# Limit of a sequence of eventually periodic points

This is a research-related question I've been thinking about for a while now -- it seems like a standard exercise in first-year analysis, but the solution eludes me.

Let $f:\mathbb{T}\to\mathbb{T}$ be any continuous, surjective, piecewise strictly monotone map from the unit circle to itself (not necessarily a homeomorphism or even a local homeomorphism). For a point $x\in\mathbb{T}$, define the full orbit of $x$ as

$[x]=\{y\in\mathbb{T} | \exists n,m\in\mathbb{N}: f^n(x)=f^m(y)\}$

Assume now that $x$ is eventually periodic (i.e. there are $p$ and $q$ such that $f^p(x)=f^q(x)$). It follows that any $y\in[x]$ is also eventually periodic. What if $y$ is not in $[x]$, but a limit point of $[x]$?

In general, such a limit point does not have to be eventually periodic (for instance, if $f$ is the standard tent map, any point has dense full orbit, but not all points of $\mathbb{T}$ are eventually periodic).

But what if I make the additional assumption that $x$ is isolated in $[x]$ (i.e. there is some neighbourhood $U$ of $x$ such that $[x]\cap U =\{x\}$)? Countless examples have led me to believe that any limit point of $[x]$ has to be eventually periodic in this case, but it seems difficult to prove...

• Doesn't it follow that all points in $[x]$ are also isolated, and that therefore $[x]$ is finite? Once you have $[x]$ finite, the eventual periodicity follows. – Alamos Apr 30 '15 at 5:24
• Well, it follows that all points in $[x]$ are isolated, but not necessarily isolated by neighbourhoods of equal size. For instance, you could have $[x]$ looking like the set $\{1/2, 1/3, 1/4, 1/5, \ldots\}$, with each point isolated, but having 0 as a limit point. (Am I using the term 'isolated' wrong?) – tschmidt Apr 30 '15 at 6:35