Let $X = \mathbb{R}$ and for $x, y \in \mathbb{R}$ define $d'(x, y)$ by
$d'(x, y) = \begin{cases} |x - y|+1 & if \space exactly \space one \space among \space x\space and\space y\space is\space (strictly)\space positive.\\ |x - y| & otherwise \\ \end{cases} $
Show that the metric space is incomplete and complete.
I noticed that if a sequence is (eventually) negative, positive or zero, then $d$ is the usual metric on $\mathbb{R}$. Thus, the sequence converges. If the sequence is say, $(\frac{1}{n})$ which converges to $0$, then it does not converge under this new metric.
My attempt a completion under this new metric was to let $X' = X \space\cup\space{+0}$ where $+0$ has the following properties:
$d(x, +0) = \begin{cases} |x| & if \space x \space \neq 0.\\ +0 & if\space x=0 \\ \end{cases} $
So, now the aforementioned series $(\frac{1}{n})$ converges to $+0$. But then I considered, $\frac{(-1)^{n}}{n}$, the subsequence of the negative terms, converge to $0$ and the subsequence of positive terms converge to $+0$.
I thought it was still convergent despite, $+0 \neq 0$ implying there exists $\epsilon_{0}$ such that $d(+0, 0) = \epsilon_{0}$. However, consider $d(+0, 0) \leq d(\frac{\epsilon_{0}}{4},+0) + d(\frac{\epsilon_{0}}{4},0) = \frac{\epsilon_{0}}{2} < \epsilon_{0}$.
I don't understand how to approach completing this metric space. Any assistance will be greatly appreciated.
Let $Z=(-\infty,0] \cup (1,\infty)$ with the usual metric $|z-w|$ induced from ${\mathbb R}$. The function $F: {\mathbb R} \to Z$ defined by $F(x)=x$ for $x \le 0$ and $F(x)=x+1$ for $x>0$, is an isometry from $ ({\mathbb R},d')$ onto $(Z, |\cdot-\cdot|)$. Since it is clear how to complete $Z$ by adding $1$, this tells us how to complete $({\mathbb R},d') $: Add a point $*$, with $d'(*,x)=1+|x|$ for $x\le0$ and $d'(*,x)= |x|$ for $x>0$.
The resulting metric space $ ({\mathbb R}\cup \{*\},d')$ is complete. Indeed, it is isometric to the completion of $Z$, namely to $(-\infty,0] \cup [1,\infty)$ with the usual metric. (The isometry is the obvious extension of $F$.)