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I want to prove the following simple lemma:

Let $T(x)=2x\mod 1$, be defined on $[0\,,1)$ to itself,suppose for any $k\ge 0$, $T^{k}(a)\notin(a\,,b)$, then we have

$\Omega=\{x\in[0\,,1):\mbox{for any k}\geq 0$, $T^{k}(x)\notin(a\,,b)\}$ is not closed. where $(a\,,b)\subset[0\,,1)$.

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  • $\begingroup$ We can consider $b-a<0.1$ as we can see, if the interval is bigger, then $\Omega$ may be just one pont, generally, if the map T is defined as a continuous map, this lemma is not true, therefore, we can consider the discontinuous points. $\endgroup$
    – Kan
    Commented Feb 21, 2014 at 11:10
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    $\begingroup$ I'm not sure what your question is but if you want to know about the doubling map with holes, I suggest my recent paper (joint with P. Glendinning): maths.manchester.ac.uk/~nikita/asymmetrical-final.pdf $\endgroup$
    – Nikita Sidorov
    Commented Feb 22, 2014 at 13:51
  • $\begingroup$ Thanks very much. Actually, I was reading your paper, and I want to study some simple and further problems which depend on your recent paper. $\endgroup$
    – Kan
    Commented Feb 23, 2014 at 9:18
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    $\begingroup$ Hi, I'm sure about your question either, but I was wandering what do you actually want. Do you want Ω to be open? dense? or another property stronger that not close? Also, I think that your lemma is not true if a=0 since the corresponding subshift is conjugated to a β-shift, see J. Nilsson. On numbers badly approximable by dyadic rationals. Israel J. Math., 171:93-110, 2009. Also, is there any hypothesis on $b$? $\endgroup$
    – Rafael Alcaraz Barrera
    Commented Feb 23, 2014 at 23:03
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    $\begingroup$ Shameless self-promotion... You might also be interested in arxiv.org/abs/1112.5390 (published in Nonlinearity). $\endgroup$
    – user25199
    Commented Feb 24, 2014 at 7:45

1 Answer 1

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I thought about my previous comment and I can assure that in general your lemma is false. If $a > \frac{1}{2}$ and you consider the hole $(a,1)$, $(\Omega, T\mid_{\Omega})$ is conjugated to a $\beta$ shift see J. Nilsson. On numbers badly approximable by dyadic rationals. Israel J. Math., 171:93-110, 2009. Also, Theorem 3.5 of S. Bundfuss, T. Kruger, and S. Troubetzkoy. Topological and symbolic dynamics for hyperbolic systems with holes. Ergodic Theory Dynam. Systems, 31(5):1305-1323, 2011, gives a negative answer to your question. Here, the authors do not consider $T$ as a continuous map.

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  • $\begingroup$ Yes. I know this paper:(Topological and symbolic dynamics for hyperbolic systems with holes. ), in general. this lemma is false, because we can consider the hole is very big,and the only orbit can escape the hole is 0, for example $(a\,,b)=(0.1\,,0.8)$ $\endgroup$
    – Kan
    Commented Feb 24, 2014 at 8:32
  • $\begingroup$ My question is that if the length of $(a\,,b)$ is small, can we say something about $\Omega$, we should note the importance of the assumption.Sidorov prove that we only need to consider $(a\,,b)\subset(0\,,\dfrac{1}{2})$ or $(a\,,b)\in (\dfrac{1}{4}\,,\dfrac{1}{2})\times(\dfrac{1}{2}\,,\dfrac{3}{4})$ in his paper: The doubling map with asymmetrical holes. $\endgroup$
    – Kan
    Commented Feb 24, 2014 at 8:41
  • $\begingroup$ I will consider your comment carefully, Thanks a lot. $\endgroup$
    – Kan
    Commented Feb 24, 2014 at 8:41

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