Die metrische Dimension analytischer Mengen. (The metric dimension of analytic sets) (Q1112371)

Let X be a separable, complete metric space. Define, for \(B\subset X\), \(q>0\), \(\alpha >0\), \[ N(B,q)=\inf \{k\in {\mathbb{N}}:\quad B\subset \cup^{k}_{1}B_ i,\quad diam B_ i\leq q\}, \] \[ {\mathcal U}_{\alpha}=\{A\subset X:\quad \overline{\lim}_{q\to 0+} N(A,q)q^{\alpha}=0\},\quad {\mathcal L}_{\alpha}=\{A\subset X:\quad \underline{\lim}_{q\to 0+}N(A,q)q^{\alpha}=0\}, \] \[ m- \overline{\dim}(B)=\inf \{\alpha >0:\quad B\subset \cup^{\infty}_{1}A_ i,\quad A_ i\in {\mathcal U}_{\alpha}\},\quad m-\underline{\dim}(B)=\inf \{\alpha >0:\quad B\subset \cup^{\infty}_{1}A_ i,\quad A_ i\in {\mathcal L}_{\alpha}\}. \] The author proves the following theorems: (1) If \(A\subset X\) is an analytic set such that m-\underbar{dim}(A)\(<+\infty\), then there is a \(\sigma\)- compact set \(K\subset A\) such that m-\underbar{dim}(A)\(=m\)- \underbar{dim}(K). (2) The same with m-\(\overline{\dim}\) instead of m- \underbar{dim} for every analytic set \(A\subset X\) (possibly having m- \(\overline{\dim}(A)=+\infty)\).