Inverse limit space and $C^1$ topology Let $f_n$ be a sequence of $C^1$-maps on closed manifold $M$. If $f_n \to f$ in $C^1$-topology. Does $M_{f_n}$ converges to $M_f$ in the Hausdorff distance?
We define $M(f)=\{\bar{x}=(x_j) \in M^{\mathbb{Z}}\mid f(x_{j-1})=x_j\}$. To define the Hausdorff distance in $(M^{\mathbb{Z}},d)$, define
$\bar{d}((x_j),(y_j))=\sum_{j \in \mathbb{Z}}\dfrac{d(x_j,y_j)}{2^{|j|}}$, where $d$ is a distance on $M$, and finally 
$D_{H}(M(f_n),M(f))=\max\{\sup_{\bar{x}}\inf_{\bar{y}}\bar{d}(\bar{x},\bar{y});\sup_{\bar{y}}\inf_{\bar{x}}\bar{d}(\bar{x},\bar{y})\}$.
 A: I think that an example of non-convergence can be constructed as follows:
Let $f\colon [0,1]\to [0,1]$ be a $C^\infty$ map such that
$g:=f\big|_{[0,1/4]}\colon [0,1/4]\to[0,1/2]$ is an increasing
diffeomorphism; $f\big|_{[1/4,3/4]}\equiv 1/2$ and
$f\big|_{[3/4,1]}\colon [3/4,1]\to[1/2,1]$ is an increasing
diffeomorphism, too.
Now, let $f_n\colon [0,1]\to[0,1]$ be a sequence of increasing
$C^\infty$-diffeomorphisms such that $f_n\to f$ in the
$C^\infty$-topology; and we can assume $f_n(1/2)=1/2$, for every $n$.
Then, for each $t\in [1/4,1/2)$ the inverse limit set $M(f)$ contains
the point $\overline x(t)=(x_j(t))$ given by $x_0(t)=t$, $x_j(t)=1/2$,
for every $j\geq 1$, and $x_j(t)=g^{j}(t)$, for every $j\leq -1$.
On the other hand, for every $n\in\mathbb{N}$, the only point
$\overline x\in M(f_n)$ suc that $x_1=1/2$, is the point $x_j=1/2$,
for every $j\in\mathbb{Z}$. And moreover, any point $\overline y\in
M(f_n)$, such that $y_1$ is close to $1/2$ will have $y_0$ close to
$1/2$ too, whenever $n$ is sufficiently big, because $f_n$ is a
diffeomorphism and $1/2$ is a fixed point. 
So, $M(f_n)$ does not converge to $M(f)$ in the
Hausdorff distance.
A: After the revision of you question, I still think the answer is negative but did not work out the example completely.
Let $f_n,f:\mathbb{T}^1\to\mathbb{T}^1$ be defined by
$$ f(x) =3x \mod 1 \quad f_n(x)=3x+\frac1n \mod 1.$$
The inverse limit spaces are abstractly homeomorphic, but they should not be close one to another as subsets of the product $\mathbb{T}^\mathbb{Z}$. Indeed, the small gap due to the translation will be multiplied by $3$ at each forward iteration, more than compensating for the coefficient $2^{-|j|}$. 

previous version suited to a previous question
No: $f_n$ could be $C^1$-close to $f$ and they could have no common value (i.e. $f_n(x)\neq f(x)$ for all $x\in M$). In this case an element $(x_j)$ of $M(f)$ and an element $(y_j)$ of $M(f_n)$ cannot coincide for more than two values, i.e. they must be at distance at least $1/2$.
This example shows that your choice of distance on the product space generates much too fine a topology. You could try with a metric that involves a distance on $M$ (endowing it with an arbitrary Riemannian metric say), but I don't think you should expect good result even in this case, as indicated by the following example.
Let $M=\mathbb{T}^1$ be the circle and take $f$ the identity, $f_n$ small $C^1$ perturbations of the identity obtained by adding a quite flat but possibly long bump, where $f_n$ is above the identity (thus creating a large fixed-point free interval $I$).
The inverse limit space of $f$ is the diagonal $\{(x_i) \mid \forall i,j: x_i=x_j\}$, while for each $x\in I$ the point $(x)_{i\in\mathbb{Z}}$ should be far from any point in $M(f_n)$.
