## 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.