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Hereditarily normal, mutually separated subsets

  1. Dec 2, 2008 #1
    Show that the following statements are equivalent for any topological space [tex](X, \tau)[/tex].

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

    (b) [tex](X, \tau)[/tex] is hereditarily normal.

    Background:

    Definition- Sets [tex]H[/tex] and [tex]K[/tex] are mutually separated in a space [tex]X[/tex] if and only if [tex]H \cap \overline{K}[/tex] [tex]= \overline{H} \cap K = \emptyset[/tex]
     
  2. jcsd
  3. Dec 2, 2008 #2
    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 [tex]X - (\bar A \cap \bar B)[/tex].
     
  4. Mar 4, 2010 #3
    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 [tex]X[/tex] has property (a), it is said to be completely normal. [tex]X[/tex] is hereditarily normal provided that every subset of [tex]X[/tex] 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 [tex]X[/tex] is completely normal. Let [tex]S \subset X[/tex]. We need to show that [tex]S[/tex] is normal. Let [tex]A, B[/tex] be disjoint closed subsets of [tex]S[/tex]. ([tex]A, B[/tex] are not necessarily closed in [tex]X[/tex].) We must show that there are disjoint open subsets [tex]U, V[/tex] of [tex]S[/tex] with [tex]A \subset U[/tex] and [tex]B \subset V[/tex]. ([tex]U, V[/tex] are not necessarily open in [tex]X[/tex].) Certainly [tex]A = A \cap S[/tex], and it is easy to see that [tex]B = S \cap \overline{B}[/tex]. Thus, [tex]A \cap \overline{B} = A \cap S \cap \overline{B} = A \cap B = \emptyset[/tex]. Likewise, we see that [tex]\overline{A} \cap B = \emptyset[/tex]. Hence, [tex]A, B[/tex] are mutually separated sets in [tex]X[/tex]. Since [tex]X[/tex] is completely normal, there are disjoint open sets [tex]O, P[/tex] in [tex]X[/tex] with [tex]A \subset O[/tex] and [tex]B \subset P[/tex]. Define [tex]U = O \cap S [/tex] and [tex]V = P \cap S [/tex]. [tex]U, V[/tex] are clearly disjoint open subsets of [tex]S[/tex] with [tex]A \subset U[/tex] and [tex]B \subset V[/tex].
     
  5. Mar 4, 2010 #4
    Here is part 2:

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