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Math Help - connectedness

  1. #1
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    connectedness

    show if A and B are connected in (M,d) with AnB!=empty set, AUB is also connected.

    that just seems obvious.....how would I go about proving this one?
    thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    show if A and B are connected in (M,d) with AnB!=empty set, AUB is also connected.

    that just seems obvious.....how would I go about proving this one?
    thanks
    Let's prove it in general. Let X be a topological space and \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} be a collection of connected subspaces of X such that \bigcap_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothi  ng. Then, \bigcup_{\alpha\in\mathcal{A}}U_\alpha=\Omega is connected. To see this, suppose not and that (E\cap \Omega)\cup (G\cap\Omega)=\Omega is a separation with E,G\subseteq X open. Then, since each E\cap \Omega,G\cap \Omega\ne\varnothing we must have that a part of some U_{\alpha_1} and U_{\alpha_2} are in each. But, U_{\alpha_1}\cap U_{\alpha_2}\ne \varnothing this in particular implies that E\cap U_{\alpha_1},E\cap U_{\alpha_2}\ne\varnothing they are open in U_{\alpha_1} and evidently (E\cap U_{\alpha_1})\cup (G\cap U_{\alpha_1})=U_{\alpha_1} but this clearly contradicts that U_{\alpha_1} is connected. The conclusion follows.
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  3. #3
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    I don't quite understand the proof, sorry

    Why is a separation?
    Do we get since any intersection of connected sets in X aren't empty?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    I don't quite understand the proof, sorry

    Why is a separation?
    I'm making a contradiction and so I'm assuming it's a seperation.
    Do we get since any intersection of connected sets in X aren't empty?
    Yes, if it were to be a separation E\cap \Omega,G\cap \Omega they must be non-empy open and disjoint. I forgot to mention disjointness but since E\cap \Omega,G\cap \Omega are disjoint evidently so are E\cap U_{\alpha_1},G\cap U_{\alpha_1}
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Here, I think I can say it so you'll better understand.


    Let E,G\subseteq X be open and be such that \left(E\cap \Omega \right)\cap\left(G\cap \Omega\right)=\varnothing and \left(E\cup \Omega\right)\cup\left(E\cap\Omega\right)=\Omega. We claim that at least one of E,G must be empty. Since the union of their intersections with \Omega is non-empty (assuming \Omega is

    non-empty, but then the conclusion is immediate) we must have that at least one of them is non-empty, assume WLOG that it's E. Then, we have that U_{\alpha_1}\cap E\ne\varnothing

    for some \alpha_1\in\mathcal{A}. But, it must be true that U_{\alpha_1}\subseteq E\cap\Omega otherwise we'd have that some point of U_{\alpha_1} is not in E\cap\Omega and since E\cap\Omega,G\cap\Omega cover \Omega it must be that that point is in G\cap\Omega. In other

    words, G\cap U_{\alpha_1} and E\cap U_{\alpha_1} are both non-empty. But, they are apparently disjoint, open in U_{\alpha_1} and their union is U_{\alpha_1}. This clearly contradicts that U_{\alpha_1} from where it follows that U_{\alpha_1}\subseteq E\cap \Omega.

    Now, the rest is easy. For, suppose that G\cap \Omega is non-empty. Then, by the previous analysis it follows that U_{\alpha_2}\cap (G\cap \Omega)\ne\varnothing and so U_{\alpha_2}\subseteq G\cap \Omega. But, this is a contradiction since G\cap\Omega and

    E\cap\Omega are disjoint and U_{\alpha_1}\cap U_{\alpha_2} is non-empty. It follows that G\cap\Omega must be empty and thus no separation of \Omega is possible. The conclusion follows.
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  6. #6
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    I think I kind of get it....

    But if I just had 2 sets, like A and B, I would say suppose AUB is disconnected.

    So there exist sets X and Y such that
    XnAUB != YnAUB != empty
    XnAUB U YnAUB = AUB
    XnAUB n YnAUB = empty

    so some part of A, B is in X, and also some other part of A, B is in Y since XnAUB, YnAUB is not empty and their intersection is empty (so the same part of A, B can't be in both Y and X)
    so XnA, YnA is not empty
    so XnA U YnA = A??
    So A is disconnected because XnA and YnA are disjoint?
    And same goes for B

    Is this correct?
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    I think I kind of get it....

    But if I just had 2 sets, like A and B, I would say suppose AUB is disconnected.

    So there exist sets X and Y such that
    XnAUB != YnAUB != empty
    XnAUB U YnAUB = AUB
    XnAUB n YnAUB = empty

    so some part of A, B is in X, and also some other part of A, B is in Y since XnAUB, YnAUB is not empty and their intersection is empty (so the same part of A, B can't be in both Y and X)
    so XnA, YnA is not empty
    so XnA U YnA = A??
    So A is disconnected because XnA and YnA are disjoint?
    And same goes for B

    Is this correct?
    Clean it up, but yeah!
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  8. #8
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    would the intersection or A and B be connected?

    I'm kind of stuck on this because I don't know if the empty set is connected or not. For example, if A=(0,1) and B=(2,3) then their intersection would be empty....but is that connected?
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    would the intersection or A and B be connected?

    I'm kind of stuck on this because I don't know if the empty set is connected or not. For example, if A=(0,1) and B=(2,3) then their intersection would be empty....but is that connected?
    The empty set is sometimes connected sometimes not depending on the author.

    But, we can prove a nice little theorem.

    Theorem: Let E,G\subseteq\mathbb{R} be connected and E\cap G\ne\varnothing then E\cap G is connected.

    Proof:This follows since the intersection of two intervals is an interval.

    In fact this is true in any linear continuum. But, this is a very special case.

    Permit me if you will to describe a picture for you.

    Imagine two crescent rolls (look here) it is easy to think of their general shape projected into \mathbb{R}^2, right? Now, think about taking two of them and having the concave sides face each other. Now, move them until just their tips are touching. Clearly each crescent roll is connected but their intersection will be the area in the overlapping tips, but this is clearly disconnected.
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  10. #10
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    But if we did not have the condition that , then the intersection of E and G would not be connected, correct?

    I'm looking at your crescent roll example for that: the intersection of the crescent rolls is empty....kind of like the intersection of two open sets, so the empty set in this case would be disconnected.
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  11. #11
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    I'm looking at your crescent roll example for that: the intersection of the crescent rolls is empty....kind of like the intersection of two open sets, so the empty set in this case would be disconnected.
    No, the two do intersect.

    I guess an easier example would be two circles in \mathbb{R}^2 that intersect at two points. Clearly both are connected but their intersection are two disconnected poitns.
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