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The topological property of the irregular sets on the lengths of basic intervals in beta-expansions

Published 6 Apr 2016 in math.DS | (1604.01470v5)

Abstract: Let $\beta > 1$ be a real number and $(\epsilon_1(x, \beta), \epsilon_2(x, \beta), \ldots)$ be the $\beta$-expansion of a point $x \in (0, 1]$. For all $x \in (0,1]$, let $A(D(x))$ be the set of accumulation points of $\frac{-\log_\beta |I_n(x)|}{n}$ as $n \rightarrow \infty$, where $|I_n(x)|$ is the length of the basic interval of order $n$ containing $x \in (0, 1]$. In this paper, we prove that $A(D(x))$ is always a closed interval for any $x \in (0,1]$. Furthermore, if $\lambda(\beta)>0$, the extremely irregular set containing points $x \in [0, 1]$ whose upper limit of $\frac{-\log_\beta |I_n(x)|}{n}$ equals to $1+\l(\beta)$ is residual, where $1+\l(\beta)$ is a constant depending on $\beta$. As a consequence, the irregular set with $x\in [0, 1]$ whose limit of $\frac{-\log_\beta |I_n(x)|}{n}$ does not exist is residual for every $\lambda(\beta)>0$.

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