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    Re: Easy Lunars in 1790
    From: Alexandre Eremenko
    Date: 2006 Apr 28, 19:45 -0400

    It has to be double checked yet,
    but I suspect that in my favorite Volkovyski
    example (cited in our paper in Iberoamericana),
    all singularities are true (have branches which do not
    return when you continue on a Jordan curve),
    and there are uncountably many of them.
    So they are not K.
    Thus it seems that no good conjectures remain.
    1. True (in the sense above) cannot be on the boundary
    of the completely invariant domain.
    2. Direct implies true. (So you fixed the bug in our paper
    with Liubich).
    3. True does not imply direct. (Simple example).
    3. True implies linearly accessible.
    4. But not vice versa, by your wonderful example.
    5. True can be uncountably many (and thus true does not
    imply K).
    6. K does not imply true (Take a K-singularity described in Goldberg's
    paper. Topologically it is like (sin z)/z).
    What else to ask?

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