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I need to prove that

$$B(a,r) = B(a_1, r)\times \cdots \times B(a_n,r)$$

in $M=M_1\times\cdots \times M_n$

where $M_i$ is a metric space and the metric is $d''(z,z') = \max\{d_i(x_i,y_i), i \in \{1,\ldots,n\}\}$ where $d_i$ is the metric for $M_i$.

I need to prove the result above also for closed balls, but I can't understand what does this means. Can somebody show me in $\mathbb{R}^2$?

Andrés E. Caicedo
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Guerlando OCs
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  • I will just give a brief description. Note that on your metric, the open (closed) balls are squares. So, in $\mathbb{R}^2$, an open (closed) square is the product of open (closed) intervals, it extends to any metric space. Try to prove by induction... For the first case you know what to do. Suppose that it is true for $M_1\times\ldots M_{n-1}.$ You know that by the induction hypothesis, any ball is of the way you expect, but one open ball on $M_n$ is a square, then $(M_1\times\ldots M_{n-1})\times M_n$ is equal the case where you have two factors... – L.F. Cavenaghi Mar 17 '16 at 18:13
  • @user wow, too abstract for me :c don't even have an idea on how to prove it – Guerlando OCs Mar 17 '16 at 18:16
  • I am sorry, I have a better proof. I will answer it. – L.F. Cavenaghi Mar 17 '16 at 18:28

2 Answers2

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$B(a,r) \subset B(a_1,r)\times \ldots \times B(a_n,r):$ Take $x \in B(a,r).$ Then $\max_i d_i(x_i,a) < r.$ But then, $d_i(x_i,a) < r$ for each $i \in \{1,\ldots,n\}.$ Then $x_i \in B(a_i,r),$ so $x \in B(a_1,r)\times \ldots \times B(a_n,r).$

$B(a_1,r)\times \ldots \times B(a_n,r) \subset B(a,r):$

Take $x \in B(a_1,r)\times \ldots \times B(a_n,r),$ then $d_i(x_i,a) < r$ for each $i.$ What means that $\max_i d(x_i,a) < r$, and then, $x \in $B(a,r).$

L.F. Cavenaghi
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Example in $R^2 :\;$ With the usual metric on $R$, we have $B(x,r)=\{y\in R:|y-x|<r\}.$

For any points $p=(a_1,a_2)$ and $p'=(a'_1,a'_2)$ in $R^2$ let $d(p,p')=\max (|a_1-a'_1|,|a_2-a'_2|).$

Let $r\in R$ and $p=(a_1,a_2)\in R^2$. Then for any $b=(b_1,b_2)\in R^2$ we have $$b\in B(p,r)\iff d(p,b)<r\iff \max (|a_1-b_1|,|a_2-b_2|)<r\iff$$ $$\iff (\;|a_1-b_1|<r\land |a_2-b_2|<r\;)\iff$$ $$\iff (\;b_1\in B(a_1,r)\land (b_2\in B(a_2,r)\;)\iff$$ $$\iff b=(b_1,b_2)\in B(a_1,r)\times B(a_2,r).$$

DanielWainfleet
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