\subseteq B$, with $A,B,A^{\prime },B^{\prime }$ all closed in $Y$, $A\cap B=A^{\prime }\cap B^{\prime }=\mathrm{\varnothing }$, then $G(A,B)\subseteq G(A^{\prime },B^{\prime })$: Since $B^{\prime }\subseteq B$, we have $\overline{B^{\prime }}\subseteq \overline{B}$, so $G(\{a\},\overline{B})\subseteq G(\{a\},\overline{B^{\prime }})$ for all $a\in A$. Thus, $G(A,B)\subseteq G(A,B^{\prime })$. In turn, it is clear that $G(A,B^{\prime })\subseteq G(A^{\prime },B^{\prime })$. Thus, $G(A,B)\subseteq G(A^{\prime },B^{\prime })$.