Find a complete metric on $\mathbb{R} \setminus \cup_{n\ge 1}\{\frac{1}{n}\}$ that gives the natural topology How can I set a metric on $\mathbb{R} \setminus \cup_{n\ge 1} \{\frac{1}{n}\}, n \ge 1$, so that the topology with this metric is the natural topology, and the space is complete?
Update:
I don't understand what will happen with $\mathbb{R}$ after this substraction. I suppose that $\mathbb{R}\backslash\cup_{n\ge 1} \{\frac{1}{n}\}$ with the standard metric is an incomplete metric space, since it is not a closed subspace of $\mathbb{R}$. 
 A: Let's first construct a complete metric on $\mathbb{R}\backslash\{0\}$. This subset of $\mathbb{R}$ is homeomorphic to the hyperbola $\{ (x, y) \ | x y- 1 =0\}$ by the map $x \mapsto (x, \frac{1}{x})$. Consider the restriction of the metric from $\mathbb{R}^2$ makes the hyperbola a complete subspace, since it's closed. We get  a complete metric on 
$\mathbb{R} \backslash \{0\}$, 
$$d(x_1, x_2) = \sqrt{(x_1- x_2)^2 + (\frac{1}{x_1} - \frac{1}{x_2})^2 }$$
Suppose now that we have a closed subset $A$ of $\mathbb{R}$ and we want a complete metric on $\mathbb{R} \backslash A$. Let $f$ a continuous function on $\mathbb{R}$ with $0$ level set $A$. $\mathbb{R} \backslash A$ is homeomorphic to the closed subset of $\mathbb{R}^2$ $\{ (x,y)\ | f(x) \cdot y - 1 = 0\}$ by the map $x \mapsto (x, \frac{1}{f(x)})$. Take the pull-back of the metric from $\mathbb{R}^2$. 
$$d(x_1, x_2) = \sqrt{(x_1-x_2)^2 + (\frac{1}{f(x_1)} - \frac{1}{f(x_2)})^2 } $$
In general, let $A_n$ a family of closed subsets. Let $f_n$ a continuous real valued function with $0$ level $A_n$. The space  $\mathbb{R}\backslash \cup_n A_n$ is homeomorphic to the closed subset of the complete metric space $\mathbb{R}^{\mathbb{N}}$
$$\{ (x, y_1, y_2, \ldots ) \ | f_n(x) y_n -1 = 0 \textrm{ for all } n \ge 1 \}$$ by the map 
$$x \mapsto ( x,\frac{1}{f_1(x)}, \frac{1}{f_2(x)},\ldots \}$$
The induced metric on $\mathbb{R}\backslash \cup_n A_n$ is complete.
In our case,  metric can be
$$d(x_1,x_2) = |x_1-x_2| + \sum_{n=1}^{\infty} \frac{1}{2^n} \cdot\frac{|\frac{1}{nx_1-1}- \frac{1}{nx_2-1}|}{1+ |\frac{1}{nx_1-1}- \frac{1}{nx_2-1}|} $$
