Contraction mapping and periodic orbit

Question

Consider a map $f:X \rightarrow X$ where $X$ is a space that may be described as the union of a finite number of intervals so that $X = I_1 \cup I_2 \cup ... \cup I_n$. The map $f$ is a contraction mapping, so for every iteration distances between close points contract. Further, for all $x \in I_i$, then $f(x)$ will deterministicly map into an interval $I_j$, where $i, j$ need not be distinct.

I need to prove that iteration of $f$ over the space $X$ will eventually converge to a periodic orbit (or fixed point). That the map will converge is clear because it is a contraction mapping, however I am unclear on how to rigorously prove that there exists a sequence of intervals $f$ will map into so that it converges to a periodic orbit.

That the map will converge to a periodic orbit seems like a trivially obvious fact because there are a finite number of intervals, but a method by which I can rigorously show this is unclear. Is there some standard "go to" argument for such situations that will show this? I cannot help to make analogy to finite multiplicative (cyclic) groups, where every element has a finite order (period).

Could this be proved using such group-theoretic argument? Because $f$ is a contraction mapping we know it would be sufficient to show that $f$ must return to some interval $I_i$ more than once, but because there are a finite number of intervals there must exist some interval $I_i$ such that $x, f^k(x) \in I_i$ for finite $k$? How can such an argument be made rigorous so it constitutes a proof?

Answer 1

You have a condition that says:

... for all $x \in I_i$, $f(x)$ will deterministically map into an interval $I_j$.

Here's a translation of that condition:

there exists a function $\sigma : \{1,2,...,n\} \to \{1,2,...,n\}$ such that if $x \in I_i$ then $f(x) \in I_{\sigma(i)}$.

So your question comes down to the behavior of a map from a finite set to itself. You should be able to prove easily the existence of $i \in \{1,2,...,n\}$ and $k \ge 1$ such that $\sigma^k(i)=i$. It follows that $f^k(I_i)=I_i$.