On the dimensional structure of hereditarily indecomposable continua (Q2782669)

Spaces considered are separable and metrizable. A continuum \(X\) is called indecomposable if for every pair \(A\), and \(B\) of subcontinua of \(X\) such that \(X=A\cup B\), either \(A\subset B\) or \(B\subset A\). Of course, one says that \(X\) is hereditarily indecomposable if every subcontinuum of \(X\) is indecomposable. It is known from work of \textit{R. H. Bing} [Trans. Am. Math. Soc. 71, 267-273 (1951; Zbl 0043.16901)] that there exist hereditarily indecomposable continua of every finite dimension. NEWLINENEWLINENEWLINENEWLINENEWLINENEWLINESuppose we are given an \(n\)-dimensional hereditarily indecomposable continuum \(X\). For \(1\leq r\leq n\), we use \(B_r\) to denote the set of points in \(X\) that belong to some \(r\)-dimensional subcontinuum but do not lie in any non-trivial subcontinuum of dimension less than \(r\). The main result of the paper states that if \(N\subset X\) and \(\dim N\leq 0\), then, \(\dim(B_n\cup(B_r\smallsetminus N))=n-(r-1)\). In particular, \(\dim B_n=1\), and one may derive that, \(\dim B_r=n-(r-1)\). NEWLINENEWLINENEWLINEAnother result records that \(\dim (B_n\smallsetminus C)=1\) for any \((n-2)\)-dimensional \(\sigma\)-compact set \(C\) in such an \(X\).