Monotone normality is hereditary

From Topospaces
Jump to: navigation, search
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)
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
Get more facts about monotonically normal space |Get facts that use property satisfaction of monotonically normal space | Get facts that use property satisfaction of monotonically normal space|Get more facts about subspace-hereditary property of topological spaces

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

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').