Suppose $B$ is a $d$-dimensional ball (for some $d \geq 1$) and $T$ is a homeomorphism from $B$ to itself. Suppose also that $T$ is not of finite order (that is, for no $n \geq 1$ is it the case that $T^n(x)=x$ for all $x$ in $B$). Can we conclude that a nonperiodic point exists (that is, for some $x$ in $B$ it is the case that $T^n(x) \neq x$ for all $n \geq 1$)?


Note that it need not be the case that for each $n \geq 1$ the set of $x$ in $B$ with $T^n(x) \neq x$ is dense. So my question is not just an application of the Baire category theorem (at least under the most simpleminded approach).


Note that the answer becomes "No" if we replace the ball $B$ by a general compact set. For instance, consider the action on $\mathbf{R}/\mathbf{Z} \times \{0,1/2,1/3,1/4,1/5,…\}$ that sends ($x$, $y$) to ($x+y$ mod 1, $y$).


This post is related to [Nonperiodic points of piecewise-linear homeomorphisms](https://mathoverflow.net/questions/190500/nonperiodic-points-of-piecewise-linear-homeomorphisms#190536) .



Nonperiodic points of homeomorphisms of a ball
Montgomery showed in 1938 that a "pointwise periodic" homeomorphism of a manifold without boundary is actually periodic (this is in Montgomery and Zippin, page 223, or thereabouts), so if your ball is a closed ball, the result you want follows from this by doubling, and if it's the open ball, you don't even need to double.


