Hausdorff Dimension of Symmetric Perfect Sets

We consider nowhere dense perfect subsets of [0, 1] that are symmetric but have no additional nice properties. We prove that if E = ∩En is a symmetric perfect set and the length of the basic intervals in En is denoted by ln then the Hausdorff dimension of E is \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $$s = \mathop {\lim inf}\limits_{n \to \infty } \{ s_n : 2^n l_n^{9_n } = 1\} = \mathop {\lim inf}\limits_{n \to \infty } \frac{{\log 2^n }}{{ - \log l_n }}$$ \end{document}. The argument we use also shows that using natural covers of E; i.e., covers consisting of the 2n closed, equal length intervals of the nth stage, yield an estimate for the s-dimensional Hausdorff measure within a factor of four.