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Math Help - Connectedness

  1. #1
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    Connectedness

    1) Let (X, d) be a metric space and let A, B be two closed subsets of X such that the union and intersection of A and B are connected. Prove that A is connected.

    2) Let (X, d) be a connected metric space and let A be a connected subset of X. Assume that the complement of A is the union of two separated sets B and C. Prove that the union of A and B are connected.

    Thanks in advance!
    Last edited by h2osprey; May 5th 2009 at 08:30 AM.
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  2. #2
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    1. Suppose for contradiction that A is not connected. Then you can write it as the union of 2 non-empty separated sets.

    2. So A^C is not connected. Now suppose  A \cup B is not connected. Then you can write it as the union of 2 non empty separable sets.
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  3. #3
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    I've tried, but I didn't manage to get anywhere.
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  4. #4
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    Suppose that A is not connected. Then there is a separation of A = H \cup K.
    Where \begin{gathered}  \emptyset  \ne H \subseteq A\;\& \;\emptyset  \ne K \subseteq A \hfill \\  \overline H  \cap K = \emptyset \;\& \;\overline K  \cap H = \emptyset  \hfill \\ <br />
\end{gathered}.

    If B \cap H = \emptyset then H \cup \left( {B \cup K} \right) is a separation of A\cup B, contradiction.

    So \left( {B \cap H} \right) \cup \left( {B \cap K} \right) = B \cap A is also a separation of A\cap B.

    Now you have to do the other.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Suppose that A is not connected. Then there is a separation of A = H \cup K.
    Where \begin{gathered}  \emptyset  \ne H \subseteq A\;\& \;\emptyset  \ne K \subseteq A \hfill \\  \overline H  \cap K = \emptyset \;\& \;\overline K  \cap H = \emptyset  \hfill \\ <br />
\end{gathered}.

    If B \cap H = \emptyset then H \cup \left( {B \cup K} \right) is a separation of A\cup B, contradiction.

    So \left( {B \cap H} \right) \cup \left( {B \cap K} \right) = B \cap A is also a separation of A\cap B.

    Now you have to do the other.
    For 2) I managed to get  A = ((X\setminus B) \cap D1) \cup ((X\setminus B) \cap D2) and also, since  B \cap C = \emptyset, (X\setminus B) \cup (X\setminus C) = X
    Hence  X = (D1 \cup (X\setminus B)) \cup D2 , but I got stuck since assuming (without loss of generality) ((X\setminus B) \cap D1) is empty doesn't guarantee that (D1 \cup (X\setminus B)) \cup D2 are separated (thus drawing the contradiction since X is connected).

    Am I using the wrong approach?
    Last edited by h2osprey; May 5th 2009 at 03:14 PM.
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  6. #6
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    Quote Originally Posted by Plato View Post
    Suppose that A is not connected. Then there is a separation of A = H \cup K.
    Where \begin{gathered}  \emptyset  \ne H \subseteq A\;\& \;\emptyset  \ne K \subseteq A \hfill \\  \overline H  \cap K = \emptyset \;\& \;\overline K  \cap H = \emptyset  \hfill \\ <br />
\end{gathered}.

    If B \cap H = \emptyset then H \cup \left( {B \cup K} \right) is a separation of A\cup B, contradiction.

    So \left( {B \cap H} \right) \cup \left( {B \cap K} \right) = B \cap A is also a separation of A\cap B.

    Now you have to do the other.
    For 2) I managed to get  A = ((X\setminus B) \cap D1) \cup ((X\setminus B) \cap D2) and also, since  B \cap C = \emptyset, (X\setminus B) \cup (X\setminus C) = X
    Hence  X = (D1 \cup (X\setminus B)) \cup D2 , but I got stuck since assuming (without loss of generality) ((X\setminus B) \cap D1) is empty doesn't guarantee that (D1 \cup (X\setminus B)) \cup D2 are separated (thus drawing the contradiction since X is connected).

    Am I using the wrong approach?
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  7. #7
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    2. Suppose  A \cup B is not connected. Then  A \cup B = H \cup K for separable sets  H and  K . Now define  A_1 = H \cap A and  A_2 = K \cap A . Then  A = A_1 \cup A_2 where  A_1 and  A_2 are separable. This implies that  A is not connected. Contradiction.
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  8. #8
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    Quote Originally Posted by manjohn12 View Post
    2. Suppose  A \cup B is not connected. Then  A \cup B = H \cup K for separable sets  H and  K . Now define  A_1 = H \cap A and  A_2 = K \cap A . Then  A = A_1 \cup A_2 where  A_1 and  A_2 are separable. This implies that  A is not connected. Contradiction.
    How would you prove that  A_1 = H \cap A and  A_2 = K \cap A are non-empty?
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  9. #9
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    Suppose  A \cup B is not connected. We know that  X-A = B \cup C is a separation of  X-A . Let  A \cup B = H \cup K be a separation of  A \cup B . Then  \overline{B} \subset X-C ,  \overline{C} \subset X-B ,  \overline{H} \subset X-K and  \overline{K} \subset X-H . Now  A \cup B \cup C = X = C \cup H \cup K . Now  A is completely contained in either  H or  K . WLOG suppose  A \subset H . Then  K = H \cup K-H \subset A \cup B-A \subset B . Also  \overline{K} \subset \overline{B} \subset X-C . Then  \overline{C \cup H} = \overline{C} \cup \overline{H} \subset (X-B) \cup (X-K) = (C \cup A) \cup (C \cup H) \subset C \cup H . Also  \overline{K} \subset (X-C) \cap (X-H) = K . So  C \cup H = \overline{C \cup H} and  K = \overline{K} are closed. Hence  X = (C \cup H) \cup K which is a separation of  X . Contradiction.
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