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Math Help - arcwise connected open set

  1. #1
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    arcwise connected open set

    Prove:
    Any connected open subset S of E^n is arcwise connected.

    Definition: A set in E^n is said to be connected if it is impossible to split S into two disjoint sets A and B, neither one empty, without having one of the sets contain a boundary point of the other.

    Assumption: Any two points in a spherical neighborhood of a point in S can be connected by a straight line.

    Let A consist of a point x in S and all other points arcwise connected to x. Let B be all other points of S (S-A).

    Consider a spherical neighborhood of a boundary point of A and B (Definition). Let z be in A and w in B. There is a straight line connecting z and w (assumption) because the boundary point and spherical neighborhood are in S. Therefore w in A- contradiction. Therefore B is empty.

    Reference: Thread "arcwise connected set"
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    The other thread was because not only was it getting heated up, the question was settled. There is no point in posting the same question twice. It is against forum rules. Reread reply #6 in the other thread. That is a clear solution. If you do not agree, tell us exactly where your disagreement is with it.
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    opalg was dorrect

    opalg was correct. My apologies. Thanks

    Opalg's proof:

    "Start as Taylor does, by fixing a point , and defining A to be the set of points in S that can be reached by a path from x. Let B be the complement of A in S. Then A is open. Reason: suppose that ; since S is open, there is a ball B(y,e) centred at y and contained in S. Then every point in that ball belongs to A, because you can connect it to y by a straight line which lies within the ball, and then from y there is a path to x.

    A very similar argument shows that B is open: suppose that ; there is a ball centred at z and contained in S. If that ball contains a point then there is a path in S consisting of a straight line from z to w and then from w to x. So – contradiction. Therefore the ball contains no points in A and hence lies entirely in B, showing that B is open.

    Thus S is the disjoint union of the open sets A and B. But if S is connected then one of those sets must be empty. It can't be A, because . Therefore B is empty, which means that S is arcwise-connected."

    The crux of it, which I see now, is that any ball in A or B is also a ball in S.

    Sorry about tag, I couldn't edit dorrect.
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    whoops

    There is a problem with opalg's proof.

    "A very similar argument shows that B is open: suppose that ; there is a ball centred at z and contained in S. If that ball contains a point then there is a path in S consisting of a straight line from z to w and then from w to x. So – contradiction."

    The underlined sentence is the problem. You can't assume that the ball contains a point of A- that, in effect, assumes what you are trying to prove. You have to make use of the definition of a connected set to get a boundary point between A and B.

    Back to Hartl's proof above.
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    missed edit for last post.

    In other words, you can't assume that B consists of points not in A and then assume B contains a point of A.
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    Quote Originally Posted by Hartlw View Post
    "A very similar argument shows that B is open: suppose that ; there is a ball centred at z and contained in S. If that ball contains a point then there is a path in S consisting of a straight line from z to w and then from w to x. So – contradiction."
    The underlined sentence is the problem. You can't assume that the ball contains a point of A- that, in effect, assumes what you are trying to prove.
    Do you understand how mathematicians/logicians use the word if?
    In the proof it says, if some ball centered at a point of p\in B also contains a point of q\in A then because balls are pathwise connected that means that
    p is connect to x. So that p\in A which is a contradiction.

    The if…then is a hypothetical, it does not assume that the ball contains a point of A. It simply says if it did we get a contradiction. Recall that S is open so B is open.
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    Quote Originally Posted by Plato View Post
    Do you understand how mathematicians/logicians use the word if?
    In the proof it says, if some ball centered at a point of p\in B also contains a point of q\in A then because balls are pathwise connected that means that
    p is connect to x. So that p\in A which is a contradiction.

    The if…then is a hypothetical, it does not assume that the ball contains a point of A. It simply says if it did we get a contradiction. Recall that S is open so B is open.
    The fact that S is open does not mean that B is open. See definition of connected set.

    To test an assumption, you have to be consistent with your assumption and then show it leads to a contradiction. You can't, for example, prove that there are rational numbers between any two real numbers by starting off with the assumption there are no rational numbers between any two real numbers and then assuming there is a rational number in order to prove a contradiction.
    Last edited by Hartlw; January 21st 2011 at 09:44 AM. Reason: remove "no" from last sentence
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  8. #8
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    Quote Originally Posted by Hartlw View Post
    The fact that S is open does not mean that B is open.
    I can see why Opalg gave up on you (and he is the most long-suffering and forgiving helper here). Are you being a troll here? If so please be honest about it and we will close this thread.

    Yes B is indeed open. It is true that B\subseteq S. S is open so if y\in B\subseteq S then there is a ball \mathcal{B}(y;\delta)\subseteq S, because S is open. Can  \mathcal{B}(y;\delta) contain a point of A~?.
    If not then  \mathcal{B}(y;\delta) \subseteq (S\setminus A)=B. So B is open.
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    Quote Originally Posted by Plato View Post
    I can see why Opalg gave up on you (and he is the most long-suffering and forgiving helper here). Are you being a troll here? If so please be honest about it and we will close this thread.

    Yes B is indeed open. It is true that B\subseteq S. S is open so if y\in B\subseteq S then there is a ball \mathcal{B}(y;\delta)\subseteq S, because S is open. Can  \mathcal{B}(y;\delta) contain a point of A~?.
    If not then  \mathcal{B}(y;\delta) \subseteq (S\setminus A)=B. So B is open.

    In other words, A is open so if B is open S is the union of two disjoint open sets which it can't be because it is connected so B is empty. Why bother with all the ofuscation?
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    Quote Originally Posted by Hartlw View Post
    In other words, A is open so if B is open S is the union of two disjoint open sets which it can't be because it is connected so B is empty. Why bother with all the ofuscation?
    Sorry, I was being lazy. If you explain what \ means, I'll go through your last post.
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  11. #11
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    Quote Originally Posted by Hartlw View Post
    In other words, A is open so if B is open S is the union of two disjoint open sets which it can't be because it is connected so B is empty. Why bother with all the ofuscation?
    I have no idea. Like Opalg said, Taylor is a good clear author, but I find his proof of this lacking clarity.
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