I am currently doing a problem involving the proof of convergence of sequence in $\Bbb R^{2}$. If we have a sequence in $\Bbb R^{2}$ $(e^{\frac{1}{n}}, \frac{1}{n})$, $n=1,2,\ldots$. How can we prove that it converges or not in the taxicab metric.
where metric defined in $\Bbb R^{2}$ as $\sum_{i=1}^{2}|x_{i}- y_{i}|$ We know that fact that the convergence requires:
$\forall \varepsilon >0; \exists N \in \Bbb N, \forall m, m>N : d(x_{m}, x_{0}) < \varepsilon $ where $X_{0}= (x_{0},y_{0}) $ is the limit of the sequence. Can we show that for individual coordinates converge so it should be the case that sequence converge to limits of those individual coordinates ?
Yes, for this metric it is necessary and sufficient that the coordinate sequences converge; note that $e^{\frac{1}{n}} \to 0$ in $\Bbb R$ and $\frac{1}{n} \to 0$ as well, so we use $(0,0)$ as the candidate limit.
Given $\varepsilon >0$ we find $N_1$ such that
$$\forall n > N_1: |e^{\frac{1}{n}}| < \frac{\varepsilon}{2}$$
and also $N_2$ such that
$$ \forall n > N_2: |\frac{1}{n}| < \frac{\varepsilon}{2}$$
from the two mentioned convergences in $\Bbb R$.
Then for all $n > \max(N_1,N_2)$:
$$d(\left( e^{\frac{1}{n}}, \frac{1}{n}\right), (0,0)) = | e^{\frac{1}{n}}| + |\frac{1}{n}| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$
so we have total convergence of the sequence to $(0,0)$.
You can see this idea will work for all sequences for which we know the component limits, in all (finite) dimensions.