Explicit value for $\delta$ in the euclidean metric.

Question

Let $Q=\{(x,y)\in\mathbb{R}^2|y>x\}$. Using the euclidean metric give and explicit value for $\delta_(x,y)$ so that $B_\delta(x,y)$ is contained in $Q$.

My work so far...

Using standard calculus techniques I can compute the minimal distance from a point $(x,y)$ in $Q$ to the line $y=x$ to be $\delta=|x-y|/\sqrt{2}$. Therefore I am trying to show that if $(x,y)\in B_{\delta}(x_0,y_0)$, then $(x,y)\in Q$. Therefore I need to show that if \begin{align*} \sqrt{(x_0-x)^2+(y_0-y)^2}<\frac{|x_0-y_0|}{\sqrt{2}} \end{align*} then, \begin{align*} y>x. \end{align*}

I have pages and pages of inequalities and I'm not sure which line of reasoning is going to get me where I need to go. Any direction is greatly appreciated.

Answer 1

Let $y>x$. If $(x',y') \in B_{\delta} (x,y)$ then $|x-x'| <\delta$ and $|y-y'| <\delta$. Let us denote $y-x$ by $r$. (Note that $r>0$). Now we have the inequalities $y'>y-\delta=x+r-\delta>(x'-\delta)+r-\delta >x'$ provided $r-2\delta >0$ which means $\delta <\frac r 2=\frac {y-x} 2$. So you can take any $\delta \in (0, \frac {y-x} 2$).

Answer 2

You may proceed as follows using the fact that a circle with radius $\delta$ around point $(x,y)$ is given by

• $(x+ \delta \cos t,y + \delta \sin t)$ for $t \in [0,2\pi]$

So, consider $$y + \delta \sin t > x+ \delta \cos t \Leftrightarrow (y-x) + \delta(\sin t - \cos t) > 0$$

We know that $|\sin t - \cos t| \leq \sqrt{2}$. Hence, choose $\boxed{0<\delta < \frac{y-x}{\sqrt{2}}}$. Then you have $$y-x + \delta (\sin t - \cos t) \geq y-x +\delta (-\sqrt{2})> 0$$

So, for any $0 < \delta < \frac{y-x}{\sqrt{2}}$ you have $\boxed{B_{\delta}(x,y) \subset Q}$.

Answer 3

Let r be half of that minimal distance.
To show the ball of radius r centered at c = (x$_0$,y$_0$) is a subset Q, let p = (a,b) be a point in that ball. The closest p can be to the diagonal line L is when p is on the line from c perpendicular to L. That distance is r.

Alternatively, if d is the distance from p to L,
then 2r <= r + d, r <= d, so p is in Q.

Answer 4

If $(x_0,y_0)\in Q$ then the delta you want will but exactly equal to the shortest distance from $(x_0,y_0)$ to the border of $Q$ which is the line $y =x$.

Now the problem reduces to nothing to do with analysis and just a standard analytic geometry problem.

The border, $y = x$ is a line with a slope of $1$. The shortest distance for any point $(x_0, y_0)$ to the line $y=x$ is via a perpendicular to the line.

If you draw a picture this is a diagonal line. The $x$ value goes over by a distance of $k$ and $y$ value goes down by a distance of $k$ and the distance between the two points is $\sqrt 2\cdot k$.

So the point of intersection is $(x_0 + k, y_0 -k)$ but it's on the border $y=x$ so $k = \frac {y_0-x_0}2$ and $\delta =\frac {\sqrt 2}2(y_0- x_0)$.