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

 

[blue_raver22]

It is possible to prove the existence of x(k) without using a counter example, by using general considerations on topology. As mentioned, if two metrics are equivalent, they induce the same topology and have the same convergence properties. In this case, we can show that d1 and d∞ are not equivalent by showing that they do not induce the same topology.

To show this, we can assume that they are equivalent and use the fact that equivalent metrics have the same convergence properties. Then, we can show that the sequence x(k) does not converge to x by d∞, which contradicts the assumption that d1 and d∞ are equivalent.

To do this, we can use the definition of convergence in a metric space: a sequence x(k) converges to x if for every ε>0, there exists an N such that for all k≥N, d(x(k),x)<ε. Now, let's consider the sequence x(k) defined in the problem. We can see that for any ε>0, there exists an N such that for all k≥N, d∞(x(k),x)<ε. However, for d1(x(k),x) to be less than ε, we would need k to be greater than 1/ε, which can never be satisfied for all ε>0. Therefore, x(k) does not converge to x by d∞, which contradicts the assumption that d1 and d∞ are equivalent.

This shows that the existence of x(k) in the counter example is not a result of the considerations on topology, but rather a result of the specific properties of d1 and d∞. The condition Ada(x,y)≤db(x,y)≤Bdb(x,y) is not a necessary condition for two metrics to be equivalent, as there are other ways to show that two metrics are not equivalent.

In conclusion, the existence of x(k) in the counter example is a result of the specific properties of d1 and d∞, and not a result of the considerations on topology. The condition Ada(x,y)≤db(x,y)≤Bdb(x,y) is not a necessary condition for two metrics to be equivalent, as there are other ways to show that two metrics are not equivalent. The original problem from Tao's Real Analysis can be solved without using a counter example, by using general considerations on topology.