Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.


    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].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook