# More than one recurrence point (Birkhoff)

Birkhoff's recurrence theorem states that for a compact metric space $$X$$ and a continuous function $$T: X\rightarrow X$$, there is a recurrence point $$x\in X$$; the latter means that for any neighbourhood $$U$$ of $$x$$, there is some $$n\in \mathbb{N}$$ such that the $$n$$-th iteration of $$T$$ at $$x$$, usually denoted $$T^n(x)$$, is in $$U$$.

There are instances of this theorem where there is only one recurrence point. My question is whether there are nice/well-known conditions under which there is more than one recurrence point?

The Poincare recurrence theorem states that for any $$f$$-invariant probability measure $$\mu$$ on $$X$$ and any set $$A$$ with $$\mu(A) > 0$$, $$\mu$$-a.e. point $$x \in A$$ returns to $$A$$, i.e. there exists $$n > 0$$ s.t. $$f^n(x) \in A$$.
If $$X$$ is second countable and Hausdorff and $$\mu$$ is Borel, then you can remove the sets of zero measure from all elements in a countable neighborhood basis to immediately show that $$\mu$$-a.e. point is recurrent. And by the Bogoliuboff-Kryloff theorem, there is always at least one such measure $$\mu$$; take a weak-* limit point of averages of the form $$\frac{1}{n} \sum_{i=0}^{n-1} f^i \nu$$ for an arbitrary (not necessarily invariant) measure $$\nu$$.
So the only case where there could be only a single recurrent point $$x$$ is if $$\delta_x$$ is the only $$f$$-invariant Borel measure, which is equivalent to saying that $$\frac{1}{n} \sum_{i=0}^{n-1} f^i \delta_y$$ converges weak-* to $$\delta_x$$ for all $$y \in X$$. This, in turn, should be equivalent to saying that for any $$y \in X$$, the orbit $$(f^n(y))$$ converges to $$x$$ except for a set of $$n$$ of zero density.