Lemma (p13 (D))

Let (X,d) be a metric space.  Then
d' = \frac{d}{1 + d} is a compatible
metric.  Furthermore, (X, d) is complete
implies (X, d') is complete.

Proof.
Let x \in X.  Let B_d(x, \epsilon) = \{ y \in X : d(x,y) < \epsilon \}.
Now d(x, y) < \epsilon implies d'(x,y) < \epsilon.
So B_{d}(x, \epsilon) \subset B_{d'}(x, \epsilon).
Also, if \epsilon < 1/2, d'(x,y) < \epsilon implies
d(x,y) < \frac{\epsilon}{1 - \epsilon} < 2\epsilon, so
B_{d'}(x, \epsilon / 2) \subset B_{d}(x, \epsilon).

Now, assume that (X, d) is complete.  Then
let (x_n) be a Cauchy sequence in (X, d'). Let \epsilon > 0, \epsilon < 1/2.
Then, (\exists N)(m,n \ge N \Rightarrow d'(x_m,x_n) < \epsilon).

Then, d(x_n,x_m) < 2\epsilon, so that (x_n) is Cauchy in (X, d),
hence x_n \to x in (X, d), and since d' < d, x_n \to x
in (X, d').  Thus, (X, d') is complete.

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