Distance making $(\mathbb{R} \setminus \mathbb{Z},d) $ complete and homeomorphic to $\mathbb{R}\times \mathbb{Z}$

Question

As in the title I have to construct a metric $d$ on $\mathbb{R} \setminus \mathbb{Z}$ s.t. $( \mathbb{R} \setminus \mathbb{Z}, d)$ is complete and homeomorphic to $ \mathbb{R} \times \mathbb{Z}$.

I don't know how to proceed!

Answer 1

$$\mathbb{R}\setminus \mathbb{Z} = \bigcup_{n \in \mathbb{Z}} (n, n+1)$$

Now make sure that points in $(n,n+1)$ have their usual distance among each other (so that is a homeomorphic copy of $\mathbb{R}$) but points in different $(n,n+1)$ and $(m,m+1)$ (so $n \neq m$) are very far apart from each other. Then we get a so-called sum space of countably many copies of the reals, which is homeomorphic to $\mathbb{R} \times \mathbb{Z}$.

To get a complete metric replace the usual metric on $(n,n+1)$ with $|h(x) -h(y)|$ where $h$ is any homeomorphism between this interval and $\mathbb{R}$.