Sets admissing metrics which make them complete having the same subspace topology

Question

I am thinking on the following exercise:

Show that the following subspaces admit a metric (inducing the subspace topology), with respect to which they are complete.
(a) $X=\mathbb{R} \backslash\{0\}$,
(b) $Y=\mathbb{R} \backslash \mathbb{Z}$,
(c) $Z=\mathbb{R}^{n} \backslash A$ where $A$ is a closed set.

I know that the discrete metric space is always complete and hence setting the metric for $X$ to be the discrete metric gives a complete space. But the discrete metric space does not give same subspace topology as induced on $X$ by $\mathbb{R}$. It would be great if someone provides a hint or a solution. Thanks in advance.

Answer 1

The metric $$d'(x,y)= |x-y| + \left|\frac{1}{d(x,A)} - \frac{1}{d(y,A)}\right|$$ on $(X,d)$ with $A \subseteq X$ closed will work. As usual $d(p,A)= \inf\{d(p,a): a \in A\}$ and if $(X,d)$ is complete, so is $(X\setminus A, d')$ and the topology induced by $d'$ is the same as the subspace topology from $d$.

This encompasses all three cases.