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Let $Q=\{(x,y)\in\mathbb{R}^2|y>x^2\}$. Using the square metric provide a value for $\delta$ so that for any point $(x_0,y_0)\in Q$ the ball $B_{\delta}(x_0,y_0))\subset Q$.

The square metric is given by, \begin{align*} d((x,y),(a,b))=max\{|x-a|,|y-b|\}. \end{align*} I am just so unfamiliar working with the square metric I am having trouble getting a handle on how to compute a specific $\delta$. Any help would be greatly appreciated.

Walt
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    Can you do it with the usual metric? – José Carlos Santos Sep 11 '19 at 14:42
  • This metric is more commonly known as $L_\infty$. In order for the distance between two points to be less than $\delta$, each coordinates' values must be less than $\delta$ away from each other.

    A point in $Q$ can be represented by $(x,x^2+\epsilon)$ for some $x\in\mathbb R, \epsilon\in\mathbb R^+$. Can you take it from here?

     – Rushabh Mehta Sep 11 '19 at 14:47

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