Two metrics:
$\begin{equation} d_P(x,y) := \begin{cases} ||x-y||_2 & \text{if x and y lie on the same ray from the origin }\\ ||x||_2+||y||_2 & \text{otherwise} \end{cases} \end{equation}$
and
$\begin{equation} d_D(x,y) = \begin{cases} 1 & \text{if x ≠ y}\\ 0 & \text{if x = y} \end{cases} \end{equation}$
How to prove:
If $(x_n)$ is a sequence in $\mathbb{R^2}$ then: $x_n$ converges to $x_0 \in \mathbb{R^2}$ with respect to $d_D$ $\Rightarrow$ $x_n$ converges to $x_0 \in \mathbb{R^2}$ with respect to $d_P$
I tried with: $\lim\limits_{n\to\infty}d_D(x_n,x_0)=0 \to \lim\limits_{n\to\infty}d_P(x_n,x_0)=0$
But i don't know if it works. Also, what if $x≠y$?
$d_D$ is the discrete metric. If $(x_n)$ converges to $x_0$ in the discrete metric, it means the sequence is eventually $x_0$, which certainly implies the sequence also converges in $d_p$.