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I need to prove that a continuous function $f$ which has a 5-cycle $\{a_1,a_2,a_3,a_4,a_5\}$ with $a_1<a_2<a_3<a_4<a_5$ and $f(a_1)=a_2, f(a_2)=a_3, f(a_3)=a_4, f(a_4)=a_5, f(a_5)=a_1$ has two 3-cycles.

I've already proven it has one 3-cycle as follows:

We may assume that $ f([a_1,a_2])=[a_2,a_3], f([a_2,a_3])=[a_3,a_4], f([a_3,a_4])=[a_4,a_5], f([a_4,a_5])=[a_1,a_5]. $ We can find an interval $B_1\subseteq [a_4,a_5]$ with $f(B_1)=[a_4,a_5]$ and an interval $B_2 \subseteq [a_4,a_5]$ met $f(B_2)=[a_3,a_4]$. We can also find an interval $K_1 \subseteq B_1$ with $f(K_1)=B_2$ such that $ f^3(K_1) = f^2(B_2) = f([a_3,a_4])=[a_4,a_5] \supseteq B_1 \supseteq K_1. $ So $f^3$ has a fixpoint in $K_1$ which is not a fixpoint of $f(x)$. So we have found our first 3-cycle.

Can somebody help me with the proof of the second 3-cycle? I don't see how I can modify this proof for a second 3-cycle.

MathGirl
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1 Answers1

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You already found out that the standard argument for this problem is the mapping of intervals and sub-intervals. If one draws all these relations one obtains the following graph

interval relations in 5-cycle

where the arrow $I\to J$ means that $I$ contains a sub-interval $I_1$ so that $f(I_1)=J$. For the quest of finding 3-cycles one can now use all the loops present in this graph. I see 3 such possibilities

  • $[a_4,a_5]$ looping thrice into itself
  • one self-loop of $[a_4,a_5]$ combined with the loop to $[a_3,a_4]$ and
  • the long loop $[a_2,a_3]$ to $[a_3,a_4]$ to $[a_4,a_5]$ and back.

You found the second variant, the argument for the others works the same way. The question Find a $3$-cycle for a continuous function where $f(a) = b, f(b) = c, f(c)= d, f(d) = e, f(e) = a$. used the last variant. In the first case it is (almost?) certain that the fixed point of $f^3$ found this way is already a fixed point of $f$.

Lutz Lehmann
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