On limits of manifolds This question should be fairly elementary. I’d just like to check I’m not missing anything.
Let $\{M_n\}_{n\ge 0}$ be an inverse system of smooth manifolds with transition maps $f_{t,s} : M_t\to M_s$, $t\ge s$, that are local diffeomorphisms.
Let $M$ be the topological inverse limit.

For every $x\in M$, does there exist an open neighborhood $U\subset M$ of $x$, such that $U$ is homeomorphic to some smooth manifold that is diffeomorphic to an open subset of $M_n$ for some $n$?

 A: In general, this will be false. Examples are found among solenoidal manifolds, defined by Sullivan. For example, 1-dimensional solenoids. 
Many of these are obtained by taking the inverse limit of finite-sheeted covers of a fixed manifold. The universal such example is obtained by taking a manifold $M$ with infinite residually finite fundamental group $\pi_1(M)$, and taking its profinite completion $\widehat{\pi_1(M)}$. Let $\tilde{M}$ denote the universal cover of $M$. Then $(\tilde{M}\times \widehat{\pi_1(M)})/ \pi_1(M)$ is a solenoid, where the action is by covering translation on the left and coset action on the right where $\pi_1(M)\subset \widehat{\pi_1(M)}$. 
This space is locally homeomorphic to open subsets of $\mathbb{R}^n \times \widehat{\pi_1(M)}$. For an infinite residually finite group, $\widehat{\pi_1(M)} \cong \mathcal{C}$, the Cantor set. So there is a neighborhood basis of sets which are homeomorphic $\mathbb{R}^n \times \mathcal{C}$ where $\mathcal{C}$ is totally disconnected, since $\mathcal{C}$ has a neighborhood basis of sets homeomorphic to itself. 
