For $x = (x_1, x_2, \ldots, x_n)$ and $y = (y_1, y_2, \ldots, y_n)$ in $\mathbb{R}^n$.

Let $d_p(x, y) = \Bigg(\sum\limits_{i=1}^n |x_i-y_i|^p\Bigg)^\frac{1}{p}$ for $1 \leq p < \infty$ and

$d_\infty = \max\lbrace |x_i-y_i|: j= 1, 2,\ldots, n\rbrace$. Let $B_p= \lbrace x \in \mathbb{R}^n: d_p(x, 0)< 1\rbrace.$

Which of the following are correct?

a) $B_1$ is open in $d_\infty$ metric. $\qquad$ $\qquad$ $\quad$ b) $B_2$ is open in $d_\infty$ metric.

c) $B_1$ is not open in $d_2$ metric. $\qquad$ $\qquad$ d) $B_2$ is not open in $d_2$ metric.

How to tackle these kind of problems?

  • $\begingroup$ Notice the proper was to typeset $d_\infty = \max\lbrace |x_i-y_i|: j= 1, 2,\ldots, n\rbrace$, as in my edit to the question. One feature of \max is the positions of subscripts when used in a displayed setting, thus: $$ \max_{x\in\mathcal X} f(x). $$ Another feature is proper spacing in things like $3\max S$. $\qquad$ $\endgroup$ – Michael Hardy Jun 19 '16 at 18:20

You can use that those metrics are equivalent (because they derived from norms). Then, you just have to verify if the sets $B_i$ are open or not in the $d_i$ metric.


Hint: On finite dimensional spaces, two metrics are equivalent.


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