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# NO CLUE!!! Tough metric spaces question!!! watch

1. hello there,
I am desperate for help with the following question. I was absent when all this material was covered!!! The question is as follows:

"Let d_1 be the metric on R^2 defined by d_1(x,y) = |x_1 - y_1| + |x_2 - y_2| . Prove that a sequence (x^n) [from n =1 to infinity] converges to x E R^2 in the metric space (R^2,d_1) iff both the sequence x_1^n tends to x_1 as n tends to infinity and the sequence x_2^n tends to x_2 as n tends to infinity. (Here x^n is the nth element of the sequence (x^n) [from n =1 to infinity] and x^n = (x_1^n, x_2^n) E R^2, i.e. x_1^n and x_2^n are the coordinates of x^n. "

Any help would be more than appreciated!!! As I say, I'm desperate for help on this one!!!

Thanks a lot for your time!!!
2. Well, let's see what we've got. Take a sequence {xi} in R2 (with xi=(xi,yi) say) and let's assume it tends to a limit x=(x,y) in R2. So, we have: given an ε>0, there exists an N such that for all n>N d1(xn,x) < ε. Then, by the definition of d1 we have:

|x - xn| + |y - yn| < ε

So given this ε>0, for all n>N we certainly have:

|x - xn| < ε
|y - yn| < ε

So, looks like we're done in one direction. Now what about the other? Well for {xi} given an ε1>0, there exists an N1 such that for all n>N1 d(xn,x) < ε1, similarly for for {yi} given an ε2>0, there exists an N2 such that for all n>N2 d(yn,y) < ε2. Now, if we take N = max(N1,N2), ε/2 = max(ε12). Then given ε>0 for all n>N we have:

|x - xn| + |y - yn| < ε

And wahey - things are looking good. Hope that helped.
3. thanks a lot for your help mate!!! And thanks for your time. Cheers, Beast.

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