### On the mole metric

Posted: **Fri Jul 07, 2017 6:41 am**

by **Riemann**

Let us define the following function:

$$f(x) = \left\{\begin{matrix}

x & , & 0 \leq x <1 \\

1& , & x >1

\end{matrix}\right.$$

as well as $d_m(x, y) = f ( \left| x -y \right| )$.

- Show that $d_m$ is a metric on $\mathbb{R}$.
*You may call it the mole metric. If points are close (closer than one meter), their distance is the usual one, but are they far apart (more than one meter) we do not distinguish between their distances; they are just far apart.*
- Show that $\mathbb{R}$ endowed with the above metric is complete and bounded but not compact. Is it totally bounded? Why / Why not?

### Re: On the mole metric

Posted: **Sat Jul 15, 2017 6:59 pm**

by **Papapetros Vaggelis**

Hi Riemann.

We observe that \(\displaystyle{0\leq f(x)\leq 1\,,\forall\,x\geq 0\,,0\leq f(x)\leq x\,,\forall\,x\geq 0}\)

and that \(\displaystyle{f}\) is strictly increasing at \(\displaystyle{\left[0,+\infty\right)}\). Also,

\(\displaystyle{f(x+y)\leq f(x)+f(y)\,,\forall\,x\,,y\geq 0}\) and \(\displaystyle{f(x)=0\iff x=0}\).

Now, \(\displaystyle{d_{m}(x,y)=f(|x-y|)\geq 0\,,\forall\,x\,,y\in\mathbb{R}}\) and

\(\displaystyle{d_{m}(x,y)=0\iff f(|x-y|)=0\iff |x-y|=0\iff x=y}\). Also,

\(\displaystyle{d_{m}(x,y)=f(|x-y|)=f(|y-x|)=d_{m}(y,x)\,,\forall\,x\,,y\in\mathbb{R}}\)

Finally, if \(\displaystyle{x\,,y\,,z\in\mathbb{R}}\), then \(\displaystyle{|x-z|\leq |x-y|+|y-z|}\), so,

\(\displaystyle{d_{m}(x,z)=f(|x-z|)\leq f(|x-y|+|y-z|)\leq f(|x-y|)+f(|y-z|)=d_{m}(x,y)+d_{m}(y,z)}\).

Therefore, the function \(\displaystyle{d_m}\) is a metric on \(\displaystyle{\mathbb{R}}\) and then

\(\displaystyle{\left(\mathbb{R},d_{m}\right)}\) is bounded since \(\displaystyle{0\leq d_{m}(x,y)\leq 1\,,\forall\,x\,,y\in\mathbb{R}}\).

The metric space \(\displaystyle{\left(\mathbb{R},d_{m}\right)}\) is not compact.

Indeed, suppose that \(\displaystyle{\left(\mathbb{R},d_{m}\right)}\) is compact. Consider the

sequence \(\displaystyle{\left(x_n=n\right)_{n\in\mathbb{N}}}\). There exists a strictly increasing

sequence of natural numbers \(\displaystyle{\left(k_n\right)_{n\in\mathbb{N}}}\) such that the

subsequence \(\displaystyle{\left(x_{k_{n}}=k_n\right)_{n\in\mathbb{N}}}\) converges.

If there exists \(\displaystyle{n\in\mathbb{N}}\) such that \(\displaystyle{0\leq k_{n+1}-k_{n}<1}\)

then,

\(\displaystyle{k_{n+1}=k_{n}}\), a contradiction, so,

\(\displaystyle{k_{n+1}-k_n\geq 1\,,\forall\,n\in\mathbb{N}\implies d_{m}(k_{n},k_{n+1})=1\,,\forall\,n\in\mathbb{N}}\)

which means that \(\displaystyle{(k_{n})_{n\in\mathbb{N}}}\) is not a \(\displaystyle{d_{m}}\)

Cauchy sequence, a contradiction, since \(\displaystyle{(k_{n})_{n\in\mathbb{N}}}\). converges.

**Comment : I can't prove the completeness.**

The metric space \(\displaystyle{\left(\mathbb{R},d_{m}\right)}\) is not totally bounded since if

\(\displaystyle{\left(\mathbb{R},d_{m}\right)}\) is totally bounded and also is complete, then we have

compactness, a contradiction.