Limit of Absolute Values and Metric Spaces

[ozkan12]

Let $\lim_{{k}\to{\infty}}d\left({x}_{m\left(k\right)},{x}_{m\left(k\right)-1}\right)=\varepsilon$ and $\lim_{{k}\to{\infty}}d\left({x}_{n\left(k\right)},{x}_{m\left(k\right)}\right)=\varepsilon$...Can we say that $\lim_{{k}\to{\infty}}d\left({x}_{n\left(k\right)},{x}_{m\left(k\right)-1}\right)=\varepsilon$ by using$\left| d\left({x}_{n\left(k\right)},{x}_{m\left(k\right)-1}\right)-d\left({x}_{n\left(k\right)},{x}_{m\left(k\right)}\right) \right|\le d\left({x}_{m\left(k\right)},{x}_{m\left(k\right)-1}\right)$...Thank you for your attention...Best wishes :)

 

[Sudharaka]

Let $\lim_{{k}\to{\infty}}d\left({x}_{m\left(k\right)},{x}_{m\left(k\right)-1}\right)=\varepsilon$ and $\lim_{{k}\to{\infty}}d\left({x}_{n\left(k\right)},{x}_{m\left(k\right)}\right)=\varepsilon$...Can we say that $\lim_{{k}\to{\infty}}d\left({x}_{n\left(k\right)},{x}_{m\left(k\right)-1}\right)=\varepsilon$ by using$\left| d\left({x}_{n\left(k\right)},{x}_{m\left(k\right)-1}\right)-d\left({x}_{n\left(k\right)},{x}_{m\left(k\right)}\right) \right|\le d\left({x}_{m\left(k\right)},{x}_{m\left(k\right)-1}\right)$...Thank you for your attention...Best wishes :)

Is $d$ corresponds to a metric on a metric space ?

 

[ozkan12]

Yes, $d$ correspond to metric on $(X,d)$ metric space.