Q:Consider the metric space $(R, d)$ where $d$ is the distance defined by $d(x, y) = min\{1,|x−y|\}$,∀x, y∈R. Let $(x_n)_1^\infty$ be a sequence in R, and l∈R. Show that if the sequence $(x_n)_1^\infty$ converges to $l \in(R, d)$, then $(x_n)_1^\infty $converges to l in R with the Euclidean/usual distance.
The question is above now and then my thoughts are we can just use the defintion:
$(x_n) \rightarrow x$ iff $\forall \epsilon > 0 \exists N \in \mathbb{N}$ s.t. $$d(x_n,x) < \epsilon \ when \ n \geq N$$
if we do this for the first case with specified space we get:
$d(x_n) = min \{1,|x_n - x| \} < \epsilon$ whenever $n \geq N$.
Once I have this I'm a bit stuck on how to progress further maybe I could use the fact the minimum would always be $>0$. Since distance are by definition larger than $0$ and then somehow relate it to the fact
$|x_n - x| < \epsilon \ whenever \ n\geq N$
But overall, I'm stuck and I would appreciate some guidance on this thank you!
Hint. If $(x_n)_{n\in\Bbb N}$ convergesto $l$ with respect to $d$ OR converges to $l$ with respect to the usual metric then there exists $n_0\in\Bbb N$ such that $$\forall m,n >n_0\,(|x_n-x_m|<1 \;\land\; |x_n-l|<1).$$