Hereditarily indecomposable subcontinua of the product of two Souslin arcs are metric (Q1004035)

A continuum is a compact connected Hausdorff space. A continuum \(X\) is hereditarily indecomposable provided that for each pair of subcontinua \(A\) and \(B\) of \(X\) such that \(A\cap B\not=\emptyset\), we have that either \(A\subset B\) or \(B\subset A\). A Hausdorff arc is a continuum with exactly two nonseparating points. A Souslin line is a nonseparable linearly ordered topological space in which each collection of pairwise disjoint open sets is countable. A Souslin arc is a Hausdorff arc that is a compactification of a connected Souslin line. A partially ordered set \(X\) is said to be an antichain if no two elements of \(X\) have a common lower bound. A partially ordered set \(X\) is said to satisfy the countable chain condition (ccc), if every antichain in \(X\) is countable. The author uses the \(ccc\) condition to prove the following: Theorem 1. Let \(X\) and \(Y\) be two Hausdorff arcs, \(Z=X\times Y\) and suppose that there is a hereditarily indecomposable continuum \(M\) lying in \(Z\) so that the projections \(\pi_1(M)\) and \(\pi_2(M)\) are onto. Then if \(X\) is Souslin arc, then \(Y\) is metric. As a consequence of the \(ccc\) condition and Theorem 1, the author obtains: Theorem 2. If \(X\) and \(Y\) are two Souslin arcs and \(M\) is a hereditarily indecomposable subcontinuum of \(X\times Y\), then \(M\) is a metric continuum. The author makes the following comment: In fact the proof (of Theorem 2) shows that not only is the continuum \(M\) of the theorem metric, but homeomorphic to a planar continuum. Question. What about higher dimensions? We conjecture that if \(\{S_j\}_{j=1}^n\) is a finite sequence of Souslin arcs and \(M\subset\prod_{j=1}^nS_j\) is a hereditarily indecomposable continuum, then \(M\) is a metric continuum.