# Thread: under standing metric spaces and cauchy sequences... (analysis/topology)

1. ## under standing metric spaces and cauchy sequences... (analysis/topology)

hmm so i have this metric defined by x,y in R^k such that d_1(x,y)=max{abs (x_j-y_j):j=1,2,...,k} and d_2(x,y)= sigma (from j=1 to k) abs (x_j-y_j). i did the first part of the question which was to verify that these are indeed metrics for R^k. the second part of the question asks to prove that d_1 and d_2 are complete. from the book definition, the metric space (S,d) is said to be complete if every cauchy sequence in S converges to some element in S. the hint given is to consider d_1(x,y)<=d(x,y)<=sqrt(k)d_1(x,y) where d is the standard metric defined by d(x,y)=sqrt (sigma from j=1 to k) (x_j-y_j)^2. can i get some help? thanks !

2. Originally Posted by squarerootof2
hmm so i have this metric defined by x,y in R^k such that d_1(x,y)=max{abs (x_j-y_j):j=1,2,...,k} and d_2(x,y)= sigma (from j=1 to k) abs (x_j-y_j). i did the first part of the question which was to verify that these are indeed metrics for R^k. the second part of the question asks to prove that d_1 and d_2 are complete. from the book definition, the metric space (S,d) is said to be complete if every cauchy sequence in S converges to some element in S. the hint given is to consider d_1(x,y)<=d(x,y)<=sqrt(k)d_1(x,y) where d is the standard metric defined by d(x,y)=sqrt (sigma from j=1 to k) (x_j-y_j)^2. can i get some help? thanks !
1)$\displaystyle d_1$ is complete. Let $\displaystyle \{ \bold{x}_n \}$ be a Cauchy sequence, so it means, $\displaystyle d(\bold{x}_n , \bold{x}_m) < \epsilon$ whenever $\displaystyle n,m>N$. Thus, $\displaystyle \max \{ |x^{(i)}_n - x^{(i)}_m | \} < \epsilon$. This means that, $\displaystyle | \bold{x}_n - \bold{x}_m| < \sqrt{k}|\max\{ |x^{(i)}_n - x^{(i)}_m| \} < \epsilon \sqrt{k}$. Thus, we see that if $\displaystyle \{ \bold{x}_n\}$ is a Cauchy with respect to $\displaystyle d_1$ then it is Cauchy with respect to the ordinary Euclidean metric. But Euclidean metric spaces are complete, thus $\displaystyle d_1$ is also complete.