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$ \beta([0,\infty))$ does not contain nondegenerate hereditarily indecomposable continua


Author: Michel Smith
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
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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 [Enhancements On Off] (What's this?)

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Additional Information

DOI: https://doi.org/10.1090/S0002-9939-1987-0902559-8
Keywords: Stone-Čech compactifications, nonmetric continua, indecomposable continua, hereditarily indecomposable continua
Article copyright: © Copyright 1987 American Mathematical Society

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