Nonperiodic points of piecewise-linear homeomorphisms Suppose $K$ is a compact polytope and $T$ is a piecewise-linear homeomorphism from $K$ 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 $K$). Can we conclude that a nonperiodic point exists (that is, for some $x$ in $K$ 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 $K$ 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).
Ideally I would like to draw stronger conclusions from the stated hypotheses (e.g. that the set of nonperiodic points has positive $d$-dimensional measure where $d$ is the dimension of $K$). Some of the hypotheses on $K$ and $T$ are likely to be red herrings, but I include them since they apply for the class of examples that motivated this question.
I have posted a variant of this question  in which the piecewise-linearity constraint is dropped; see Nonperiodic points of homeomorphisms of a ball. Montgomery's theorem solves that problem and therefore solves this one, though I'm still hoping for (and will award a bounty to) the best self- contained, clear, and complete proof for the piecewise-linear case.
 A: The answer is yes. In other words, if the order of the orbit $x_n=T^n(x)$ of each point $x\in K$ is finite then $T^n$ is identity map for some $n$.
Choose a simplex $\triangle$ of maximal dimension $m$.
Set 
$$E_n=\{\,x\in\triangle\mid T^n(x)=x\,\}.$$
By Baire theorem, $E_n$ has nonempty interior for some $n$;
fix such a value $n_1$.
Note that $E_{n_1}$ is a polytope in $\triangle$.
If $E_{n_1}\ne \triangle$,
we can choose an $m$-simplex $\triangle'\subset \triangle$ such that
(1) its base in $E_{n_1}$ 
(2) the remaining part is outside of $E_{n_1}$ and (3) $T^{n_1}$ is linear on $\triangle'$.
Given $\ell_1$, we can choose a simplex $S_1$ in $\triangle'$ so that (1) $T^{n_1\cdot m}(x)\ne x$ if $x\in S_1$ and $m<\ell_1$ and (2) for some $m_1>\ell_1$, the set $E_{m_1\cdot n_1}\cap S_1$ has nonempty interior but $E_{m_1\cdot n_1}\not\supset S_1$. 
Indeed (1) follows if entire $S_1$ lies close to the base. 
Further by Bair theorem $S_1\cap E_{m_1\cdot n_1}$ has nonempty interior for some $m_1>\ell_1$. 
Note that $x\notin E_{m_1\cdot n_1}$ if $x$ lies terribly close to the base (depending on $m_1$) and we can arrange that there is such point in $S_1$.  
Set $n_2=m_1\cdot n_1$ and iterate the construction for an increasing sequence of numbers $\ell_i$.
We get a nested sequence of compact set $S_1\supset S_2\supset\dots$ such that $T^n(x)\ne x$ for any $x\in S_n$. 
If $x\in\bigcap_nS_n$ then $T^n(x)\ne x$ for any $n$, a contradiction.
We proved that the restriction $T^n|_\triangle$ is identity map for some $n$.
Now remove the interior of $\triangle$ from $K$ and 
pass to the map $T^n$.
Repeat this procedure recursively till no simplex is left in $K$.
P.S. The same holds for any continuous map, see the answer in here.
