Abstract
Suppose f is a map of a continuum X onto itself. A periodic continuum of f is a subcontinuum K of X such that f n [ K ] = K for some positive integer n. A proper periodic continuum of f is a periodic continuum of f that is a proper subcontinuum of X. A proper periodic continuum of f is maximal if and only if X is the only periodic continuum that properly contains it. In this paper it is shown that the maximal proper periodic continua of a map of a hereditarily decomposable chainable continuum onto itself follow the Sarkovskii order, provided the maximal proper periodic continua are disjoint. The case in which the Sarkovskii order does not hold reduces to the scenario in which the map's domain is the union of two overlapping period-two continua, each of which is maximal.
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