Hereditarily normal, mutually separated subsets

[mathsss2]

Show that the following statements are equivalent for any topological space (X, \tau).

(a) Whenever A, B are mutually separated subsets of X, there exist open disjoint U, V such that A \subseteq U and B \subseteq V.

(b) (X, \tau) is hereditarily normal.

Background:

Definition- Sets H and K are mutually separated in a space X if and only if H \cap \overline{K} = \overline{H} \cap K = \emptyset

 

[adriank]

What have you tried?

For others: hereditarily normal is apparently another name for completely normal.

A hint from my textbook (Munkres): if X is hereditarily normal, consider X - (\bar A \cap \bar B).

 

[rickhev]

I am breaking this reply into two parts because I get a "Database Error" when I combine them and do a "Preview Post."

Here is part 1:

If X has property (a), it is said to be completely normal. X is hereditarily normal provided that every subset of X is normal. That (a) implies (b) is relatively straightforward. That (b) implies (a) is trickier--for that we will use the hint from Munkres.

Suppose that X is completely normal. Let S \subset X. We need to show that S is normal. Let A, B be disjoint closed subsets of S. (A, B are not necessarily closed in X.) We must show that there are disjoint open subsets U, V of S with A \subset U and B \subset V. (U, V are not necessarily open in X.) Certainly A = A \cap S, and it is easy to see that B = S \cap \overline{B}. Thus, A \cap \overline{B} = A \cap S \cap \overline{B} = A \cap B = \emptyset. Likewise, we see that \overline{A} \cap B = \emptyset. Hence, A, B are mutually separated sets in X. Since X is completely normal, there are disjoint open sets O, P in X with A \subset O and B \subset P. Define U = O \cap S and V = P \cap S. U, V are clearly disjoint open subsets of S with A \subset U and B \subset V.

 

[rickhev]

Here is part 2:

Now suppose that X is hereditarily normal. For any set Y \subset X we define \tilde{Y} = X - Y. Let A, B be mutually separated sets in X. We must show that there are disjoint open sets U, V in X with A \subset U and B \subset V. Clearly, A \subset \tilde{\overline{B}} and B \subset \tilde{\overline{A}}. Define S = \tilde{\overline{A}} \cup \tilde{\overline{B}} (Munkres). We see that A, B \subset S. Now define C to be the closure of A in S and D to be the closure of B in S. It is easy to see that C = \overline{A} \cap S, whence C = \overline{A} \cap (\tilde{\overline{A}} \cup \tilde{\overline{B}}) = \overline{A} \cap \tilde{\overline{B}}. Likewise, D = \overline{B} \cap \tilde{\overline{A}}. Therefore, C \cap D = \emptyset. Since X is hereditarily normal, S must be normal. Thus, since C, D are disjoint closed subsets of S, there are disjoint open subsets U, V of S with C \subset U and D \subset V. But S is open in X, so U, V are also open in X. Clearly, A \subset U and B \subset V.