For $x,y \in \mathbb R^n$ let $$ d_p(x,y) = \left(\sum_{i=1}^n \def\abs#1{\left|#1\right|}\abs{x_i - y_i}^p\right)^{1/p}$$ for $1 \le p < \infty$ and $$ d_\infty(x,y) = \max\{\abs{x_i -y_i} \mid i = 1, \ldots, n\} $$ Let $B_p = \{x\in \def\R{\mathbb R}\R^n \mid d_p(x,0) < 1\}$, $1 \le p \le \infty$. 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.

Help me with the geometrical approach to solve this. How to construct those sets and all those ?? and explain the ruled out options in detail

  • $\begingroup$ 1. cannot read indices in the answer. 2. what you mean with ?? $\endgroup$ – Emanuele Paolini Jun 11 '14 at 9:04
  • $\begingroup$ I suggest that you type the question yourself in LateX $\endgroup$ – Vishal Gupta Jun 11 '14 at 9:05
  • $\begingroup$ Try doing this in $\mathbb{R}^{2}$ first. See that ball in $d_{1}$ metric is a rhombus, in $d_{2}$ metric, a ball is a circle and in $d_{\infty}$ metric, it is a square. This is the most geometrical way I can see. Then you can generalize to higher dimensions. $\endgroup$ – Vishal Gupta Jun 11 '14 at 9:23
  • $\begingroup$ As Emanuele Paolini pointed out in his answer, all topologies are same and you can see this because you can always get a circle inside a rhombus and vice versa. Simlarly circle inside a square and vice versa and a square inside a rhombus and vice versa. $\endgroup$ – Vishal Gupta Jun 11 '14 at 9:24
  • $\begingroup$ @martini Thanks for the edit. $\endgroup$ – Vishal Gupta Jun 11 '14 at 9:32

Prove that for any $p,q$ the ball $B_p$ is contained in some rescaling of the ball $B_q$. As a consequence all the topologies are equal each other hence statements like "$X$ is open with respect to $d_p$" do not depend on $p$.


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