1973
DOI: 10.2140/pjm.1973.45.643
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Non-aposyndesis and non-hereditary decomposability

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Cited by 15 publications
(4 citation statements)
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“…But surprisingly the characterization is not valid in 3-space and Example 1 is a continuum not separated by any subcontinuum, not containing any region-containing indecomposable subcontinuum yet not of type A'. Interestingly enough though, if M is hereditarily decomposable it is of type A' and this is proved in Theorem 2 using a result of Schlais [6]. But, whereas the former condition is too weak to obtain a characterization as Example 1 shows, the latter condition is too strong as easy examples show.…”
mentioning
confidence: 98%
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“…But surprisingly the characterization is not valid in 3-space and Example 1 is a continuum not separated by any subcontinuum, not containing any region-containing indecomposable subcontinuum yet not of type A'. Interestingly enough though, if M is hereditarily decomposable it is of type A' and this is proved in Theorem 2 using a result of Schlais [6]. But, whereas the former condition is too weak to obtain a characterization as Example 1 shows, the latter condition is too strong as easy examples show.…”
mentioning
confidence: 98%
“…Conditions for type A' continua. The first theorem in this section depends on a theorem due to Schlais [6]. Schlais states his theorem for a point x but the proof goes through without any difficulty if x is replaced by a continuum H with void interior and will be stated here in this more general setting.…”
mentioning
confidence: 99%
“…For each point x of a continuum M, F. B. Jones [2] defines K(x) to be the set consisting of all points y of M such that M is not aposyndetic at x with respect to v. H. E. Schlais [4] proved that if M is a hereditarily decomposable continuum, then for each point x of M, no nonempty open set in M is contained in K(x). A continuum M is said to be A connected if any two of its points can be joined by a hereditarily decomposable continuum in M. Here we prove that if M is a A connected plane continuum, then for each point x of M, the set K(x) does not contain a nonempty open subset of M. Suppose that M is a plane continuum.…”
mentioning
confidence: 99%
“…We call a nondegenerate metric space that is compact and connected a continuum. For each point x of a continuum M, F. B. Jones [2] defines K(x) to be the set consisting of all points y of M such that M is not aposyndetic at x with respect to y. H. E. Schlais [4] proved that if M is a hereditarily decomposable continuum, then for each point x of M, no nonempty open set in M is contained in K(x). A continuum M is said to be A connected if any two of its points can be joined by a hereditarily decomposable continuum in M. Here we prove that if M is a A connected plane continuum, then for each point There exists a sequence of disjoint circular regions {Zi} in E2 that converges to p such that for each positive integer i, the region Zi contains Cl Ui, the set MnZi is in K(x), and the point x does not belong to Cl Zi.…”
mentioning
confidence: 99%