Question on the number of equilibria

Question

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?

Answer 1

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.

Answer 2

As Robert Israel has pointed out: by the definition of hyperbolic fixed point, the number of fixed points is finite (or countable).

A natural generalization of hyperbolicity for non-isolated equilibria is that of normally hyperbolic invariant manifold. Several results that are known for hyperbolic fixed points carry over to normally hyperbolic invariant manifolds, in particular when the invariant manifold is compact.

An excellent book about this matter, which also treats the case when the invariant manifolds is non-compact, is Normally hyperbolic invariant manifolds — the noncompact case by J. Eldering. The pre-proof version is freely available online.