Results 1 to 9 of 9

Thread: Connectedness

  1. #1
    Member
    Joined
    Apr 2008
    Posts
    123

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2008
    Posts
    156
    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 $\displaystyle A \cup B $ is not connected. Then you can write it as the union of 2 non empty separable sets.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2008
    Posts
    123
    I've tried, but I didn't manage to get anywhere.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    Suppose that $\displaystyle A$ is not connected. Then there is a separation of $\displaystyle A = H \cup K$.
    Where $\displaystyle \begin{gathered} \emptyset \ne H \subseteq A\;\& \;\emptyset \ne K \subseteq A \hfill \\ \overline H \cap K = \emptyset \;\& \;\overline K \cap H = \emptyset \hfill \\
    \end{gathered}$.

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

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

    Now you have to do the other.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2008
    Posts
    123
    Quote Originally Posted by Plato View Post
    Suppose that $\displaystyle A$ is not connected. Then there is a separation of $\displaystyle A = H \cup K$.
    Where $\displaystyle \begin{gathered} \emptyset \ne H \subseteq A\;\& \;\emptyset \ne K \subseteq A \hfill \\ \overline H \cap K = \emptyset \;\& \;\overline K \cap H = \emptyset \hfill \\
    \end{gathered}$.

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

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

    Now you have to do the other.
    For 2) I managed to get $\displaystyle A = ((X\setminus B) \cap D1) \cup ((X\setminus B) \cap D2) $ and also, since $\displaystyle B \cap C = \emptyset, (X\setminus B) \cup (X\setminus C) = X$
    Hence $\displaystyle X = (D1 \cup (X\setminus B)) \cup D2 $, but I got stuck since assuming (without loss of generality) $\displaystyle ((X\setminus B) \cap D1)$ is empty doesn't guarantee that $\displaystyle (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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Apr 2008
    Posts
    123
    Quote Originally Posted by Plato View Post
    Suppose that $\displaystyle A$ is not connected. Then there is a separation of $\displaystyle A = H \cup K$.
    Where $\displaystyle \begin{gathered} \emptyset \ne H \subseteq A\;\& \;\emptyset \ne K \subseteq A \hfill \\ \overline H \cap K = \emptyset \;\& \;\overline K \cap H = \emptyset \hfill \\
    \end{gathered}$.

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

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

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

    Am I using the wrong approach?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2008
    Posts
    156
    2. Suppose $\displaystyle A \cup B $ is not connected. Then $\displaystyle A \cup B = H \cup K $ for separable sets $\displaystyle H $ and $\displaystyle K $. Now define $\displaystyle A_1 = H \cap A $ and $\displaystyle A_2 = K \cap A $. Then $\displaystyle A = A_1 \cup A_2 $ where $\displaystyle A_1 $ and $\displaystyle A_2 $ are separable. This implies that $\displaystyle A $ is not connected. Contradiction.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Apr 2008
    Posts
    123
    Quote Originally Posted by manjohn12 View Post
    2. Suppose $\displaystyle A \cup B $ is not connected. Then $\displaystyle A \cup B = H \cup K $ for separable sets $\displaystyle H $ and $\displaystyle K $. Now define $\displaystyle A_1 = H \cap A $ and $\displaystyle A_2 = K \cap A $. Then $\displaystyle A = A_1 \cup A_2 $ where $\displaystyle A_1 $ and $\displaystyle A_2 $ are separable. This implies that $\displaystyle A $ is not connected. Contradiction.
    How would you prove that $\displaystyle A_1 = H \cap A $ and $\displaystyle A_2 = K \cap A $ are non-empty?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2008
    Posts
    156
    Suppose $\displaystyle A \cup B $ is not connected. We know that $\displaystyle X-A = B \cup C $ is a separation of $\displaystyle X-A $. Let $\displaystyle A \cup B = H \cup K $ be a separation of $\displaystyle A \cup B $. Then $\displaystyle \overline{B} \subset X-C $, $\displaystyle \overline{C} \subset X-B $, $\displaystyle \overline{H} \subset X-K $ and $\displaystyle \overline{K} \subset X-H $. Now $\displaystyle A \cup B \cup C = X = C \cup H \cup K $. Now $\displaystyle A $ is completely contained in either $\displaystyle H $ or $\displaystyle K $. WLOG suppose $\displaystyle A \subset H $. Then $\displaystyle K = H \cup K-H \subset A \cup B-A \subset B $. Also $\displaystyle \overline{K} \subset \overline{B} \subset X-C $. Then $\displaystyle \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 $\displaystyle \overline{K} \subset (X-C) \cap (X-H) = K $. So $\displaystyle C \cup H = \overline{C \cup H} $ and $\displaystyle K = \overline{K} $ are closed. Hence $\displaystyle X = (C \cup H) \cup K $ which is a separation of $\displaystyle X $. Contradiction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Path connectedness and connectedness
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Sep 19th 2011, 09:16 PM
  2. connectedness
    Posted in the Differential Geometry Forum
    Replies: 10
    Last Post: Apr 25th 2010, 07:06 PM
  3. Connectedness
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Mar 15th 2010, 09:06 AM
  4. Connectedness
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Sep 28th 2009, 07:00 AM
  5. connectedness
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: Apr 29th 2009, 04:59 PM

Search Tags


/mathhelpforum @mathhelpforum