For $x=(x_1,x_2,\dots, x_n)$ and $y=(y_1,y_2,\dots, y_n)$ in $\mathbb{R}^n$, let $$d_p(x,y)=\Big(\sum_{i=1}^n|x_i-y_i|^p\Big)^\frac{1}{p} \mbox{ for $p\geq 1$, and } d_\infty(x,y)=\text{max}\{|x_i-y_i|: i=1,2,\dots,n\}.$$

Let $B_p =\{ x \in \mathbb{R}^n: d_p(x,0)<1 \}$ with $1\leq p \leq \infty$ .

Which of the following are correct?

  1. $B_1$ is open in the $d_{\infty}$- metric

  2. $B_2$ is open in the $d_{\infty}$- metric

  3. $B_1$ is not open in the $d_2$ metric

  4. $B_2$ is not open in the $d_2$ metric

My attempts: I know that a set $A$ in a metric space $(X,d)$ is open if for each $x \in A$ there is a number $r >0$ such that $B_r(x)\subset A$, but here I do not know how to visualize the diagram of open ball in $d_{\infty}$- metric and $d_2$-metric.

Please help me and tell me the solution. I would be thankful

  • $\begingroup$ let say $d_2$ metric is a standard Euclidean one, while $d_{\infty}$ is a taxi-cab $\endgroup$ – dEmigOd Sep 27 '17 at 10:04
  • $\begingroup$ im not getting how can i show B1 and B2 is open in d2 and taxicab metric@ dEmigOd $\endgroup$ – user396850 Sep 27 '17 at 10:10
  • $\begingroup$ from you definition $B_2$ is an open ball in our standard human Euclidean space. It just should be open. Draw it on the piece of paper, pick a point and draw a small ball around it inside $B_2$. While you didn't ask, but a ball in the $d_1$ is a 45 deg rotated square. As all your balls are open in their respectful metrics, they are also open in the asked ones [draw small ball or square around any picked point]. $\endgroup$ – dEmigOd Sep 27 '17 at 10:15

Hint for question 1: $B_1$ is open in the $d_{\infty}$- metric?

It has been shown here (the question is yours) that $\forall x\in \mathbb{R}^n$ the following inequalities hold $$\|x\|_{\infty}\leq \|x\|_1\leq n\|x\|_{\infty}.$$ Let $x_0\in B_1$ and let $x$ be such that $\|x-x_0\|_{\infty}< r$ for some $r>0$. Then $$\|x\|_1\leq \|x-x_0\|_{1}+\|x_0\|_{1}\leq n\|x-x_0\|_{\infty}+\|x_0\|_{1}< nr+\|x_0\|_{1}.$$ Is there any positive value for $r$ such that $nr+\|x_0\|_{1}<1$, i.e. $x\in B_1$?

In that case the set $B_1$ is open with respect to the metric $d_{\infty}$.

P.S. For $B_2$ the following inequalities should be useful $$\|x\|_{\infty}\leq \|x\|_2\leq \sqrt{n}\|x\|_{\infty}.$$

  • $\begingroup$ thanks a lots @ Robert Z ,,,,but i have one doubt question that,,according to my previous question that u have been answerd,,,,, |X|2 ≤ |X|1,, , i mean B1 is not open in d2 metrics because B1 is not contained in d-2 metrics....am i right ? pliz tell me $\endgroup$ – user396850 Sep 27 '17 at 12:07
  • $\begingroup$ @kalomlego $\|x\|_2\leq \|x\|_1$ implies that $B_1\subseteq B_2$. $\endgroup$ – Robert Z Sep 27 '17 at 12:34
  • $\begingroup$ @kalomlego Take a look here: en.wikipedia.org/wiki/Norm_(mathematics)#Properties $\endgroup$ – Robert Z Sep 27 '17 at 12:36
  • $\begingroup$ thanks a lots ,,,,once again ,,you are my best friends @Robert Z ,,,i will vote up now $\endgroup$ – user396850 Sep 27 '17 at 12:39

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