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Say I have the dynamical system given by

$f(x)=2x \mod 1$

For $x \in (I=[0, 1], \text{Euclidean metric})$.

$f$ as defined is not minimal (because it has periodic points). That is not all the orbits of $f$ are dense.

My question is does restricting $f$ to the set of irrational numbers of $I$ make $f$ minimal? (the codomain of this restriction is also the irrationals of $I$)

If the answer is no, is there any restriction similar to the one above that gives a minimal dynamical system?

pops
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1 Answers1

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Taking the set of all irrational numbers does not work. Indeed, there are irrational $x$ whose orbit under $f$ is not dense.

The short answer is that there are many choices of $X \subset [0,1]$ such that $f$ restricted to $X$ is minimal. It is possible for instance to construct invariant Cantor sets that are minimal.

Albert
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  • Thank you. Do you know where I can find details on how to construct such a Cantor set? – pops Mar 10 '20 at 10:49
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    this may be a little complicated to read, but you should find everything you want in this paper https://people.cas.uab.edu/~ablokh/jamie4.pdf – Albert Mar 10 '20 at 10:58