problem 1 page # 126 (a) in R n , define d\'(X,Y)=abs(x 1 -x 2 )+....+abs(x n -y
ID: 2939605 • Letter: P
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problem 1 page # 126 (a) in Rn , define d'(X,Y)=abs(x1-x2)+....+abs(xn-yn) show that d' is a metric that induces the usual topology ofRn. sketch the basis elements under d' when n=2. (b) more generally, given p>= 1, define d'(X,Y)=[abs(xi-yi))p]1/p for X,Y belongs to R^n . assume that d' is a metric . showthat it induces the usual topology on R^n. problem 1 page # 126 (a) in Rn , define d'(X,Y)=abs(x1-x2)+....+abs(xn-yn) show that d' is a metric that induces the usual topology ofRn. sketch the basis elements under d' when n=2. (b) more generally, given p>= 1, define d'(X,Y)=[abs(xi-yi))p]1/p for X,Y belongs to R^n . assume that d' is a metric . showthat it induces the usual topology on R^n. for X,Y belongs to R^n . assume that d' is a metric . showthat it induces the usual topology on R^n.Explanation / Answer
For part (a) We need to show that d defines a metric on R^n. This involvesshowing three things: For x,y,z in R^n (1) d(x,y)>0 with d(x,y)=0 iff x=y (2) d(x,y)=d(y,x) (3)d(x,y)<(or equal to) d(x,z)+d(z,y) For (1), d(x,y)>0 since the expression is just the sum ofabsolute values and it is equal to zero if and only if every xi=yifor i=1,2,..,n . In other words, when x=y For (2) d(x,y)=|x1-y1|+|x2-y2|+...+|xn-yn|=|y1-x1|+|y2-x2|+...+|yn-xn|=d(y,x) (since|-a|=|a|) For (3) d(x,y)=|x1-y1| +|x2-y2|+...+|xn-yn|=|x1-z1+z1-y1|+|x2-z2+z2-y2|+...+|xn-zn+zn-yn| <(or equal to)|x1-z1|+|z1-y1|+|x2-z2|+|z2-y2|+...+|xn-z2|+|zn-yn|=d(x,z)+d(z,y) (from the triangle inequality |a+b|<(or equal to) |a|+|b|) For part(b) You have to use Minkowski's Inequality(here is a link to ithttp://en.wikipedia.org/wiki/Minkowski_inequality) It's pretty much the same formulation as in part(a), exceptthat third part(3) you need to use the inequality.Related Questions
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