\(\mathcal{T}\)-closed sets (Q524352)

A continuum is a compact, connected metric space. Let \({\mathcal T}\) denote Jones' set function. A subset \(A\) of a continuum \(X\) is called a \({\mathcal T}\)-closed set if \({\mathcal T}(A) = A\). \textit{D.~Bellamy} et al. in [Topology Appl. 195, 209--225 (2015; Zbl. 1333.54022)] proved that if \(A\) is a subset of a continuum \(X\) and \(A\) is a \({\mathcal T}\)-closed set, then every component of \(X \setminus A\) is open and continuumwise connected. Then they asked (Question 4.6) if the converse is true if we additionally assume that the set function \({\mathcal T}\) is idempotent on \(X\) (i.e. \({\mathcal T}({\mathcal T}(A)) = {\mathcal T}(A)\) for any subset \(A\) of \(X\)). In this paper the authors provide a negative answer to that question. They construct a continuum \(X\) on which the set function \({\mathcal T}\) is idempotent, and a subcontinuum \(A\) of \(X\) such that every component of \(X \setminus A\) is open and continuumwise connected but \({\mathcal T}(A) \neq A\). Moreover they give a characterization of \({\mathcal T}\)-closed sets.