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Math Help - arcwise connected set

  1. #16
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    Quote Originally Posted by Hartlw View Post
    To get really nasty about this, how do you know that A isn't empty? Let x be a point in S with a spherical neighborhood. How do you know you can connect from x to any point in the spherical neighborhood without assuming what you are trying to prove?
    A isn't empty because x \in A (the one Opalg took when he said "fix x \in S").

    Another way to look at Opalg's proof is this. Assume that S is indeed connected, and towards a contradiction, assume that S is not path connected, ie. there exist points x,y \in S such that there is no path between x and y. Then take the same sets A and B as before - the last assumption means that B is not empty, and as we've seen A is not empty. Then S = A \cup B where A,B are disjoint (by definition), nonempty and both open, so we get a contradiction to S being a connected set.

    I see you agree that A is open. Now take some point, say b \in B and look at a ball B(b,r) \subseteq S around b (which exists since S is open). Now,
    1) We know that if c \in B(b,r) then you can construct a path from b to c, namely the straight line from b to c.
    2) If there is any a \in B(b,r) such that a \in A, then by the definition of A there is a path from x to a. But a \in B(b,r), and by (1) there exists a path from a to b. By "gluing" the two paths together, we get a path from x to b. but b \in B, which means that there exists no path from x to b - contradiction.

    Can you point what step, exactly, is not clear in this proof?
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  2. #17
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    x alone is not a member of A. A is all points connectible to x. You have to prove such a point exists.

    A and B are not both open sets. See previous posts.
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  3. #18
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    There is a path from x to x, namely, f:[0,1] \to S, & f(t)=x, and so x \in A.

    With which part of Opalg's proof do you not agree?
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  4. #19
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    A straight line is defined by distinct end points.

    A and B cannot both be open sets. See previous posts.
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  5. #20
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    The previous posts don't show that the proof is wrong. To make it clear, I am using the definition that a set S is connected if there are no A,B \subseteq S such that A \cap B = \varnothing , \ A \cup B = S and both A and B are open.

    Also, recall that the definition of a path between x and y is a continuous function f:[0.1] \to S such that f(0) = x, \ f(1) = y.
    The function f(t) \equiv x is constant and therefore continuous, and f(0) = f(1) = x so it is indeed a path between x and x, so  x \in A.

    Please read over the proof again and tell me what step fails to hold. I am convinced that Opalg has shown both A and B are open, but I might be mistaken - where exactly do you think his approach fails?
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  6. #21
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    A continuous function cannot consist of a single point.

    Opalg's proof fails becaues he assumes both A and B are open, which is impossible by the definition of a connected set.
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  7. #22
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    Quote Originally Posted by Hartlw View Post
    A continuous function cannot consist of a single point.

    Opalg's proof fails becaues he assumes both A and B are open, which is impossible by the definition of a connected set.
    A continuous function cannot consist of a single point? What does that even mean?!

    Opalg does not assume anywhere that both A and B are open. He proves that they are both open, resulting in the fact that one of them must be empty (this is because S is connected). Since x \in A, it must be that B is empty and so we get that A=S and therefore every point in S is path connected to x.

    ---

    To prove that B contains no accumulation points of A: Assume it does, and let b \in B be such a point. Then there is some r>0 such that B(b,r) \cap A \neq \emptyset. Let a \in B(b,r) \cap A. Then the function \gamma :[0,1] \to S, & \gamma(t) = tb + (1-t)a shows us that there is a path from a to b (since the open ball is convex). Now, since a \in A, there exists a continuous \beta :[0,1] \to S such that \beta(0)=x, \beta(1)=a.

    Now, define \alpha :[0,1] \to S as follows:

    \alpha (t) = \left\{<br />
\begin{array}{lr}<br />
\beta(2t) & 0 \leq t \leq \frac{1}{2}\\<br />
\gamma(2t-1) & \frac{1}{2} < t \leq 1<br />
\end{array}<br />
\right.

    So, since \gamma, \beta are continuous and \beta(1) = \gamma(0) = a we get that \alpha is continuous and \alpha(0)=x, \alpha(1)=b. This gives us that there is a path from x to b, and so b \in A which is, of course, absurd.
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  8. #23
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    I have never heard of a continuous function defined on a domain consisting of a single point. A domain has to be an open set. Look up definition of continuity.

    Consider A is [0,1) and B is [1,0]. Can they be connected by a line, ie, across the boundary? Intuitiveley I would say yes, but I have to think about a proof. If they can, I would be inclined to consider your proof.
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  9. #24
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    Where did I say it is defined on a single point? f:[0,1] \to S defined by f(t) = x for each t \in [0,1] is defined on [0,1] and its image is the singleton \{x\}; the same can be said for every constant function. Surely you've seen a constant function?

    Furthermore, I do not understand your notation. What do you mean by B =[1,0]? How do you connect two sets by a line?
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  10. #25
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    Quote Originally Posted by Defunkt View Post
    Where did I say it is defined on a single point? f:[0,1] \to S defined by f(t) = x for each t \in [0,1] is defined on [0,1] and its image is the singleton \{x\}; the same can be said for every constant function. Surely you've seen a constant function?

    Furthermore, I do not understand your notation. What do you mean by B =[1,0]? How do you connect two sets by a line?
    If you won't accept that it takes two points to define a line, we have to part ways on this. Personally, I think you lose credibility by trying to defend the indefensible on the basis of ego. I know, I would do the same thing.

    [0,1) is the interval on a line closed on the left and open on the right. [0,1] is closed at both ends. 1 is a boundary point between both (every neighborhood of 1 has a point in A and B). If you start a line (continuous) in A and move to the right, can you cross the boundary point? Though it looks like you can, I wouldn't know off-hand how to prove it.
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  11. #26
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    Quote Originally Posted by Hartlw View Post
    If you won't accept that it takes two points to define a line, we have to part ways on this. Personally, I think you lose credibility by trying to defend the indefensible on the basis of ego. I know, I would do the same thing.
    Not to be brusque, but I believe it is you that is here on the math site and it's Defunkt who's trying to help you. If there were things that you might not understand you wouldn't be here. In fact, he is one-hundred percent correct. A constant function is a path.


    [0,1) is the interval on a line closed on the left and open on the right. [0,1] is closed at both ends. 1 is a boundary point between both (every neighborhood of 1 has a point in A and B). If you start a line (continuous) in A and move to the right, can you cross the boundary point? Though it looks like you can, I wouldn't know off-hand how to prove it.
    Can you possibly rephrase this?
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  12. #27
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    Quote Originally Posted by Drexel28 View Post
    Not to be brusque, but I believe it is you that is here on the math site and it's Defunkt who's trying to help you. If there were things that you might not understand you wouldn't be here. In fact, he is one-hundred percent correct. A constant function is a path.




    Can you possibly rephrase this?
    If you won't accept that 2 points determine a line, we are not speaking the same language. Also, I don't want to get into a non-tecnical political debate with you.
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  13. #28
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    Quote Originally Posted by Hartlw View Post
    If you won't accept that it takes two points to define a line, we have to part ways on this. Personally, I think you lose credibility by trying to defend the indefensible on the basis of ego. I know, I would do the same thing.
    Given some fixed x \in S I will define A = \{a \in S : \text{there exists }  f:[0,1] \to S \text{ that is continuous and }  f(0)=x, f(1)=a\}. Surely by this definition x \in A, regardless of how you or I define a line.

    [0,1) is the interval on a line closed on the left and open on the right. [0,1] is closed at both ends. 1 is a boundary point between both (every neighborhood of 1 has a point in A and B). If you start a line (continuous) in A and move to the right, can you cross the boundary point? Though it looks like you can, I wouldn't know off-hand how to prove it.
    I don't know how this has any relevance to the question at hand. What is a boundary point between two sets? If I understand correctly you took A=[0,1), B=[0,1]. If so, any point y \in [0,1] intersects both B and A in any neighborhood, I don't see how 1 is any different in this sense. What do you mean by start a line? What are you trying to connect? Can you at least be more rigorous?
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  14. #29
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Hartlw View Post
    If you won't accept that 2 points determine a line, we are not speaking the same language. Also, I don't want to get into a non-tecnical political debate with you.
    I'm speaking the language of mathematics, what are you speaking?? I'm not one of those internet guys who just wants to screw with you about semantical issues, but what was said is the technical definition. I'd like to help you understand, or clear up anything, so is there any way you could rephrase your second question please?

    This is pointless
    Last edited by Plato; January 20th 2011 at 05:00 PM.
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