Let $X$ bet a metric space with metric $d$.
Prove that $p(x,y) = \frac{d(x,y)}{1 + d(x,y)}$ defines a metric on $X$.
Show $p(x,y)$ is is bounded by $1$. (i.e. $p(x,y) \le 1$ for all $x,y$).
Given $x \in X$, determine $B_1^p$.
I can prove it is a metric, I am just wondering about the other parts.
Here's what I have -
Want to show $p(x,y) = \frac{d(x,y)}{1 + d(x,y)} \le 1$ for all $x,y$. Let $n = d(x,y)$ and we have $p(x,y) = \frac{n}{1 + n} = \frac{1}{\frac{1}{n} + 1}$
Let $n \to \infty$ and $\frac{1}{\frac{1}{n} + 1} \to 1$ So $p(x,y)$ is bounded by $1$. Is that sufficient or is there something else I need to show?
Then for part 2. I would say that $p(x,y)$ is in fact strictly less than $1$ as it only equals $1$ when $x$ and $y$ are infinitely far apart. Hence the open ball $B_1^p(x)$ will contain all points in $X$ as every point $y \in X$ will be less than $1$ from $x$.