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Math Help - (path) connected

  1. #1
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    (path) connected

    I am having a little trouble with path connectedness.

    I am trying to determine whether or not the set R^2\{(0,0)} is path connected.

    My first thought was that since R^2\{(0,0)} is a subset of R^2 and R^2, being the cartesian product RxR, is path connected, it follows that R^2\{(0,0)} cannot be both open and closed and thus cannot be path connected.

    However, when I imagine it, it seems that given any two non-zero points in R^2, I can connect them by the straight-line path between them (if it does not go through the origin) or I can choose a third point with which to form a broken line connection, perfectly satisfactory as a path.

    So that would imply that R^2\{(0,0)} is in fact path connected.

    So I think I need clarification of the following iff statement:

    A space X is connected iff the only subsets of X that are both open and closed in X are the empty set and X itself.

    So, do I understand correctly that this statement says that every path connected set is open and closed? And if so, does this statement not say that no subset of a path connected set can be path connected? It doesn't seem to make sense. What does this statment mean for my space R^2\{(0,0)}?

    Any help on the statement as well as the properties (open, closed, connected, path connected) of R^2\{(0,0)} would be greatly appreciated.
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  2. #2
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    There is a standard theorem: Every open connected subset of R^n is path connected.

    Now in this case, R^2, your approach bypasses that theorem.
    It is easy to construct a path.
    As you note, if the line segment between two points does not contain (0,0) you are done.
    It is does we can use a semicircular arc around the origin.
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  3. #3
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    Quote Originally Posted by TXGirl View Post
    I am having a little trouble with path connectedness.



    My first thought was that since R^2\{(0,0)} is a subset of R^2 and R^2, being the cartesian product RxR, is path connected, it follows that R^2\{(0,0)} cannot be both open and closed and thus cannot be path connected.
    I think this is what you are thinking of

    The only clopen sets (Both open and closed) are the space X and the emptyset.

    This is equivelent to X is conncected

    \mathbb{R}^2\  \{ 0,0 \}
    is a subset of R^2 but is an open set.

    All Euclidean spaces satisfiy the above property.

    I hope this helps.
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    I think this is what you are thinking of

    The only clopen sets (Both open and closed) are the space X and the emptyset.

    This is equivelent to X is conncected

    \mathbb{R}^2\  \{ 0,0 \}
    is a subset of R^2 but is an open set.

    All Euclidean spaces satisfiy the above property.

    I hope this helps.
    So, am I correct that \mathbb{R}^2 is connected and thus clopen. If so, how is it that there is a subset, namely \mathbb{R}^2\  \{ 0,0 \} that is connected, and thus clopen?
    Last edited by TXGirl; May 11th 2008 at 03:55 PM.
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  5. #5
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    Quote Originally Posted by TXGirl View Post
    So, am I correct that \mathbb{R}^2 is connected and thus clopen. If so, how is it that there is a subset, namely \mathbb{R}^2\  \{ 0,0 \} that is connected, and thus clopen?

    I think I see my error...

    I was thinking of it as a subspace of \mathbb{R}^2

    Our Space X=\mathbb{R}\ \{0,0\}

    Since this is the whole space its compliment must be empty

    X^c=\emptyset also \emptyset^c=X

    before I claimed that X^c=\{0,0\} I think this is incorrect

    X is open because for every point x \in X
    There exits and \epsilon > 0 such that a Ball centered at x of radius epsilon
    B_{\epsilon}(x) \subset X

    The Emptyset is also open

    Since they are compliments of each other they are also closed.

    I'm sorry if I haven't been very helpful. I guess I don't remember this as well as I thought. Time for more review.

    I hope this helps.

    Good luck.
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  6. #6
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    Quote Originally Posted by TheEmptySet View Post
    I think I see my error...

    I was thinking of it as a subspace of \mathbb{R}^2

    Our Space X=\mathbb{R}\ \{0,0\}

    Since this is the whole space its compliment must be empty

    X^c=\emptyset also \emptyset^c=X

    before I claimed that X^c=\{0,0\} I think this is incorrect

    X is open because for every point x \in X
    There exits and \epsilon > 0 such that a Ball centered at x of radius epsilon
    B_{\epsilon}(x) \subset X

    The Emptyset is also open

    Since they are compliments of each other they are also closed.

    I'm sorry if I haven't been very helpful. I guess I don't remember this as well as I thought. Time for more review.

    I hope this helps.

    Good luck.
    No, you were correct, we are considering the subspace X=\mathbb{R}^2\ \{0,0\} of \mathbb{R}^2

    Any other ideas?? Thanks again for your help. I'm in Germany and we're on holiday until Tuesday, so I cannot go see the prof until then...and for now, I seem to have hit a brick wall in my understanding....
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  7. #7
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    Quote Originally Posted by TXGirl View Post
    Any other ideas??
    Did you even see my reply?
    In fact, the other replies have nothing to do with this question.
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  8. #8
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    Quote Originally Posted by Plato View Post
    Did you even see my reply?
    In fact, the other replies have nothing to do with this question.
    Yes, I saw it; however, I could see that \mathbb{R}^2\  \{ 0,0 \} is connected; my problem was, and is, that

    there is a theorem saying:

    A space X is connected iff the only clopen subsets of X are X and the empty set.

    How then is it that if \mathbb{R}^2 is connected and thus clopen, there is a subset, namely \mathbb{R}^2\  \{ 0,0 \}, that is connected, and thus clopen?

    Somewhere I seem to be misunderstanding the theorem. The way I understand it is seems to say that a subspace of a connected space, cannot be connected; however, this does not seem sensible and, in fact, the problem at hand stands as a contradiction, so I was hoping that someone could explain the theorem, particularly as it relates to the to sets \mathbb{R}^2 and \mathbb{R}^2\  \{ 0,0 \}
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