Results 1 to 6 of 6

Math Help - Connected sets

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    7

    Connected sets

    Let A be connected subset of X and let AB ⊂ cl(A). Show that B is connected and hence, in particuar, cl(A) is connected.

    Hint: (Use) Let G∪H be a disconnection of A and let B be a connected subset of A then we see that either BH= or B∩G=∅, and so either BG or B⊂H.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,801
    Thanks
    1691
    Awards
    1
    Quote Originally Posted by AcCeylan View Post
    Let A be connected subset of X and let AB ⊂ cl(A). Show that B is connected and hence, in particuar, cl(A) is connected.
    Suppose that G\cup H=B is a disconnection of B.

    That means that G\cap B\ne\emptyset~\&~H\cap B\ne\emptyset.

    Also G\cap\overline{H}=\emptyset~\&~H\cap\overline{G}=\  emptyset

    Using the fact that B\subseteq\overline{A} prove that contradicts the given that A is connected.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    7
    Do you mean that I must show cl(A) is disconnected and this will imply A is also disconnected. hence,the result will give a contradiction..? am I right?
    if not how can I do this , can you explain, please thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,801
    Thanks
    1691
    Awards
    1
    Quote Originally Posted by AcCeylan View Post
    Do you mean that I must show cl(A) is disconnected and this will imply A is also disconnected. hence,the result will give a contradiction..? am I right?
    THEOREM The closure of a connected set is a connect set.
    That is what you are really asked to prove.

    As a lemma (previous theorem) you should have proved that if A is a connected set and A\subset G\cup H where G~\&~H are separated sets then  A\subset G or  A\subset H .
    Two set are said to be separated if neither is empty and neither contains a point nor a limit of the other.

    Now it is quite easy to prove the lemma. If the conclusion is false then (A\cap G)\cup(A\cap H)=A is a separation of A.
    But A is connected. So that is a contradiction.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2011
    Posts
    7
    Thanks for everything but I have one more question..you said that
    "Using the fact that B ⊂ cl(A) prove that contradicts the given that A is connected"
    How can I show this?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,801
    Thanks
    1691
    Awards
    1
    Quote Originally Posted by AcCeylan View Post
    Thanks for everything but I have one more question..you said that
    "Using the fact that B ⊂ cl(A) prove that contradicts the given that A is connected"
    How can I show this?
    Suppose that G\cup H=B is a separation of B.
    Because A\subseteq B and A is connected we have A\subseteq G or A\subseteq B.
    Say A\subseteq G. Now is it possible for H to be nonempty?
    If the answer is no, then you have a contradiction. And that means that B is connected.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Connected sets
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 25th 2009, 03:36 PM
  2. Show sets not connected
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 27th 2009, 09:05 PM
  3. Proof that union of two connected non disjoint sets is connected
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 27th 2009, 08:22 AM
  4. Connected sets
    Posted in the Calculus Forum
    Replies: 0
    Last Post: December 11th 2008, 07:12 AM
  5. connected sets
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 26th 2008, 09:26 AM

Search Tags


/mathhelpforum @mathhelpforum