In this lecture we define a new ordering $\preceq$ of the natural numbers, called the Sharkovsky ordering. This ordering has the property that:

  • $1$ is the smallest number. (No surprises here.)
  • $2$ is the second smallest number. (Or here.)
  • $3$ is the largest number. (Wait, wut?)

The motivation for this ordering is the result we proved last lecture: if $f \colon [0,1] \to [0,1]$ has a periodic point of period 3, then $f$ has periodic points of all orders.

The famous Sharkovsky Theorem states that if $f \colon [0,1] \to [0,1]$ has a periodic point of period $n$ and $ k \preceq n$ then $f$ also has a periodic point of period $k$.

We prove this today.



Comments and questions?