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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?

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  • $\begingroup$ Aware that this exposes my ignorance, could you please define a hyperbolic point? Thanks. $\endgroup$ Sep 11, 2014 at 22:43
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    $\begingroup$ en.wikipedia.org/wiki/Hyperbolic_equilibrium_point $\endgroup$
    – user31317
    Sep 11, 2014 at 22:45
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    $\begingroup$ Hyperbolic fixed points are isolated, are they not? That implies the number in any compact set must be finite. $\endgroup$ Sep 12, 2014 at 0:36

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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.

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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.

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    $\begingroup$ Why the hyperbolic fixed points are isolated? In case $f$ is diffeomorphism, then i think we can use Hartman-Grobman. If $f$ is not a diffeomorphism? Thanks $\endgroup$
    – user31317
    Sep 12, 2014 at 4:47
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    $\begingroup$ Wow, thanks for the free advertisement! ;-) $\endgroup$ Sep 12, 2014 at 9:02
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    $\begingroup$ @user31317: At a hyperbolic fixed point $x$, the total derivative $Df(x)$ is non-degenerate, so by the implicit function theorem there is an open neighborhood around $x$ that does not contain any other fixed points. $\endgroup$ Sep 12, 2014 at 9:03
  • $\begingroup$ @JaapEldering and user31317 now I understood better the question, thanks for the clarification. I was thinking that the definition of fixed point already included that the point were isolated. $\endgroup$
    – Tadashi
    Sep 12, 2014 at 17:20
  • $\begingroup$ @JaapEldering You're welcome, although I didn't had time yet to read until the end, I really liked your style of writing :) $\endgroup$
    – Tadashi
    Sep 12, 2014 at 17:23

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