I prefer the proof that takes f:n->n and gets you a labeled tree plus an ordered pair of vertices of the tree. It works as follows: If i is preperiodic for f, then connect i to f(i). Now put the periodic points in a line in increasing order, and permute them by f, so that the periodic point x with f(x) the lowest possible is first, etc. Then connect the periodic points up into a path in that order, and your ordered pair is the two ends of the path in order.
amazing simple sweet and intutive explanation thanks a lot!!😀
I prefer the proof that takes f:n->n and gets you a labeled tree plus an ordered pair of vertices of the tree. It works as follows: If i is preperiodic for f, then connect i to f(i). Now put the periodic points in a line in increasing order, and permute them by f, so that the periodic point x with f(x) the lowest possible is first, etc. Then connect the periodic points up into a path in that order, and your ordered pair is the two ends of the path in order.
Thank you for sharing that alternate method eyalminsky!