Let $(X,d_1)$ and $(Y,d_2)$ be metric spaces. Consider the metric $d:(X\times Y)\times (X\times Y)\to \mathbb{R}$ defined by $d((x_1,y_1),(x_2,y_2))=max \{{d_1(x_1,x_2),d_2(y_1,y_2)}\}$.Let $p\in X$,$q\in Y$ and $r$,$s$>$0$.Show that $B(p,r)\times B(q,s)$ is an open set in $(X \times Y,d)$.
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To prove it nothing other than the definition of open set is used :); try it. – Yes Aug 25 '16 at 03:38
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For brevity let $B_{d_1}(p,r)\times B_{d_2}(q,s)=A.$
For $(x,y)\in A$ let $r_{x,y}=\min (\; r-d_1(x,p),\;s-d_2(y,q)\;),$ and let $U(x,y)=B_{d_1}(x, r_{x,y})\times B_{d_2}(y,r_{x,y}).$
Then $(x,y)\in U(x,y)=B_d (\;(x,y),r_{x,y})$, and by the triangle inequality we have $U(x,y)\subset A$. So we have $$A=\cup_{(x,y)\in A}\{(x,y)\}\subset \cup_{(x,y)\in A}U(x,y)\subset A.$$
So $A=\cup_{(x,y)\in A}U(x,y)$ is a union of open d-balls.
DanielWainfleet
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Thank you very much. People on this site are very arrogant because they know everything and they assume everyone should know as well. – user114873 Aug 25 '16 at 10:52
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2Some are like that. They forget that the users include people of every age and background. On MathOverload, which is for the professionals, they do know a lot and they're not arrogant about it. – DanielWainfleet Aug 25 '16 at 14:22
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Yeah. I consider myself average in maths.I know maths it doesn't mean I can solve every problem in it.Read the biography of Évariste Galois in Gallian's book. – user114873 Aug 25 '16 at 15:09