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.



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> 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 $n$?
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On limits of manifolds
In general, this will be false. Examples are found among [solenoidal manifolds](http://www.journalofsing.org/volume9/sullivan.pdf), defined by Sullivan. For example, 1-dimensional [solenoids](https://en.wikipedia.org/wiki/Solenoid_(mathematics)). 


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](https://en.wikipedia.org/wiki/Profinite_group#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](https://en.wikipedia.org/wiki/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. 


