Is $\mathbb{N}$ a totally bounded metric space with this metric $d(a,b) = \sqrt{1-2\frac{\gcd(a,b)}{a+b}}$?

Question

Is $X:=\mathbb{N}$ a totally bounded ( https://en.wikipedia.org/wiki/Totally_bounded_space#Definition_for_a_metric_space ) metric space with this metric $d(a,b) = \sqrt{1-2\frac{\gcd(a,b)}{a+b}}$?

This metric plays a role in the formulation of the abc-conjecture:

https://mathoverflow.net/questions/352054/the-abc-conjecture-as-an-inequality-for-inner-products

(It was shown by @quasi here ( Is $\mathbb{N}$ a complete metric space with this metric $d(a,b) = \sqrt{1-2\frac{\gcd(a,b)}{a+b}}$?) that this metric space is complete. If it was totally bounded, then it would be compact.)

Thanks for your help!

Answer 1

As shown in the referenced answer,

$\;\;\;$Is $\mathbb{N}$ a complete metric space with this metric $d(a,b) = \sqrt{1-2\frac{\gcd(a,b)}{a+b}}$?

if $a\ne b$, then $d(a,b) \ge {\large{\frac{1}{\sqrt{3}}}}$, hence for any $\epsilon\in (0, {\large{\frac{1}{\sqrt{3}}}}]$, an open ball of radius $\epsilon$ is a singleton.

It follows that $X$ is not totally bounded.

Answer 2

Consider the sequence $\{2^n\}_n$. For distinct $n$ and $m$ we have $d(2^n, 2^m)\geq 1/\sqrt{3}$ so $\mathbb{N}$ can’t be covered with finitely many $1/(2\sqrt{3})$ balls because each such ball contains at most one point in out sequence.