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Can one construct a compact metric space $X$, and a homeomorphism $f\in \text{Homeo}(X)$, such that $f$ has noncompact periodic point set? I am trying to find a homeomorphism on a cylinder $S^1\times [0,1]$, but not succeeded yet. Looking forward to your brilliant answer. Thanks

Here are some of my attemptations

Let $f\in\text{Homeo}(S^1\times [0,1])$, which sends $(x,y,h)$ to $(-x,-y, g(h))$, i.e the antipodal map on the circle, then lift $g(h)$, so it suffices to find a $g:[0,1]\to[0,1]$ which is surjective,continuous and has dense periodic points (or just fixed points is Okay, I guess).

Because no matter whether $g$ is a homeomorphism, $f$ is a bijective continuous map, thus in this case, a homeomorphism, if the periodic points of $g$ is dense in $[0,1]$, but not the whole interval, we can deduce that the periodic points set of $f$ is not even closed, thus a noncompact one, so now the problem is to construct that peculiar $g$.

Z. Liu
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Consider the map $x\to 2 x \mod 1.$ (think of it as a homeomorphism of $S^1$). Then, every rational number with odd denominator is a periodic point.

Igor Rivin
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