- #1
prophetlmn
- 18
- 0
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 positive 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
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 positive 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