Bourbaki's Mittag-Leffler theorem I've been trying to understand the Mittag-Leffler theorem (not the one with meromorphic functions, but one for metric spaces), but I stumbled upon a detail.   In Theorem 2.4.14 of this Runde's book, the author constructs new metrics besides the ones in the hypothesis and claim they are equivalent: picture; that's what I can't see why. 
Can you please help me?
(I've been reading the proof from Volker Runde - A taste of topology (page 47).)
Thank you!  
 A: The new metrics are constructed inductively. We begin with $\tilde{d}_0 = d_0$. Then (1) - (3) below are satisfied for $n = 0$.Assume that we have constructed $\tilde{d}_0,\dots,\tilde{d}_n$ such that
(1) $d_i,\tilde{d}_i$ are equivalent for $i=0,\dots,n$
(2) $(X_i,\tilde{d}_i)$ is complete for $i=0,\dots,n$
(3) $\tilde{d}_{i-1}(f_n(x),f_n(y) \le \tilde{d}_i(x,y)$ for $i=1,\dots,n$ and all $x,y \in X_n$
$\tilde{d}_{n+1}$ is defined by $\tilde{d}_{n+1}(x,y) = d_{n+1}(x,y) + \tilde{d}_n(f_{n+1}(x), f_{n+1}(y))$.
(1) We have $d_{n+1}(x,y) \le \tilde{d}_{n+1}(x,y)$ for all $x,y$, hence $\tilde{d}_{n+1}$ is stronger than $d_{n+1}$. To show that $d_{n+1}$ is stronger than $\tilde{d}_{n+1}$, consider $x \in X_{n+1}$ and $\epsilon > 0$. We have to find $\delta > 0$ such that $d_{n+1}(x,y) < \delta$ implies $\tilde{d}_{n+1}(x,y) < \epsilon$. Since $f_{n+1} : (X_{n+1},d_{n+1}) \to (X_n,\tilde{d}_n)$ is continuous at $x$, there exists $\rho > 0$ such that $d_{n+1}(x,y) < \rho$ implies $\tilde{d}_n(f_{n+1}(x), f_{n+1}(y)) < \epsilon/2$. Then $\delta = min(\rho,\epsilon/2)$ will do.
(2) Let $(x_m)$ be a Cauchy sequence in $(X_{n+1},\tilde{d}_{n+1})$. Since $d_{n+1}(x,y) \le \tilde{d}_{n+1}(x,y)$ for all $x,y$, we see that $(x_m)$ is a Cauchy sequence in $(X_{n+1},d_{n+1})$ and therefore converges with respect to $d_{n+1}$ to some $\xi \in X_{n+1}$. This means that each $\delta > 0$ admits $M \in \mathbb{N}$ such that $d_{n+1}(\xi,x_m) < \delta$ for all $m \ge M$. Given $\epsilon > 0$, choose $\delta > 0$ such that $d_{n+1}(\xi,y) < \delta$ implies $\tilde{d}_{n+1}(\xi,y) < \epsilon$. Then $\tilde{d}_{n+1}(\xi,x_m) < \epsilon$ for $m \ge M$, i.e. $(x_m)$ converges with respect to $\tilde{d}_{n+1}$ to $\xi$.
(3) This is obvious from the definition.
