TITLE: Closure by Projective Limits of the category of Coverings of a Topological Space
QUESTION [15 upvotes]: Let $X$ be a connected topological space, and $C_{finite}$ the
category of its finite coverings. Then I claim that the category $C$
of coverings of $X$ can be obtained by $C_{finite}$ taking
projective limits and disjoint unions.
Since projective limits and disjoint unions commute the statement is
reduced to say that $A:= \varprojlim A_i$, where each $A_i$ is a
finite covering, is itself a covering space of $X$. For the proof
this is my problem: Fixed a point $x\in X$, and a point $\{y_i \in
A_i\} \in A$ of the fiber over $x$, then I must find an open
neighborhood $x\in U \subset X$ such that it is homeomorphic to a
convenient neighborhood $V$ of $\{y_i\}\in A$.
The passage is problematic because, if the limits is taken over an
infinite index set $I$, then I'm not allowed anymore to restrict
this neighborhood $U$ in a convenient way for each $A_i$ (Infinite
intersection of opens can be not open).
There is a way to bypass this problem? Or my claim is simply wrong?
Probably in the solution it will be helpful (or necessary) to
restrict to a base space $X$ which admits a universal covering, i.e.
it is locally path connected and semi-locally simply connected.
Edit: I forgot to mention the following fact, which justify my claim. If $X$ admits a universal covering, then it can be obtained as projective limit of finite (normal) coverings of $X$.

REPLY [3 votes]: I don't see how your claim works, even in the simple case where $X$ is a circle.  The universal covering space is the real line, winding infinitely often around the circle.  The fiber over any point of the circle is a countably infinite discrete space.  But a projective limit of finite coverings would, as far as I can see, have fibers that are projective limits of finite sets, so the fibers would be compact.