Recall that a set $\Lambda \subset M$ is called hyperbolic with respect to some $C^1$ diffeomorphic function $f:M \to M$ if it is $f$ invariant and if there exists a $Df$-invariant splitting $T_{\Lambda}M=E^{u}\oplus E^s$ such that $Df_x$ contracts on $E^s(x)$ and expands on $E^u(x)$.
Now fix an arbitrary set $\Lambda \subset M$.
I was wondering whether one can always construct or at least assert the existence of a function $f: M \to M$ with hyperbolic set equal $\Lambda$.
The answer is no. From my point of view the simplest example is $S^1$.
First note that the requirement on invariant splittings is automatically satisfied, although you still need an argument. What I am saying that if $S^1$ is a hyperbolic set, then either all stable spaces have dimension $1$ or all unstable spaces have dimension $1$.
Recall that the dimensions of the stable and unstable spaces are locally constant (this is the simplest nontrivial property of a hyperbolic set). Being integers, covering $S^1$ by intervals where the dimensions are constant and taking a finite subcover you get the former claim.
Let's assume that all stable spaces have dimension $1$ (you can treat in a similar manner the case when all unstable spaces have dimension $1$). Then paths on $S^1$ are contracted after sufficiently many iterations (if $c$ and $\lambda$ are the usual constants in the notion of a hyperbolic set, then after $n$ iteration provided that $c\lambda^n<1$). Now take a closed path that covers $S^1$ with the same initial and final points. It has length $1$ but after $n$ iterations it has length smaller than $1$ (this follows from estimating the length using the derivative). On the other hand, the image of the path is itself.
This contradiction show that $S^1$ cannot be a hyperbolic set.