Find if a metric space complete Let $(\mathbb{R},d)$ be a metric space with a distance function $d(x,y)=|f(x)-f(y)|,$ where
$$
f(x) = 
\begin{cases} 
&\displaystyle\frac{1}{x-2}, \quad &\text{$x < 2$} \\
&\ln \ x, \quad &\text{$x\ge 2$.} 
\end{cases}
$$
Is this metric space a complete metric space? If it is not what is the completion of it?

First of all we consider our particular case. This space is not complete and an example can be the Cauchy sequence $\{-n+2\}_n$. In fact this sequence is Cauchy
if $n,m\geq M $, then $|f(-n+2)-f(-m+2)|=|\frac{1}{n}-\frac{1}{m}|\leq \frac{2}{M}$
Conversely, is not a convergent sequence on $(\mathbb{R},d)$. By contradiction, we suppose $\{-n+2\}_n$ is convergent to some $\alpha \in \mathbb{R}$. Then one get
$0=\lim_{n\to \infty}d(-n+2, \alpha)=\lim_{n\to\infty } |\frac{1}{n}-f(\alpha)|=|0-f(\alpha)|=|f(\alpha)|$
This means $f(\alpha)=0$ and this is not possible because $f^{-1}(0)=\emptyset$
Thus $(\mathbb{R},d)$ is not a complete space.
What is the obstruction to the positive answer? The problem is that the image of $f$ is not a closed subset of $\mathbb{R}$ but only the subset $(-\infty ,0)\cup [ln(2), \infty)$
If one want to complete this space it is more convenient to study the problem in a more general way, in order to understand what it happens in a general situation.
We consider a bijective function $f: X\to Y$. If $Y$ is a topological space, then it is possible to inherit a natural topological structure on $X$ by $f$ such that $X$ and $Y$ are homeomorphic topological spaces and $f$ is an homeomorphism.
Morevoer, if $(Y,d_Y)$ is a metric space inducing that topology on $Y$, then one can get a metric $d_X$ on $X$ such that the induced topology is the topology inherited on $X$ by $f$ and $f$ is not only an homeomorphism but also an isometry with respect the two metrics fixed respectively on $X$ and $Y$. The metric induced on $X$ will be clearly
$d_X(x,y):=d_Y(f(x),f(y))$
In this way one can observe $(Y,d)$ is a complete metric space if and only if   $X$ is a complete metric space.
At this point we can consider our particular case. The bijection in this case is the following
$f: \mathbb{R} \to (-\infty ,0)\cup [ln(2), \infty)\subseteq \mathbb{R}$
and so $(\mathbb{R}, d)$  can not be a complete metric space because $((-\infty ,0)\cup [ln(2), \infty), |\cdot |)$ is not complete. In fact a subset of a complete metric space ( that in this case is $(\mathbb{R}, |\cdot|)$ ) is complete if and only if it is closed and in our case $(-\infty ,0)\cup [ln(2), \infty)$ is not closed.
At this point it is clear what is a completion of $(\mathbb{R}, d)$. The completion will be simply the complete space
$cl ((-\infty ,0)\cup [ln(2), \infty))=(-\infty ,0]\cup [ln(2), \infty)$
with the metric $|\cdot |$.
