On the mole metric

General Topology
Post Reply
User avatar
Riemann
Posts: 175
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

On the mole metric

#1

Post 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 $d_m(x, y) = f ( \left| x -y \right| )$.
  1. 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.
  2. Show that $\mathbb{R}$ endowed with the above metric is complete and bounded but not compact. Is it totally bounded? Why / Why not?
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: On the mole metric

#2

Post 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.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 1 guest