Monotone normality is hereditary

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, $$G(A,B) \subseteq G(A',B')$$.