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Connected spaces in topology

  1. Jan 6, 2010 #1
    I have a question here and i'm not sure what to do as it always confuses me, any help?

    Let A,B be closed non-empty subsets of a topological space X with AuB and AnB connected.

    (i) Prove that A and B are connected.

    (ii) Construct disjoint non-empty disconnected subspaces A,B c R such that AuB is connected.
     
  2. jcsd
  3. Jan 6, 2010 #2
    Well suppose A is disconnected. Then we can find a separation of A:
    [tex]A = C \cup D[/tex]
    for disjoint, non-empty, closed sets C,D. Now since
    [tex]A \cap B = (C \cap B) \cup (D \cap B)[/tex]
    is connected and both [itex]C \cap B[/itex] and [itex]D \cap B[/itex] are closed we can assume without loss of generality that [itex]C \cap B = \emptyset[/itex] (otherwise we would have a separation of [itex]A \cap B[/itex]). Now we have that B and D are both disjoint from C, so [itex]B \cup D[/itex] is disjoint from C which gives us a separation:
    [tex]A \cup B = (B \cup D) \cup C[/tex]
    which is a contradiction.

    ii)
    Just consider what it means to be disconnected and connected. For instance,
    [tex]A = (0,2) \cup (2,3) \qquad B = (0,1)\cup (1,3)[/tex]
    [tex]A \cup B = (0,3)[/tex]
     
  4. Jan 7, 2010 #3
    Shouldn't you be considering sets which are open or closed with respect to each subspace topology? And your sets in (ii) are not disjoint.
     
  5. Jan 7, 2010 #4
    What set are you referring to. Remember that if Y is a subspace of X, then A is open in Y if and only if [itex]A \cap Y[/itex] is open in X, so if [itex]A \subseteq Y[/itex] then A is open in Y if and only if A is open in X.

    About (ii) you're right I must have misread the question as I seem to have striven for open sets. In that case take A=(0,1) U (1,2) and B={1,2}.
     
  6. Jan 7, 2010 #5
    You would be right if Y was closed or if your set didn't contain the boundary points. But the sets here are closed and necessarily contain the boundary points. An obvious counterexample is [-1,1], which has [-1,0) as an open set which is not open in R.
     
  7. Jan 7, 2010 #6
    Yes, you're right I spoke too quickly without thinking however I still don't see where this affects my answer. I only ever use the correct direction: That if [itex]A \subseteq Y[/itex] and A is open in X, then A is open in Y since [itex]A=Y\cap A[/itex]. The same holds true for closed sets. This shows A and B are closed in AUB.

    If A is closed in the closed subspace Y of X, then A can be written as [itex]A = B\cap Y[/itex] with B closed in X and therefore A is closed. This shows that closedness is transitive.

    C and D are a separation of A so C and D are closed in A, then [itex]C \cap A \cap B = C \cap B[/itex] is closed in [itex]A \cap B[/itex].

    From this we see C is closed in A and therefore also in A U B. In the same way we see D is closed in AUB. Thus [itex]B \cap D[/itex] is closed in AUB so our separation is valid.
     
  8. Jan 7, 2010 #7
    When you say transitive, do you mean reflexive? I've been trying this problem by explicity using the intersections for the open sets of each subspace topology. I came to a problem in showing that the result of plugging my separation of A into AUB satisfied the definition of AUB being disconnected. However I think the first part of your post solved my problem.
     
  9. Jan 7, 2010 #8
    I mean transitive (when I said closedness I meant the relation "is closed in"). In the sense that if we write A < B if A is closed in B and A is a subspace of B, then we have:
    A < B and B < C imply A < C.
     
  10. Jan 7, 2010 #9
    Yes but what you have written is that
    1) if A is contained in Y and contained in X, then A is closed in Y
    2) if A is closed in Y and Y is closed in X, then A is closed in X

    Ok I just saw what you meant there. Never mind I get it now.
     
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