# Hereditarily normal, mutually separated subsets

1. Dec 2, 2008

### 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 $$H \cap \overline{K}$$ $$= \overline{H} \cap K = \emptyset$$

2. Dec 2, 2008

### 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 $$X - (\bar A \cap \bar B)$$.

3. Mar 4, 2010

### 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 $$B \subset V$$.

4. Mar 4, 2010

### 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 $$B \subset V$$.

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