Results 1 to 3 of 3

Math Help - Union of connected subsets

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    17

    Union of connected subsets

    Problem: Let A_{1}, A_{2}, and A_{3} be connected subsets of X, such that A_{1} \cap A_{2} \neq \emptyset and A_{2} \cap A_{3} \neq \emptyset. Prove that A_{1} \cup A_{2} \cup A_{3} is connected.

    So far this is what I have:

    Assume A_{1} \cup A_{2} \cup A_{3} is disconnected, then (A_{1} \cup A_{2} \cup A_{3}) \cap U \neq \emptyset, (A_{1} \cup A_{2} \cup A_{3}) \cap V \neq \emptyset, (A_{1} \cup A_{2} \cup A_{3}) \cap U \cap V = \emptyset.

    Edit: Forgot to include (A_{1} \cup A_{2} \cup A_{3}) \subseteq (U \cup V)

    Therefore (A_{1} \cap U \cap V) \cup (A_{2} \cap U \cap V) \cup (A_{3} \cap U \cap V) = \emptyset, (A_{1} \cap U) \cup (A_{2} \cap U) \cup (A_{3} \cap U) \neq \emptyset, (A_{1} \cap V) \cup (A_{2} \cap V) \cup (A_{3} \cap V) \neq \emptyset

    Thus (A_{1} \cap U \cap V) = (A_{2} \cap U \cap V) = (A_{3} \cap U \cap V) = \emptyset

    Now, I'm not 100% sure if I have been approaching this correctly, but I think from here I have to show for one of the 3 connected subsets its intersection with U and Y is nonempty and thus it is disconnected meaning there is a contradiction, but I am not sure how to prove this part.
    Last edited by okor; December 5th 2010 at 07:21 PM. Reason: Forgot to include something
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by okor View Post
    Problem: Let A_{1}, A_{2}, and A_{3} be connected subsets of X, such that A_{1} \cap A_{2} \neq \emptyset and A_{2} \cap A_{3} \neq \emptyset. Prove that A_{1} \cup A_{2} \cup A_{3} is connected.

    So far this is what I have:

    Assume A_{1} \cup A_{2} \cup A_{3} is disconnected, then (A_{1} \cup A_{2} \cup A_{3}) \cap U \neq \emptyset, (A_{1} \cup A_{2} \cup A_{3}) \cap V \neq \emptyset, (A_{1} \cup A_{2} \cup A_{3}) \cap U \cap V = \emptyset.

    Therefore (A_{1} \cap U \cap V) \cup (A_{2} \cap U \cap V) \cup (A_{3} \cap U \cap V) = \emptyset, (A_{1} \cap U) \cup (A_{2} \cap U) \cup (A_{3} \cap U) \neq \emptyset, (A_{1} \cap V) \cup (A_{2} \cap V) \cup (A_{3} \cap V) \neq \emptyset

    Thus (A_{1} \cap U \cap V) = (A_{2} \cap U \cap V) = (A_{3} \cap U \cap V) = \emptyset

    Now, I'm not 100% sure if I have been approaching this correctly, but I think from here I have to show for one of the 3 connected subsets its intersection with U and Y is nonempty and thus it is disconnected meaning there is a contradiction, but I am not sure how to prove this part.
    I think you're misunderstanding the definitions at hand. Let A_1\cup A_2\cup A_3 be given the subspace topology. Assume that A_1\cup A_2\cup A_3=U\cup V where U,V are open and disjoint. We may assume WLOG that A_1\cap U\ne \varnothing. Argue then that A_1\subseteq U otherwise \left(A_1\cap U\right)\cup\left(A_1\cap V\right) is a disconnnection of A. Then, argue that A_2\subseteq U since otherwise A_2\cap V is non-empty and open in A_2 and since A_1\subseteq U and A_1\cap A_2\ne\varnothing then A_1\cap U\ne\varnothing and so \left(A_2\cap U\right)\cup\left(A_2\cap V\right) is a disconnection of A_2, thus A_2\subseteq U. Apply the exact same argument to show that A_3\subseteq U and thus V=\varnothing. Conclude that no disconnection of A_1\cup A_2\cup A_3 exists.


    Remark The same methodology (using induction) shows that if \{A_n\} is a sequence of connected subspaces of some topological space for which A_n\cap A_{n+1}\ne\varnothing then \displaystyle \bigcup_{n\in\mathbb{N}}A_n is connnected.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2010
    Posts
    17
    Quote Originally Posted by Drexel28 View Post
    I think you're misunderstanding the definitions at hand. Let A_1\cup A_2\cup A_3 be given the subspace topology. Assume that A_1\cup A_2\cup A_3=U\cup V where U,V are open and disjoint. We may assume WLOG that A_1\cap U\ne \varnothing. Argue then that A_1\subseteq U otherwise \left(A_1\cap U\right)\cup\left(A_1\cap V\right) is a disconnnection of A. Then, argue that A_2\subseteq U since otherwise A_2\cap V is non-empty and open in A_2 and since A_1\subseteq U and A_1\cap A_2\ne\varnothing then A_1\cap U\ne\varnothing and so \left(A_2\cap U\right)\cup\left(A_2\cap V\right) is a disconnection of A_2, thus A_2\subseteq U. Apply the exact same argument to show that A_3\subseteq U and thus V=\varnothing. Conclude that no disconnection of A_1\cup A_2\cup A_3 exists.


    Remark The same methodology (using induction) shows that if \{A_n\} is a sequence of connected subspaces of some topological space for which A_n\cap A_{n+1}\ne\varnothing then \displaystyle \bigcup_{n\in\mathbb{N}}A_n is connnected.
    I understood the definitions, but following through with your reasoning, I see where I approached it wrong. I forgot to include the (A_{1} \cup A_{2} \cup A_{3}) \subseteq (U \cup V) condition, which is from where I should have started. I knew I needed to pull in the intersection between A_{1} and A_{2} and the intersection between A_{2} and A_{3}, but I wasn't sure where, which your explanation clears up.

    Thank you for your assistance.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The union of non-disjoint connected is connected
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: September 17th 2010, 04:06 AM
  2. Union of subsets
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: December 6th 2009, 08:51 PM
  3. Union of Connected Sets
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: October 17th 2009, 01:28 AM
  4. Is a set the union of its finite subsets?
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: October 13th 2009, 02:09 PM
  5. Proof that union of two connected non disjoint sets is connected
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 27th 2009, 09:22 AM

Search Tags


/mathhelpforum @mathhelpforum