#2

Post
by **Papapetros Vaggelis** » Sat Jul 15, 2017 6:59 pm

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.