On the Euclidean space Rn, define the norm ||x|| = forallx = (xi1,,xin). Prove t

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Use the fact that R is a Banach space. Consider a Cauchy sequence {x_n} in R^n, We are given ||x_p - x_q|| --> 0 as m,n --> infinity that is same as saying ||x_p - x_q||^2 --> 0 as m,n --> infinity let x_p = (x_p1,x_p2,...,x_pn), similarly x_q =(x_q1,x_q2,...,x_qn) ||x_p - x_q||^2 = (x_p1 - x_q1)^2 + (x_p2 - x_q2)^2 + ... + (x_pn - x_qn)^2 all are positive terms so if LHS goes to zero each of the terms on the right should go to zero. So, that implies when {x_n} is a Cauchy seq. in R^n its n components form n cauchy sequences in R. R is complete, i.e. every Cauchy Sequence in R converges in R. So each of the n component sequences converge in R. Let their respective limits be y1,y2,...y_n. So |x_ij - yj| --> 0 as i --> infinity for all j =1,2,..n. Call (y1,y2,...,yn) as y is in R^n. ||x_p - y||^2 = (x_p1 - y1)^2 + (x_p2 - y2)^2 + ... + (x_pn - yn)^2 Each term on the RHS goes to zero as p--> infinity So LHS must also go to zero as p--> infinity. Hence starting with every Cauchy sequence in R^n it converges in R^n, hence we are done. Have a nice day !!

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