# 14. The Sharkovsky Theorem

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?