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Topology connectedness proof

  1. Jan 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Let A, B be closed non-empty subsets of a topological space X with [tex] A \cup B [/tex] and [tex] A \cap B [/tex] connected.

    Prove that A and B are connected.


    2. Relevant equations

    A set Q is not connected (disconnected) if it is expressible as a disjoint union of open sets, [tex] Q = S \cup T [/tex]


    3. The attempt at a solution

    I'm trying a proof by contradiction.
    By the above definition, a set which is not connected must be open (is this really true?). So start by assuming A is disconnected, ie [tex] A = C \cup D [/tex] for C, D open and disjoint. Then A must be open, but should also be closed. Now consider [tex] A \cup B = C \cup D \cup B [/tex] and [tex] A \cap B = C \cup D \cap B [/tex]. I want to to arrive at a contradiction. The given properties are that A is both open and closed, B is closed, C and D are open and disjoint and [tex] A \cup B = C \cup D \cup B [/tex] and [tex] A \cap B = C \cup D \cap B [/tex] are connected. This all seems very complicated.
     
  2. jcsd
  3. Jan 4, 2010 #2
    No, thankfully this isn't true. The set [tex][0,1] \cup [2,3][/tex] is disconnected and closed. The definition of connectedness requires S and T to be relatively open; that is, open in the subspace topology induced on Q. Keeping this in mind, an equivalent definition of disconnectedness of a subspace Q is that there exist disjoint open sets S and T such that [tex]Q \subset S \cup T[/tex] and [tex]Q \cap S \neq \emptyset \neq Q \cap T[/tex]. The subset relation shere removes your concern about disconnected sets being open.
     
    Last edited: Jan 4, 2010
  4. Jan 4, 2010 #3
    Ah right instead of saying "a set is not connected if..." i should have said "a topological space is not connected if...". Hmm does this mean the rest of my reasoning was wrong? I have to replace my C's and D's with intersections over X.
     
  5. Jan 4, 2010 #4
    Ok so thanks to the push in the right direction I seem to have solved it. However I didnt use the fact that A and B or closed or that [tex] A \cup B [/tex] is connected. It was done entirely on the fact that [tex] A \cap B [/tex] is connected. Could this be right?
     
    Last edited: Jan 4, 2010
  6. Jan 4, 2010 #5
    No, that's certainly not true. A counterexample is given by [tex]A = (1,2) \cup (2,3)[/tex] and [tex]B = [2,3)[/tex]. Here [tex]A \cup B = (1, 3)[/tex] and [tex]A \cap B = (2, 3)[/tex] are connected, but A is not connected. Also, you should try to think of a very simple example that shows that neither A nor B must be connected even it [tex]A \cap B[/tex] is.
     
    Last edited: Jan 4, 2010
  7. Jan 5, 2010 #6
    Yeah it took me a while to realise what I missed. I showed that if A is disconnected then [tex] A \cap B [/tex] is expressible as a union of disjoint open sets, but I need to show that they are non-empty sets.
     
  8. Jan 5, 2010 #7
    Yup. Generally homework-style questions don't give you more assumptions than you need, so if you haven't used some/most of your assumptions, you should examine your proof. This heuristic will take you far in math.
     
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