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Thread: Metric Topology

  1. #1
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    Metric Topology

    Hello everybody

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

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

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

    $\displaystyle \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|}$$\displaystyle \le \delta \left(\frac{|y_0| + |x_0|}{|y_0|}\right) \frac{1}{|y|}$

    what $\displaystyle \delta$ should be in order to make $\displaystyle \delta \left(\frac{|y_0| + |x_0|}{|y_0|}\right) \frac{1}{|y|} = \epsilon$?
    the $\displaystyle \frac{1}{|y|}$ 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 $\displaystyle \mathbb{R}^2 \longrightarrow \mathbb{R}$is continuous in $\displaystyle \mathbb{R}$ with the metric $\displaystyle d(a,b) = |a-b|$ on $\displaystyle \mathbb{R}$ and the metric on $\displaystyle \mathbb{R}^2$ given by the equation $\displaystyle \rho((x,y),(x_0,y_0)) = \max\{|x-x_0|,|y-y_0|\}$

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

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

    $\displaystyle \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|}$$\displaystyle \le \delta \left(\frac{|y_0| + |x_0|}{|y_0|}\right) \frac{1}{|y|}$

    what $\displaystyle \delta$ should be in order to make $\displaystyle \delta \left(\frac{|y_0| + |x_0|}{|y_0|}\right) \frac{1}{|y|} = \epsilon$?
    the $\displaystyle \frac{1}{|y|}$ 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 $\displaystyle (x_0,y_0)$ and using $\displaystyle \frac{1}{y}$'s (apparent...I have not actually checked) continuity to state that $\displaystyle \left|\frac{1}{y}\right|=\frac{1}{|y|}\leqslant K$ for some $\displaystyle K$?
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