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$f:\mathbb R \to \mathbb R $ continuous, with a point of odd period, implies existence of a point of even period

This is the question. I can't prove it. It's an exercise to prove Sarkovskii theorem, but it on I have to do this part and I'm ready.

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In there are useful links – Ross Millikan Oct 28 '11 at 18:17
What does "it on I have to do this part and I'm ready" mean? and what are you ready for? – Gerry Myerson Oct 28 '11 at 21:51
Strictly speaking, a fixed point is a point with odd period, so we should change "a point of odd period" to "a point of odd period that is not a fixed point". Otherwise set $f$ to be the identity, then every point is of odd period. – A. Alp Uzman Mar 16 at 18:26

1 Answer 1

The answer to this question appear in the book dynamical systems by Robert Devaney. In this book, Devaney give a proof for Sarkovskii theorem, may be you will see this proof and take the idea for you exercice. I hope this be useful.

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