Convergence of a sequence in $\mathbb{R}^n$ with the $d_p$ metric.

Question

Consider the metric space $(X=\mathbb{R}^n,d_p)$, where $d_p(\tilde{x},\tilde{y})=\left(\sum_{i=1}^{n}|x_i-y_i|^p\right)^{1/p}\;\; (p\geq 1)$. We consider a sequence of elements in this metric space, say $(\tilde{x}^{k})$. I would like to prove that convergence in the metric implies point-wise convergence (the converse I have dealt with). If $(\tilde{x}^{k})\in \mathbb{R}^n $ converges to $\tilde{x}\in \mathbb{R}^n $, then for every $\epsilon>0\;\exists n_0(\epsilon)\in\mathbb{N}:$ \begin{align*} \left(\sum_{i=1}^{n}|x^{k}_i-x_i|^p\right)^{1/p}<\epsilon\;\;\;\;\;\;\text{for}\;k\geq n_0 \end{align*} This means that \begin{align*} |x_{1}^{k}-x_1|^p+|x_{2}^{k}-x_2|^p+\cdots|x_{n}^{k}-x_n|^p<\epsilon^{p}\;\;\;\;\;\;\text{for}\;k\geq n_0 \end{align*} After this step, how do I show that point-wise convergence takes place. I am confused as to how to reduce the finite sum to one term. Perhaps we could fix the $i$?

Answer 1

For every $p>0$, and every choice of $x_i$, we have $$\max_i|\xi_i|^p\leq\sum_{i=1}^n|\xi_i|^p$$ applying this, in particular, to $\xi_i=x_i^k-x_i$, you immediately get that if the r.h.s tends to zero then for each $1\leq i\leq n$, the sequence $\{x_i^k\}_{k=1}^{\infty}$ tends to $x_i$.