An indecomposable continuum as subpower Higson corona (Q2631604)

A metric space is called proper if all its closed balls are compact. For a proper metric space $(X,d)$, the author considers in ZFC the following compactifications of $(X, d)$: the Higson compactification $h_H(X)$, the subpower Higson compactification $h_P(X)$ and the sublinear Higson compactification $h_L(X)$. Then $h_L(X)\preccurlyeq h_P(X)\preccurlyeq h_H(X)$ where $\preccurlyeq$ is the standard partial order between Hausdorff compactifications. For $A\in\{H, P, L\}$, the remainder $v_A X=h_A(X)\setminus X$ is called the Higson-type corona of $X$. In particular, $v_H X$ is the Higson corona of $X$, $v_P X$ is the subpower Higson corona of $X$, and $v_L X$ is the sublinear Higson corona of $X$. Let us denote by $[0, +\infty)$ the metric space $([0, +\infty), d)$ where $d(x,y)=\vert x-y\vert$ for all $x,y\in [0,+\infty)$. The author proves that the subpower Higson corona $v_P[0, +\infty)$ of $[0, +\infty)$ is a non-metrizable indecomposable continuum and, for each $A\in\{H, P, L\}$, there exists a continuous surjection from $v_P [0, +\infty)$ onto the compactification $h_A([0, +\infty))$. Using the example given on pages 183--184 in [\textit{Y. Iwamoto} and \textit{K. Tomoyasu}, Tsukuba J. Math. 25, No. 1, 179--186 (2001; Zbl 1017.54014)], the author shows that there is a proper metric $\rho$ on the interval $[0,+\infty)$ equivalent to $d$ such that the subpower Higson corona of the metric space $([0, +\infty), \rho)$ is a decomposable continuum. Among the open problems posed in the article, there is the following question: Is the sublinear Higson corona of $[0, +\infty)$ an indecomposable continuum? \par Reviewer's remark: Unfortunately, Corollary 2.5 is false. Perhaps, the author intended to formulate Corollary 2.5 as follows: Let $Y$ be a closed subset of a proper metric space $(X, d)$. If $Y$ is $R$-dense in $(X, d)$ for some positive real number $R$, then, for the closure $\overline{Y}$ of $Y$ in $h_P(X)$, the equality $\overline{Y}\setminus Y=v_P X$ holds, so $v_P Y$ is homeomorphic with $v_P X$.