Proving existence of metric spaces

Question

Show that the pair $(\mathbb{N}, d)$, where $d:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{R}$ is defined by $$ d(u,v)=\frac{|u-v|}{uv} $$ for $u,v\in\mathbb{N}$, is a metric space.

I have managed to prove the first two conditions of a metric (equality and symmetry). I have trouble with proving the triangular condition. Here is what I came up with $$ d(x,y) + d(y,z) = \frac{|x-y|}{xy} + \frac{|y-z|}{yz} = \frac{z\,|x-y| + x\,|y-z|}{xyz} $$ and $$ d(x,z) = \frac{|x-z|}{xz} = \frac{|x-y+y-z|}{xz} \leq \frac{|x-y|+|y-z|}{xz} $$

We can clearly see from that the following inequation is true: $$ \frac{|x-z|}{xz} \leq \frac{|x-y|+|y-z|}{xz} $$

I do not know if there is a way to continue the further prove. Am I doing it in the right way? How can I prove this third condition?

Answer 1

Note that \begin{eqnarray} d (n_1, n_2) & = & \left| \frac{1}{n_1} - \frac{1}{n_2} \right| . \nonumber \end{eqnarray} Therefore, for every $n_1,n_2,n_3\in\mathbb{N}^+$ one has \begin{eqnarray} d (n_1, n_2) & = & \left| \left( \frac{1}{n_1} - \frac{1}{n_3} \right) + \left( \frac{1}{n_3} - \frac{1}{n_2} \right) \right| \nonumber\\ & \leqslant & \left| \frac{1}{n_1} - \frac{1}{n_3} \right| + \left| \frac{1}{n_3} - \frac{1}{n_2} \right| \nonumber\\ & = & d (n_1, n_3) + d (n_3, n_2) \nonumber .\end{eqnarray}

Answer 2

Note that $d(n,m)=|\frac1n-\frac1m|$ so instead of considering $n,m \in \mathbb{N}^+$ you can consider $x,y \in \{\frac1n : n \in \mathbb{N}^+\}$ and distance function $d^*(x,y)=|x-y|$ which we know satisfies the triangle inequality. Note that we didn't do anything fancy here, it's just the substitution $x=\frac1n$ to get the new metric on this new set.

Now since $d^*(x,z) \leq d^*(x,y) + d^*(y,z)$ so that $|x-z|\leq |x-y|+|y-z|$ and substituting back we have $|\frac1n-\frac1p|\leq |\frac1n-\frac1m|+|\frac1m-\frac1p|$ which proves $d(n,p) \leq d(n,m) + d(m,p)$ on $\mathbb{N}^+$.

Answer 3

Your $(\mathbb{N}, d)$ is a metric space by transport of structure from the subset $\{1/n\mid n∈\Bbb N\}$ of $\Bbb R$, equipped with the usual distance.

More explicitely: $$ d(u,v)=\left|{\frac1v-\frac1u}\right| $$ and $$\left|\frac1z-\frac1x\right|\le\left|\frac1z-\frac1y\right|+\left|\frac1y-\frac1x\right|,$$ hence $$d(x,z)\le d(y,z)+d(x,y).$$