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Math Help - Normal space and Tietze/Urysohn

  1. #1
    CSM
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    Lightbulb Normal space and Tietze/Urysohn

    I’m having a really hard time with this one. On a conceptual level and on a technical level. I’ve got my three topology books here, but I can’t seem to make anything out of it. Can somebody enlighten me?

    Let T be a topological space. And let T be normal and \{C_1, C_2,…,\} a countable family of closed subsets of T. Suppose that every point of T has an neighbourhood U such that U\cap C_i\not=\emptyset for at most one C_i. (Note in perticular that C_i\cap C_j=\emptyset when i\not= j.)

    Prove: there are open sets U_1,U_2,… such that C_i\subset U_i for each i and such that \overline{U_i}\cap\overline{U_j}=\emptyset when i\not= j
    (Use Tietze Extension Theorem or Urysohn’s lemma)
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  2. #2
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    Quote Originally Posted by CSM View Post
    Iím having a really hard time with this one. On a conceptual level and on a technical level. Iíve got my three topology books here, but I canít seem to make anything out of it. Can somebody enlighten me?

    Let T be a topological space. And let T be normal and \{C_1, C_2,Ö,\} a countable family of closed subsets of T. Suppose that every point of T has an neighbourhood U such that U\cap C_i\not=\emptyset for at most one C_i. (Note in perticular that C_i\cap C_j=\emptyset when i\not= j.)

    Prove: there are open sets U_1,U_2,Ö such that C_i\subset U_i for each i and such that \overline{U_i}\cap\overline{U_j}=\emptyset when i\not= j
    (Use Tietze Extension Theorem or Urysohnís lemma)
    Each x\in C_1 has an open neighbourhood U_x with U_x\cap C_n = \varnothing for all n>1. Let V_1 = \textstyle\bigcup_{x\in C_1} U_x, and use Urysohn or Tietze to get an open set U_1 with C_1\subseteq U_1\subseteq\overline{U_1}\subseteq V_1.

    Now use induction. Suppose that open sets U_1,\ldots,U_n with disjoint closures have been chosen so that C_k\subseteq U_k and \overline{U_k}\cap C_r = \varnothing for 1\leqslant k\leqslant n and for all r>n. Show that each x\in C_{n+1} has an open neighbourhood U_x disjoint from C_r for all r>n, and also disjoint from \overline{U_k} (1\leqslant k\leqslant n). As for the case n=1, define open sets

    C_{n+1}\subseteq U_{n+1}\subseteq\overline{U_{n+1}}\subseteq V_{n+1} = \bigcup_{x\in C_{n+1}} U_x,

    and check that U_{n+1} has the right properties to continue the induction.
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  3. #3
    CSM
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    Quote Originally Posted by Opalg View Post
    Each x\in C_1 has an open neighbourhood U_x with U_x\cap C_n = \varnothing for all n>1. Let V_1 = \textstyle\bigcup_{x\in C_1} U_x, and use Urysohn or Tietze to get an open set U_1 with C_1\subseteq U_1\subseteq\overline{U_1}\subseteq V_1.

    Now use induction. Suppose that open sets U_1,\ldots,U_n with disjoint closures have been chosen so that C_k\subseteq U_k and \overline{U_k}\cap C_r = \varnothing for 1\leqslant k\leqslant n and for all r>n. Show that each x\in C_{n+1} has an open neighbourhood U_x disjoint from C_r for all r>n, and also disjoint from \overline{U_k} (1\leqslant k\leqslant n). As for the case n=1, define open sets

    C_{n+1}\subseteq U_{n+1}\subseteq\overline{U_{n+1}}\subseteq V_{n+1} = \bigcup_{x\in C_{n+1}} U_x,

    and check that U_{n+1} has the right properties to continue the induction.
    Thank you very much for explaining me.

    But I can't grasp it totally.
    In you first sentence you tell me to use Urysohn/Tietze to get an open set U_1 with C_1\subseteq U_1\subseteq\overline{U_1}\subseteq V_1.
    How is Tietze/Urysohn used there?

    BTW. I've got a lemma + proof in my book that says: A topological space X is normal if and only if for each closed subset E of X and each open set W containing E there exists an open set U containing E such that \overline{U}\subset W.
    Isn't that just what I need for getting that set U_1

    I thought I had to do something with functions and inverse images.... by the hint...?

    I really appreciate the time you take to help me.
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  4. #4
    CSM
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    I'd really like to understand the way you tried to solve this, nevertheless I'll show another attempt at solving it (helped by others):


    \bullet Let C=\bigcup_n C_n.
    Then C is closed. Because if you take a x not in C, we'll find a neighbourhood O_x of x intersecting at most one C_n, say C_k. Now O_x\backslash C_k is open, it contains x (for x is not in C_k because it is not in C) and misses all C_n en thus C).

    \bullet Each C_n is still closed in C, (...), and for defining a continuous function, we only need to define a continuous function f_n on every C_n and combine them ( f will be continuous again).

    \bullet Define f_n on C_n as being constant n. Extend f to X (Tietze) and use the f^{-1}(n-\frac{1}{3},n+\frac{1}{3}), n\in N as the required U_n.

    - f(C_1)=1, f(C_2)=2,... etc. So f^{-1}(\frac{2}{3},1\frac{1}{3})=C_1 and f^{-1}(1\frac{2}{3},2\frac{1}{3})=C_2 so it easy to show that their closures are disjoint.
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  5. #5
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    That looks like a very good way to do the question.
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