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I have the sequence $ x_n:=\left(1-n^{-n}, 2-e^{-n}, 3-2^{-n}\right) $ in $ \mathbb{R}^3 $ and consider the discrete metric $ d(v,w)=\begin{cases}0,\quad v=w\\1,\quad v\neq w\end{cases} $.

I want to show that this sequence does not converges to $ a:=(1,2,3)\in \mathbb{R}^3 $ under the discrete metric. My plan is to negate the definition of convergence to $$ \exists \varepsilon > 0\quad \forall N_{\varepsilon}\in \mathbb{N}\quad \exists n\geq N_{\varepsilon}: \ d(a,x_n)\geq \varepsilon $$ and show that this statement is true:

Chose $ \varepsilon = 1 $, let be $ N_{\varepsilon}\in \mathbb{N} $ arbitrary and chose $ n=N_{\varepsilon} $. Then we get $ d(a,x_{N_{\varepsilon}})=1=\varepsilon $.

Is that right? If not what went wrong?

hallo007
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    What you have done is correct. – Kavi Rama Murthy Sep 12 '20 at 00:26
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    Note that with equal ease you can prove the general result that in a metric space with the discrete metric, $\langle x_k:k\in\Bbb N\rangle$ converges to $y$ iff the sequence is eventually constant, i.e., iff there is an $n\in\Bbb N$ such that $x_k=y$ for all $k\ge n$ – Brian M. Scott Sep 12 '20 at 00:30

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