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Math Help - proving continuity of g(x, y) = x

  1. #1
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    proving continuity of g(x, y) = x

    g(x, y) = x

    I am trying to prove that g is continuous on R^2 using the epsilon-delta definition of continuity.

    I gave it a try but it doesn't look right:

    Given c \in R and \epsilon > 0, set \delta = \epsilon

    Then whenever \| x-c \| < \delta, it follows that \| g(x, y) - g(c, y) \| < \epsilon



    Thanks for any help in advance.
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  2. #2
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    |g(x,y)-g(x_0,y_0)|<\epsilon tells you that the distance between x and x_0 is less than epsilon (i.e. |x-x_0|<\epsilon).

    ||(x,y)-(x_0,y_0)||<\delta tells you that |x-x_0|\leq\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta.

    Do you see why you get continuity by picking delta as you did?
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  3. #3
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    Quote Originally Posted by Focus View Post
    |g(x,y)-g(x_0,y_0)|<\epsilon tells you that the distance between x and x_0 is less than epsilon (i.e. |x-x_0|<\epsilon).

    ||(x,y)-(x_0,y_0)||<\delta tells you that |x-x_0|\leq\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta.

    Do you see why you get continuity by picking delta as you did?
    Oh, okay. I think I get it now.

    If \delta = \epsilon,
    |x-x_0| \leq \sqrt{(x-x_0)^2+(y-y_0)^2}<\delta would imply
    |x-x_0| < \epsilon which would imply
    |g(x,y)-g(x_0,y_0)| < \epsilon

    Edit: also I'd be given (x_0, y_0) \in R^2, not just c \in R since g is in R^2
    Last edited by DPMachine; November 18th 2009 at 06:59 AM.
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  4. #4
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    Quote Originally Posted by DPMachine View Post
    Oh, okay. I think I get it now.

    If \delta = \epsilon,
    |x-x_0| \leq \sqrt{(x-x_0)^2+(y-y_0)^2}<\delta would imply
    |x-x_0| < \epsilon which would imply
    |g(x,y)-g(x_0,y_0)| < \epsilon

    Edit: also I'd be given (x_0, y_0) \in R^2, not just c \in R since g is in R^2
    Looks good to me. Good luck with your studies.
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