# Showing $p(x,y) = \frac{d(x,y)}{1 + d(x,y)}$ is bounded by $1$

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 $X$.

1. Show $p(x,y)$ is is bounded by $1$. (i.e. $p(x,y) \le 1$ for all $x,y$).

2. 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 $x$.

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Well, first I'd use something other than $n$ for notation...but you can note $0\leq1/(1+n)\leq 1$ for all $n\geq0$, always, forever, in this or any other universe. That would prove (1). –  Alex Nelson Nov 25 '12 at 1:48
We always have $d(x,y)\ge 0$. Now $\frac{d(x,y)}{1+d(x,y)}$ is a non-negative number, divided by a bigger non-negative number. Of course it is $\lt 1$. –  André Nicolas Nov 25 '12 at 2:11

Note that

$$\frac{d(x,y)}{1 + d(x,y)} \leq 1 \Longleftrightarrow d(x,y) \leq 1 + d(x,y) \Longleftrightarrow 0 \leq 1$$

Note that $\frac{1}{n} \to 0$ but $\frac{1}{n} \not\leq 0$ for any $n$, so your proof does not suffice.

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Nice the way you showed the metric is bounded by $1$ independent of the $x$ and $y$ chosen. What do I have to show regarding $\frac{1}{n} \not\leq 0$ for any $n$? –  sonicboom Nov 25 '12 at 2:04
My point with $\frac{1}{n} \not\leq 0$ was that just because a limit converges to $0$ does not mean it is bounded by $0$. Similarly, although you showed that $p(x,y) \to 1$, that is not enough to conclude that it is bounded by $1$. –  Isaac Solomon Nov 25 '12 at 2:39
$\frac{1}{n}$ is approaching from the right so it is bounded below by $0$, i.e. $0 \le \frac{1}{n}$ for all $n \in \mathbb{N}$. In my post I am approaching from the left so the implication is that $\frac{1}{\frac{1}{n} + 1}$ is bounded above by $1$. –  sonicboom Nov 25 '12 at 12:15

The fact that $\dfrac{n}{1+n}\to1$ as $n\to\infty$ does not prove that $\dfrac{n}{1+n}\le 1$. And you don't need limits for that. If $A,B$ are positive numbers and $A<B$, then $\dfrac AB\le 1$.

The usual definitions of metrics do not allow points to be infinitely far apart, so you don't really need to say that. However, in some contexts it could make sense to modify the definition. For example, in some contexts one might define the distance from $x$ to $y$ to be the greatest lower bound of all lengths of paths from $x$ to $y$, and then if there's no path from $x$ to $y$ then the distance is $\infty$ because the greatest lower bound of the empty set is $\infty$.

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