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Consider the one-sided full shift map $\sigma$ and the associated shift space of infinite sequences in two letters $\{0,1\}^\mathbb{N}$ on which the shift map acts, equipped with the usual metric. Also consider another metric space $(X,d)$, along with a continuous map $f$ that evolves the elements of $X$ (i.e. it describes dynamics on $X$).

Assume that the action of $f$ on a compact subset of $X$ is topologically conjugate to the one-sided full shift map $\sigma$ on $\{0,1\}^\mathbb{N}$.

Is that compact subset always a repelling set of $f$ on $X$?

MOTIVATION: All of the examples I know of that are actually topological conjugates of the shift map/space cause the mapped shift space to be a repelling set, like the tent map for $\mu = 3$, the $10x \text{ mod } 1$ map, and the logistic map for $r > 4$, which makes the chaos essentially "invisible" to computers. The examples for which the chaos is "visible" do not properly conjugate the shift space onto the underlying set over which the dynamics take place—for example, the $2x \text{ mod } 1$ map, the logistic map for $r=4$, and the tent map for $r = 2$. (You can check this by observing that in the latter, the shift space—which is compact—is mapped onto open sets and/or the mapping is not a homeomorphism at all.)

Apologies if the question is elementary—I'll remove it if indicated to.

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  • $\begingroup$ What does repelling mean? Is it repelling in $X=\{0,1,2\}^{\mathbb{N}}$? $\endgroup$
    – Ville Salo
    Commented Jan 13, 2021 at 20:37
  • $\begingroup$ A repelling subset of $f$ is a set that is attracting for the “temporally backwards” dynamics of $f$, and a subset of $X$ is attracting under $f$ if every point of that subset possesses an open neighborhood such that every point within that open neighborhood approaches the subset asymptotically as time goes to infinity. $\endgroup$ Commented Jan 13, 2021 at 20:56
  • $\begingroup$ You give the definition of attracting for a function, but the inverse of a function may not be a function, so the definition of repelling is still not completely clear. In any case my example clearly solves your question for any variant. $\endgroup$
    – Ville Salo
    Commented Jan 14, 2021 at 7:08

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