Let $f: C \to C$ be a smooth function and $C$ be a compact set, subset of $\mathbb{R}^n$. 
We assume that all the fixed points are hyperbolic. Is it true that the number of fixed points is finite or countable?



Question on the number of equilibria
Building upon what others have already said, the number of hyperbolic fixed points is indeed countable, but need not be finite.


First, it follows from the definition of a hyperbolic fixed point $x$ that they are isolated: its total derivative $Df(x)$ does not have any eigenvalues on the unit circle, which precludes a hyperbolic fixed point from being a limit point of fixed points.


If the set $X$ of hyperbolic fixed points were uncountable, then the collection $\{U\_x\}\_{x \in X}$ together with $C \setminus \bar{X}$ would form an open cover of $C$ without a locally finite refinement, contradicting paracompactness (details left out, and there may be a shorter argument here).


On the other hand, consider $f(x) = x + x^3 \sin(\pi/x)$ on $C = [0,1]$. All points $1/n$ are hyperbolic fixed points; the limit $0$ is fixed, but not hyperbolic, since the derivative $f'(0) = 1$.


**Edit: corrected for the fact that $f$ is a map, not a vector field.**


