I've reading the Pugh's Analysis book and I have problems with one exercise. This says:
The baker's transformation: a rectangle of dough is stretched to twice its length and folded back on itself. Is this transformation continuous? A formula for the baker's transformation in one variable is $f(x)=1-|1-2x|$ (...) The orbit of a point $x$ is defined as
$$\{x, fx,f^2x,\ldots,f^nx\ldots\}$$
e) If $x$ is rational prove that of the orbit of $x$ is a finite set.
f) IF $x$ is irrational what is the orbit?
Honestly I don't have much idea of how to attach this kind of problems. Seems intuitively for me that the Baker's map is continuous, visualizing any open [closed] set in the target and checking that its preimage must be open [closed]. But I don't think this is really what the author would expect as an answer because its lack of formality.
e) Let $x=m/n\in [0,1]$ in the lowest terms. We have to show that the orbit of $x$ is contained in $\{x\}\cup\{2k/n:\text{for all } 2k\le n\}$; since the latter is finite this will show that former must be finite as well.
We use induction to show that $f^ix$ is either $x$ or $2k/n$. Define $f^0x$ as $x$. Then the result is trivial for $i=0$. Suppose we have already proven the assertion for $i=n$. Then we have two possibilities:
For $f^nx=x$, then we have $f^{n+1}x=fx$. Either $fx$ is $2m/n$ or $2(n-m)/n$ depending if $x\le 1/2$ or $x\ge 1/2$. On the other hand for $f^nx=2k/n$, where $2k\le n$. If $2k/n\le 1/2$, $f(2k/n)=4k/n$, where $4k\le n$. If $2k/n\ge 1/2$, $f(2k/n)=2(n-2k)/n$, and the assertion follows.
Any help would be great for the part f). Thanks in advance.