MHB Limit of Absolute Values and Metric Spaces

ozkan12
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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 :)
 
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ozkan12 said:
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 ?
 
Yes, $d$ correspond to metric on $(X,d)$ metric space.
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...
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