I'm trying to figure out how to solve this problem where, Let $f:R\rightarrow$ $R$ have a cycle {${a_1, a_2, a_3, a_4, a_5}$} where $f(a_i) = a$i+1 , $i= 1,2,3,4$ and $f(a_5) = a_1$.
If $a_1 < a_2 < a_3 < a_4 < a_5 $ ;
How can I prove that $f$ has a $3$ cycle?
Note that $f[a_4,a_5] ⊇ [a_1,a_5]$ and $f[a_3,a_4] ⊇ [a_4,a_5]$
hint : find $B_1 ⊂[a_4,a_5]$ with $f(B_1) = [a_3,a_4]$ and
find $B_2 ⊂[a_4,a_5]$ with $f(B_2) = B_1$
Not sure if I am approaching this correctly:
I read an example about a Period $5$ that does not imply period $3$ but I'm not sure if this applies to this problem here that is
Let $F: [1,5] → [1, 5]$, be defined such that $F(1) = 3$, $F(2) = 5$, $F(3) = 4$, $F(4) = 2$, $F(5) = 1$ . We later find that $F^3$ must have a fixed point in $[3,4]$. With a little more work we get that
The intervals on $[1,2]$, $[2,3]$, and $[4,5]$ are impossible at $F^3$. On the interval $[3,4]$, F is defined linearly and so $F(x)= 10 -2x$. It has a fixed point $10/3$ and it is easy to see that $F^3$ has a unique fixed point, which must be $10/3$. Hence there is no point of period $3$.
From the given problem, I see $f(a_1)=a_2$, $f(a_2)=a_3$, $f(a_3)=a_4$, $f(a_4)=a_5$, $f(a_5)=a_1$
therefore $a_1→a_2 →a_3→a_4 → a_5$
I'm having trouble understanding the last two lines (the hint) including what is B1 for $f(B_2) = B_1$. Can someone explain or graphically explain what is happening?
The sets $B_1$ and $B_2$ that you're looking for are closed intervals and the key is using the Intermediate Value Theorem and the fact that if $I$ is a closed interval and $g:I \rightarrow \mathbb{R}$ is a continuous function such that $I \subset g(I)$, then $g$ has a fixed point, this follows from the Intermediate Value Theorem by looking at the function $g(x) - x$.
Supposing you have found those intervals, then $B_2$ is such that $f^2(B_2) = [a_3,a_4]$, since $a_3$ is mapped to $a_4$ and $a_4$ to $a_5$, then $B_2 \subset f^3(B_2)$, therefore, $f^3$ has a fixed point which is either a fixed point for $f$ or a $3$-cycle for $f$, but it cannot be a fixed point since this point would be in $B_2 \subset [a_4,a_5]$ whose second iterate goes to $[a_3,a_4]$ and the only possible intersection point between those two sets is $a_4$ which is not a fixed point, therefore, there exists a $3$-cycle.
To find $B_1$, notice that the image of $[a_4,a_5]$ contains $[a_1,a_5]$, in particular, it contains $[a_3,a_4]$, just take a connected component of the closed set $f^{-1}([a_3,a_4])$ that lies inside $[a_4,a_5]$, this will be $B_1$. Do the same thing to find $B_2$ since the image of $[a_4,a_5]$ contains $B_1$