How Do Different Metrics Affect Convergence and Divergence of Sequences?

[ozkan12]

Let $X=R$ and ${d}_{1}\left(x,y\right)=\frac{1}{\eta}\left| x-y \right|$ $\eta\in \left(0,\infty\right)$ and ${d}_{2}\left(x,y\right)=\left| x-y \right|$..By using ${d}_{1}$ and ${d}_{2}$ please show that ${x}_{n}=\left(-1\right)^n$ is divergent and ${x}_{n}=\frac{1}{n}$ is convergent...

 

[Euge]

Hello again ozkan12,

You should show what you've tried or at least mention you don't know where to begin so that we know you've attempted the problem. For now, here are some things to consider. The metric $d_1$ is a constant multiple of $d_2$, so a sequence converges in $(X,d_1)$ if and only if it converges it $(X,d_2)$. Hence, it suffices to work with just the metric $d_2$.

 

[ozkan12]

Dear Euge,

Yes, it suffices to work with metric ${d}_{2}$...I found somethings, I sent these things...Please can you check it ?

Firstly, we will use metric ${d}_{2}$ and discuss on sequence ${x}_{n}=\frac{1}{n}$.

Let $\varepsilon>0$ and ${n}_{0}>\frac{1}{\varepsilon}$ (i.e the smallest integer that is larger than $\frac{1}{\varepsilon}$.)

İf, $n\ge{n}_{0}$

$n\ge{n}_{0}>\frac{1}{\varepsilon}$ $\implies$ $\frac{1}{n}\le\frac{1}{{n}_{0}}<\frac{1}{\varepsilon}$

Then, $d\left({x}_{n},0\right)=\frac{1}{\eta}\left| {x}_{n}-0 \right|=\frac{1}{\eta}\left| {x}_{n} \right|=\frac{1}{\eta}\frac{1}{n}\le\frac{1}{\eta}\frac{1}{{n}_{0}}<\frac{1}{\varepsilon.\eta}$.

But, I haven't a opinion related to sequence $x_n=(-1)^n$...Please can you check first one and help me related to $x_n=(-1)^n$

 

[Euge]

Assuming $d = d_1$, what you have is almost correct -- the two last $(1/\epsilon)$'s you have should be replaced with $\epsilon$. As for the case $x_n = (-1)^n$, consider that $d(x_n,x_m) = 2/\eta$ if $n + m$ is odd and $0$ otherwise. Hence, if $\epsilon = 1/\eta$, then given $N\in \Bbb N$, $d_1(x_{N+1},x_N) = 2/\eta > \epsilon$. Therefore, the sequence $(x_n)$ is not Cauchy in $(X,d_1)$. This implies $(x_n)$ is divergent in $(X,d_1)$.

 

[ozkan12]

Dear Euge

Thank you for your attention...Best wishes...:) Have you any information related to modular metric spaces ?