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I am having trouble while solving a question,because i don't know the proper Definition of Open set in $d_{\infty}$ .I know the definition of $d_{\infty}$

$d_{\infty}(x,y) = \max \left \{|x_i-y_i| \right \}$ where $x,y \in \mathbb{R^{n}}$

Please help.

azif00
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gaurav saini
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    Same way as you define open set in any metric space. A set $U \subseteq X$ is open if and only if for every point $x \in U$, there is some open ball $B_{\epsilon}(x)$ centered at $x$ and contained in $U$. Such open balls are sets of the form $B_{\epsilon}(x) = {y \in X : d(x,y) < \epsilon}$ for $\epsilon > 0$. –  Sep 22 '19 at 04:44
  • For $n\in \Bbb N$: For $1\le r\in \Bbb R,$ the function $d_r(x,y)=(\sum_{i=1}^n|x_i-y_i|^r)^{1/r}$ is a metric on $\Bbb R^n$ and generates the same topology (i.e. has the same family of open sets) that $d_{\infty}$ generates. The reason for the notation $d_{\infty}$ is that, restricting $r$ to values in $ [1,\infty),$ we have $\lim_{r\to \infty}d_r(x,y)=d_{\infty}(x,y).$ – DanielWainfleet Sep 22 '19 at 11:11

2 Answers2

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In any metric space $(X,d)$ you define openness the same way. You define a ball of $d$ as $$B_d(x,r)= \{y \in X: d(x,y) < r\}$$

for any $x \in X$ and real number $r>0$. Then $U$ is open iff

$$\forall x \in U: \exists r>0 : B_d(x,r) \subseteq U\text{.}$$

But it's not very hard to show that $d_2$ (the standard Euclidean metric on $\Bbb R^n$, so $d_2(x,y) = \sqrt{\sum_{i=1}^n (x_i-y_i)^2}$ and $d_\infty(x,y)=\max\{|x_i - y_i|: i = 1,\ldots,n\}$ are equivalent, in the sense that any $d_2$-ball around $x$ contains a $d_\infty$-ball around $x$ and vice versa, so that the conditions for openness for these metric are equivalent too, and so they give exactly the same open sets.

More details can be found in this answer, e.g.

Henno Brandsma
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  • https://math.stackexchange.com/questions/2447353/is-b-1-and-b-2-is-open-in-d-infty-metric?rq=1 this is the question i want to ask ,the solution in this question is some complex for me so i am trying to make it simple.In this queston${B_1}$ is open in $d_{\infty}$ but not on ${d_2}$ – gaurav saini Sep 22 '19 at 05:59
  • @gauravsaini $B_1$ is open in both $d_2$ and $d_\infty$ in that question (and in $d_1$ too, or any $d_p$ for that matter..). 1 and 2 are correct, 3 and 4 are not. – Henno Brandsma Sep 22 '19 at 06:08
  • That's why i am confused here.But this question was asked in national level entrance exam . – gaurav saini Sep 22 '19 at 06:10
  • @gauravsaini You have to choose which statements are correct, and I'm right here. two are correct and two are false. That is the correct answer. What is your source for the supposedly correct answers? What entrance exam? For entering a master's program? – Henno Brandsma Sep 22 '19 at 06:40
  • okkk now i got this. – gaurav saini Sep 22 '19 at 07:02
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Because $d_\infty$ is in fact normable, and all norms on $\mathbb R^n$ are equivalent, so open sets in the $d_\infty$-topology are the open sets in euclidean topology.

Nick
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