1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Topology connectedness proof