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Connected sets

  1. Oct 8, 2008 #1
    If A is a connected subset of a disconnected set X s.t. X=MUN , M,N nonempty closed disjoint sets, how do we show, A is either contained in M or in N?

    I can start a proof, but then, I am kind of stuck.
    I would go by contradiction and say A intersection M is non empty and A intersection N is non empty. Hence A would be the union of 2 non empty disjoint sets. But since A is connected, A intersection M and A intersection N cannot be both open. So without loss of generality, say A intersection M is not open. Hence X\(A intersection M) is not closed.
    Then I get stuck. Any help?
  2. jcsd
  3. Oct 9, 2008 #2
    X is disconnected. That means, there are at least two nonempty disjoint open-and-closed sets. You are conveniently given these two sets. They are M and N.

    If A is connected, that means that, relative to the topology on X, there is only one non-empty disjoint open-and-closed set. Namely, A itself.

    Now, suppose that A intersects both M and N. What can you say about AnM and AnN? For both, is the set empty? Is it closed?
  4. Oct 9, 2008 #3
    I guess AnM and AnN are both clopen. But since A is connected, one of the AnM or AnN has to be empty. Say, AnM is empty. So ACN.

    Is that right?
  5. Oct 10, 2008 #4
    assume that AnM and AnN are non empty and A is connected. Since M and N are a disconnection of X they are open in X and by the subspace topolgoy A'=AnM and A''=AnN are open and disjoint in A and form a disconnection of A. contradiction QED
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