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Math Help - Metric Topology

  1. #1
    Senior Member
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    Metric Topology

    Hello everybody

    I want to show that the quotient operation \mathbb{R}^2 \longrightarrow \mathbb{R}is continuous in \mathbb{R} with the metric d(a,b) = |a-b| on \mathbb{R} and the metric on \mathbb{R}^2 given by the equation \rho((x,y),(x_0,y_0)) = \max\{|x-x_0|,|y-y_0|\}

    Start by given (x_0,y_0) and \epsilon >0. We need to find \delta s.t if \max\{|x-x_0|,|y-y_0|\} < \delta then \left|\frac{x}{y}-\frac{x_0}{y_0} \right| < \epsilon

    Let \max\{|x-x_0|,|y-y_0|\} < \delta.

    \left|\frac{x}{y}-\frac{x_0}{y_0} \right| =\left|\frac{y_0(x-x_0) + x_0(y_0-y)}{yy_0}\right| \le \frac{|y_0||x-x_0| + |x_0||y_0-y|}{|yy_0|} \le \delta \left(\frac{|y_0| + |x_0|}{|y_0|}\right) \frac{1}{|y|}

    what \delta should be in order to make \delta \left(\frac{|y_0| + |x_0|}{|y_0|}\right) \frac{1}{|y|} = \epsilon?
    the \frac{1}{|y|} term that make me confuse. . .

    thanks for any comment. . .
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dedust View Post
    Hello everybody

    I want to show that the quotient operation \mathbb{R}^2 \longrightarrow \mathbb{R}is continuous in \mathbb{R} with the metric d(a,b) = |a-b| on \mathbb{R} and the metric on \mathbb{R}^2 given by the equation \rho((x,y),(x_0,y_0)) = \max\{|x-x_0|,|y-y_0|\}

    Start by given (x_0,y_0) and \epsilon >0. We need to find \delta s.t if \max\{|x-x_0|,|y-y_0|\} < \delta then \left|\frac{x}{y}-\frac{x_0}{y_0} \right| < \epsilon

    Let \max\{|x-x_0|,|y-y_0|\} < \delta.

    \left|\frac{x}{y}-\frac{x_0}{y_0} \right| =\left|\frac{y_0(x-x_0) + x_0(y_0-y)}{yy_0}\right| \le \frac{|y_0||x-x_0| + |x_0||y_0-y|}{|yy_0|} \le \delta \left(\frac{|y_0| + |x_0|}{|y_0|}\right) \frac{1}{|y|}

    what \delta should be in order to make \delta \left(\frac{|y_0| + |x_0|}{|y_0|}\right) \frac{1}{|y|} = \epsilon?
    the \frac{1}{|y|} term that make me confuse. . .

    thanks for any comment. . .
    I didn't look at it too closely but have you thought about taking a compact neighborhood of (x_0,y_0) and using \frac{1}{y}'s (apparent...I have not actually checked) continuity to state that \left|\frac{1}{y}\right|=\frac{1}{|y|}\leqslant K for some K?
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