A problem about equivalent metric

[prophetlmn]

Given X=R∞ and its element be squences
let d1(x,y)=sup|xi-yi|
let d∞(x,y)=Ʃ|xi-yi|
then there exists some some x(k) which convergences to x by d1
but not by d∞ ,for example let x be the constant squence 0,
i.e xn=0 ,and let
x(k)n=(1/k2)/(1+1/k2)n
then d1(x(k),x)=1/k2
and d∞(x(k),x)=1+1/k2 by the use of geometry seire,
so as k goes to ∞, Lim d1(x(k),x)=0
Lim d∞(x(k),x)=1
my problem is that--- is it possible prove this without such counter example but merely prove the existence of such xk by general consideration on topology?
i.e if two metric are equivalent ,then they induce same topology,and same convergence properties. As a special case if there exist positive A B,such that for all x,y
Ada(x,y)≤db(x,y)≤Bda(x,y) then these two metric da and db are equivalent,it's obvious here d1(x,y)≤d∞(x,y),then we could have A=1 here,but for any postive M, there exist x,y such that
Md1(x,y)≤d∞(x,y) so no B exist
does this considerations led to the existence of x(k) in the counter example?
or Is Ada(x,y)≤db(x,y)≤Bdb(x,y) a necessary condition for da and db to be equivalent?
the original problem is from Tao‘s Real Analysis chapter12 Metric space ，the counter example is from Rudin’s Principles of mathematical analysis example 7.3 where he use a slightly different form and use it to show the limit of a squence of continuous function is not continous

 

[Office_Shredder]

but for any postive M, there exist x,y such that
Md1(x,y)≤d∞(x,y) so no B exist

To prove that this is true you have to find a sequence of x's and y's with d1(x,y) converging to zero, and d∞(x,y) not converging to zero. Which is essentially what you've done anyway

 

[micromass]

let d1(x,y)=sup|xi-yi|
let d∞(x,y)=Ʃ|xi-yi|

This is not the standard terminology. Usually, we swap these two metrics!

 

[prophetlmn]

To prove that this is true you have to find a sequence of x's and y's with d1(x,y) converging to zero, and d∞(x,y) not converging to zero. Which is essentially what you've done anyway

yes，but i mean another kind of proof，as i know from one of my friend if the metrics are compact then the condtion for A and B is necessary，while in general it‘s not，I have nearly got a proof about this yesterday，but tonight I found there is a small mistake in my proof，and l still have not got a counter example in the non compact case yet