By Barmak J.A., Minian E.G.

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**Example text**

This contradiction will show that the family J\f must be empty, and so A must be hyperbolic. 6, it suffices to prove that limn_^oo \(fn)'(xn)\ = oo for every pre-orbit (xn)n G /lo of a point x0 G Ao. Denote by F the closure of the sequence (xn)n. This is a compact invariant set. If F is properly contained in /lo then it is hyperbolic, and so lim n \{fn)'{xn)\ — oo. Otherwise, F — /lo and so there is k > 0 such that Xk £ J. Replacing x by Xk, we may suppose k = 0. Since J is adapted to /lo, it has an (associated) pre-orbit ((/)n)n such that (pn(xo) = xn for all n.

Let f : M —* M be a C1 map and A be a compact set such that lim |(/")'(*„)| = oo n—>oo for all x £ A. Then A is a hyperbolic set for f. 2 Non-critical behavior 17 A pre-orbit of an open interval / C M is a sequence (

So, by Zorn's lemma, the family J\f has some minimal element AQ. The compact set AQ can not be a union of periodic orbits: otherwise, it would be hyperbolic. We also claim that AQ — f(Ao): if /(/lo) were properly contained in /1Q, then it would be hyperbolic, which would imply the hyperbolicity of /lo. This shows that the set C\nfn(Ao) = /lo contains some non-periodic point. 7. From the existence of such an interval, we are going to deduce that /l 0 is hyperbolic. This contradiction will show that the family J\f must be empty, and so A must be hyperbolic.