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I read here that the following problem of Hausdorff is apparently still open.

Is there a maximal branch $C$ in the poset $\omega^\omega$ with the eventual domination order, such that $C$ has no $\omega_1$-gap.

My questions concerning the above problem are:

  1. What is its current status?

  2. What are the consequences of its solution? negative, independence or positive.

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I think the problem is solved. See

Francisco Kibedi: Maximal Saturated Linear Orders

See also Kibedi's PhD thesis

Maximal Saturated Linear Orders.

The problem is independence of ZFC. The question asks "is there a pantachy containing no $(\omega_1, \omega_1^*)$-gap?". See Kanovei's answer given in Hahn's Embedding Theorem and the oldest open question in set theory.

Now as the author explains, CH implies a no answer to the above equivalent question (known to Hausdorff). The author shows the consistency of the statement.

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  • $\begingroup$ I don't see how it implies that. $\endgroup$
    – Rahman. M
    Commented Apr 3, 2017 at 9:45
  • $\begingroup$ You make us have to go to the link just to find out if the answer is yes or no or maybe? $\endgroup$
    – bof
    Commented Apr 3, 2017 at 10:13
  • $\begingroup$ The abstract is not exhaustively clear. The key result is formulated as Con(ZFC + ¬CH + ∃ a maximal saturated linear order of size continuum in the space of real-valued sequences partially ordered by eventual domination), where saturated' lacks the cardinal parameter of saturation, while of size continuum' looks strange since the whole poset is of cardinality c. Also, there is no indication how the key result implies a required gapless pantachy. $\endgroup$
    – Vladimir Kanovei
    Commented Apr 4, 2017 at 19:34
  • $\begingroup$ I have just recognized that the reference to `See also Kibedi's PhD thesis Maximal Saturated Linear Orders.' does not work $\endgroup$
    – Vladimir Kanovei
    Commented Apr 5, 2017 at 12:22
  • $\begingroup$ @VladimirKanovei It works but possibly you need to use some anti-filter program (the same is for me). $\endgroup$
    – Mohammad Golshani
    Commented Apr 6, 2017 at 7:47

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