Suppose $(X,d)$ is a complete metric space and $U$ is a countable collection of open subsets of $X$. Show that $\bigcap U$ can be endowed with a metric topologically equivalent to $d$ with respect to which $\bigcap U$ is complete.
Attempt: Some Background:
(i) Theorem: Open subsets of a complete metric space can be made into complete metric spaces by judiciously altering the metric. Let $U$ be an open set in $X$. Define $f(x) = \dfrac {1}{dist_d(x,U^c)}$ where $dist$ refers to the distance function i.e. $\inf \{d(x,w)~|~w \in U^c \}$. Then, the metric $e:(a,b) \rightarrow d(a,b)+|f(a)-f(b)|$ converts $U$ into a complete space. Also,$d,e$ are topologically equivalent
EDIT: I found a solution below:
I couldn't understand why $e(a,b) \le d(a,b) + |f_,m(a) - f_m(b)|$. Is it an error?
Also, is there really a need for the summation? Thanks
You could considere the metric on $U$ :
$$\rho(a,b)=\sum\limits_{n=1}^\infty \frac{1\wedge e_n(a,b)}{2^n}$$
It is known as Alexandroff-Mazurkiewicz theorem : Proof of Mazurkiewicz theorem