Given X=R(adsbygoogle = window.adsbygoogle || []).push({}); ^{∞}and its element be squences

let d_{1}(x,y)=sup|x_{i}-y_{i}|

let d_{∞}(x,y)=Ʃ|x_{i}-y_{i}|

then there exists some some x^{(k)}which convergences to x by d_{1}

but not by d_{∞},for example let x be the constant squence 0,

i.e x_{n}=0 ,and let

x^{(k)}n=(1/k^{2})/(1+1/k^{2})^{n}

then d_{1}(x^{(k)},x)=1/k^{2}

and d_{∞}(x^{(k)},x)=1+1/k^{2}by the use of geometry seire,

so as k goes to ∞, Lim d_{1}(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 x^{k}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

Ad_{a}(x,y)≤d_{b}(x,y)≤Bd_{a}(x,y) then these two metric d_{a}and d_{b}are equivalent,it's obvious here d_{1}(x,y)≤d_{∞}(x,y),then we could have A=1 here,but for any postive M, there exist x,y such that

Md_{1}(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 Ad_{a}(x,y)≤d_{b}(x,y)≤Bd_{b}(x,y) a necessary condition for d_{a}and d_{b}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 continous function is not continous

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# A problem about equivalent metric

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