1. ## 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. . .

2. Originally Posted by dedust
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 $K$?