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 
$D\_{H}(M(f\_n),M(f))=\max\{\sup\_{\bar{x}}\inf\_{\bar{y}}\bar{d}(\bar{x},\bar{y});\sup\_{\bar{y}}\inf\_{\bar{x}}\bar{d}(\bar{x},\bar{y})\}$.



Inverse limit space and $C^1$ topology
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.


