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Math Help - Connected spaces

  1. #1
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    Lightbulb Connected spaces

    I'm having trouble understanding how deleting a point from R2 leaves it a connected space, while deleting a point from R doesn't do so.

    Can someone help me with this?

    Thanks,
    MAX
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  2. #2
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    Quote Originally Posted by MAX09 View Post
    I'm having trouble understanding how deleting a point from R2 leaves it a connected space, while deleting a point from R doesn't do so.
    This is one characterization of connect sets:
    A set is connected if it is not the union of two non-empty separated sets.
    Two sets are separated if neither contains a point or a limit point of the other.

    Now \Re \backslash \{ a\}  = \left( { - \infty ,a} \right) \cup \left( {a,\infty } \right), clearly the union of two separated sets.

    Try that with  \Re  \times \Re \backslash \{ (a,b)\} . What two separated sets would be possible?
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    Lightbulb

    That clarity about R is quite obvious.. I don't know how I missed it.

    Yet, I have one question.
    # how do you express RXR/{(a,b)} in terms of set notation?
    # Please walk me through the solution...

    thanks,
    MAX09
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  4. #4
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    Quote Originally Posted by MAX09 View Post
    # how do you express RXR/{(a,b)} in terms of set notation?
    There are several different notations.
    \Re ^2  = \Re  \times \Re  = \left\{ {\left( {x,y} \right):x \in \Re  \wedge y \in \Re } \right\}
    So \Re ^2 \backslash \left\{ {(a,b)} \right\} = \left\{ {\left( {x,y} \right) \in \Re ^2 :x \ne a \wedge y \ne b} \right\}
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  5. #5
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    Lightbulb

    so, when we delete a point (a,b) from R2 it's the set

    R2-(a,b) = {(x,y) such that x !=a & y! =b}

    Are we supposed to establish the result using a contradiction?

    if a separation existed for the above mentioned set, then...???
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