Completeness of a metric space $(\mathbb{R},d_2)$.

Question

I'm trying to prove that $(\mathbb{R},d_2)$, where $d_2 (x,y) = |\phi(x) - \phi(y)|, \forall x,y \in \mathbb{R}$ is not complete. We define $\phi: \mathbb{R} \to \mathbb{R}$ as: $$\phi(t) = \frac{t}{1+|t|}, \forall t \in \mathbb{R}.$$ I was trying to find a Cauchy and not convergent sequence but I couldn't find an example to show it. Also, I know that a normed space $X$ is complete if, and only if, every absolutely convergent series is convergent, but I don't know if it can be useful here.

Answer 1

The reason why $d_2$ is not complete is that you can have e.g. a sequence $(a_n)_{n\geq 1}$ diverging to $+\infty$ in $\mathbb{R}$, then $\phi(a_n)$ would converge to $1$ on $[0,1]$ equipped with the Euclidean metric. However, the image of $\mathbb{R}$ under $\phi$ is not $[0,1]$, but $(0,1)$. So, since $(\phi(a_n))_{n\geq 1}$ a sequence converging to $1$ in $\mathbb{R}$ (in the Euclidean sense), it is Cauchy, but since there is no $x \in \mathbb{R}$ with $\phi(x) = 1$, it can not be convergent. So, one could take $a_n = n$ as an example.

This can be "fixed" if we consider the extended real line $\overline{\mathbb{R}}$ and define $\phi(-\infty) = -1, \phi(\infty) = 1$. What you obtain then is precisely the two-point compactification of the real line.