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, \(\displaystyle{d_{m}(x,y)=f(xy)\geq 0\,,\forall\,x\,,y\in\mathbb{R}}\) and
\(\displaystyle{d_{m}(x,y)=0\iff f(xy)=0\iff xy=0\iff x=y}\). Also,
\(\displaystyle{d_{m}(x,y)=f(xy)=f(yx)=d_{m}(y,x)\,,\forall\,x\,,y\in\mathbb{R}}\)
Finally, if \(\displaystyle{x\,,y\,,z\in\mathbb{R}}\), then \(\displaystyle{xz\leq xy+yz}\), so,
\(\displaystyle{d_{m}(x,z)=f(xz)\leq f(xy+yz)\leq f(xy)+f(yz)=d_{m}(x,y)+d_{m}(y,z)}\).
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.
