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I am working on dynamical systems (more specifically Sharkovskii) and I have to show there exists a $3$-cycle for a continuous function with $f(a) = b, f(b) = c, f(c)= d, f(d) = e, f(e) = a$ where $a<b<c<d<e$.

Now I wonder if my approach works. My idea is, since $f$ is continuous, that we know intervals map to the next interval (except for the last one), and thus we can make use of subsections of each interval. Thus I do the following:

$\exists B \subseteq [b,c]$ with $ f(B) = [c,d]$, also $\exists C \subseteq [c,d]$ with $ f(C) = [d,e]$ and since $f[d,e] = [a,d]$, there also exists some $D \subseteq [d,e]$ with $f(D) = [b,c]$.

Hence there is some subset of $D$, which we name $E$, such that $$f^{3}(E) = f^{2}[b,c] = f[c,d] = [d,e] \supseteq E,$$

which yields we have a $3$-cycle.

Now since my experience is limited, I wonder if the above holds and I would really appreciate some feedback.

  • Compare with the more extended question in https://math.stackexchange.com/questions/3841445/how-to-prove-that-a-continuous-function-with-a-5-cycle, where the same argument was employed. – Lutz Lehmann Sep 27 '20 at 12:57
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    @TheSilverDoe but this function does not meet the set criteria. Since there is no "loop" back, which $f(d) = f(a)$ implies... –  Sep 27 '20 at 12:58
  • @LutzLehmann; I would argue the argument is different, since in that solution one maps twice to the same interval, in my case being $[d,e]$, while I do not do that in the above... –  Sep 27 '20 at 13:06
  • @Steven : Are you sure that it is not the converse, i.e. that given a $3$-cycle, you have to construct a $5$-cycle ? $3$ is less than $5$ in the order of Sharkovskii, so it should work the opposite. – TheSilverDoe Sep 27 '20 at 13:09
  • On second glance, yes. So your cycle is the second one for that task. – Lutz Lehmann Sep 27 '20 at 13:09
  • @TheSilverDoe : That is true in general, that's why there is the monotone ordering of the cycle as necessary condition for the converse. – Lutz Lehmann Sep 27 '20 at 13:10
  • @LutzLehmann Yes. I should wear my glasses and read all of that more carefully. – TheSilverDoe Sep 27 '20 at 13:11

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Your last chain is wrong in this way. You only get inclusions, as the images of intervals may be larger than the intermediate value theorem requires.

So you get $$ f^3(D)=f^2([b,c])\supseteq f^2(B)=f([c,d])\supseteq f(C)=[d,e]\supseteq D $$ This means that $(f^3|_D)^{-1}$ maps $D$ into itself. Now make sure that $f^3$ is monotonic on $D$ then the inverse is continuous and thus has a fixed point, giving the 3-cycle. Or argue that the nested sequence of pre-images $(f^3|_D)^{-n}(D)$ has a non-empty limit.

Lutz Lehmann
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  • I see what you mean, thanks. And good point about the monotonicity, I would have argued this was set by the criteria, but obviously it is not. –  Sep 27 '20 at 17:52
  • I'm not sure that monotonicity can be assumed, weird oscillations are possible in continuous functions, I think of the kind $x+0.5x\sin(1/x)$. The set based approach circumvents that. – Lutz Lehmann Sep 27 '20 at 17:59