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Math Help - Prove a set is connected

  1. #1
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    Prove a set is connected

    Would someone please check my work?

    Problem:
    Let B = {(0,y) : -1 <= y <= 1}
    C = {(x,sin(1/x)) : 0<x<=1}
    and A = BUC

    Prove A is connected.

    My proof:

    Suppose to the contrary that A is not connected, there exists operators W and V such that: 1) W interset A and V interset A are not empty.
    2) [A interset W] U [A interset V] = A.
    3) (A interset W) interset (A interset V) = 0.

    Now, (0,0) is in either W or V. So suppose WOLOG that (0,0) in W.
    Then there exists r>0 such that Nr(0,0) C W.
    Now, there exists p in C in Nr(0,0).
    B is pathwise connected, so B is connected. Implies W,V cannot separate B, in which implies B is either in W or V.
    Since (0,0) is in W, so B is in W.

    Similarly, C is connected, because its range is an interval and domain is continuous. In which also implies W and V cannot separate C => C is in W or V.

    Now, if C is in W, then A = (BUC) is a subset of W and V interset A would be empty, in which is contradicting our assumption (1). Which implies C is in V.

    Now, p is in Nr(0,0), so p is in W; however, p is also in C, that is in V from our original assumption. Then we have p in both W and V, in which contradict our assumption (3) (A interset W) interset (A interset V) is empty.

    Thus we conclude that A is connected.

    Q.E.D.

    You think this is right? Thanks.
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  2. #2
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    You proof is a bit long but it works.
    The easy way is to show that B is a subset of C closure.
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  3. #3
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    So a closure set is always connected?

    What is a closure set? Is it the same thing is being closed?
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  4. #4
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    C is connected, RIGHT?
    The closure of a connected is also connected.
    You did that one before.
    http://www.mathhelpforum.com/math-he...t-problem.html
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