$\beta ([0,\infty ))$ does not contain nondegenerate hereditarily indecomposable continua
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- by Michel Smith PDF
- Proc. Amer. Math. Soc. 101 (1987), 377-384 Request permission
Abstract:
Bellamy has shown that if $A = \left [ {0,\infty } \right )$, then $\beta A - A$ is an indecomposable continuum and every nondegenerate subcontinuum of $\beta A - A$ can be mapped onto every metric continuum. Thus, it follows that every nondegenerate subcontinuum of $\beta A - A$ contains a nondegenerate indecomposable continuum. We show, however, that no nondegenerate subcontinuum of $\beta A - A$ is hereditarily indecomposable. Thus, every nondegenerate subcontinuum of $\beta A - A$ contains a decomposable continuum as well as a nondegenerate indecomposable continuum.References
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Additional Information
- © Copyright 1987 American Mathematical Society
- Journal: Proc. Amer. Math. Soc. 101 (1987), 377-384
- MSC: Primary 54D40; Secondary 54F20
- DOI: https://doi.org/10.1090/S0002-9939-1987-0902559-8
- MathSciNet review: 902559