# 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 $G$.

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