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Endow $R^2$ with the metric $d(a,b)$ ={ $max{|a_1-b_1|,|a_2-b_2|}$} where $a$=$(a_1,a_2)$ and $b$=$(b_1,b_2)$.

Show that $S$={${a \in R^2|a_1^2+a_2^2<1}$} is open in $R^2$ with this metric.

$S$ is definitely open with respect to the euclidean metric in $R^2$,but how can i show that it is open with respect to the metric given in this question?

johny
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2 Answers2

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Hint: Around each point, can you construct a square, thereby proving under your metric the set is open?

Clayton
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  • $0$ will definitely be an interior point of the set $S$,so if i construct a square around $0$ which is an arbitrary point of the set $S$ will i be done ? – johny Oct 22 '13 at 12:47
  • @johny $(0,0)$ is not an arbitrary point of $S$. – Pedro Oct 22 '13 at 13:04
  • @PedroTamaroff Yes you are right, it is not.My mistake. – johny Oct 22 '13 at 13:10
  • @johny: Take an arbitrary point $(x,y)\in S$. How far away from the boundary of $S$ is the point under the Euclidean metric? Can you use that information to determine an appropriate distance so that the point is inside a square? – Clayton Oct 22 '13 at 13:16
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For each point $x$ in $S$, the distance from $x$ to the boundary of the circle is actually $|a|-|x|$. Put it in another way. We draw a circle $C$ centered at $x$ with radius $|a|-|x|$. For each point $y$ in $C$, $|y|\le |y-x|+|x| \le |a|-|x|+|x|=|a|$. This means $y$ lies in $S$, and $C$ is contained in $S$. Therefore, $S$ is an open set.

hb12ah
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