choose the correct option regarding complete metric space

Question

let $X= \mathbb{N}$,the set of postive integer consider the metrics $d_1,d_2 $ on $X$ given by
$d_1(m,n) =|m-n|$ , $m , n \in X $

$d_2(m,n) =|\frac{1}{m} - \frac{1}{n}|$ , $m ,n \in X $

let $X_1,X_2$ denotes the metric space $(X,d_1),(X,d_2)$ respectively .Then choose the coorect option

$1.$ $X_1$ is complete

$2.$ $X_2$ is complete

$3.$ $X_1$ is totally bounded

$4.$ $X_2$ is totally bounded

My attempt : I got option $1)$ and option $3)$ beacuse the sequence $x_n = n \in \mathbb{N}$ is a Cauchy sequence for $d$. And $(x_n)$ don't converge to a limit.

is its true

Any hints/solution will be appreciated thanks u

Answer 1

1. is OK, as Cauchy sequences are eventually constant.

2. is false, indeed, as $x_n= n$ is a non-convergent Cauchy sequence.

3. is false, as we cannot cover $X$ by finitely many balls of radius $1$.

4. is true, as we can compactify it by adding a single point, e.g.

Answer 2

For point $4.$, to explicitly see that $X_2$ is totally bounded, fix $\epsilon > 0$. Choose $N > 3/\epsilon.$ Then for any $M > N$, we see $$d_2(N,M) = \left \lvert \frac 1 N - \frac 1 M \right \rvert \le \frac 1 N + \frac 1 M \le 2\epsilon/3 < \epsilon.$$ Thus the set $\{ m \in \mathbb N \, : \, m \ge N\}$ is contained in $B_\epsilon(N)$. Thus the collection $\{B_\epsilon(n)\}^N_{n=1}$ forms a finite cover of $\mathbb N$ by balls of radius $\epsilon$.

The rest of the points can be elucidated with simple examples, as in the answer by @HennoBrandsma.