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Let $A$ be an infinite subset of $\omega$ such that $\omega\setminus A$ is also infinite.

Under the Continuum Hypothesis is there a sequence $(A_\alpha)_{\alpha<\omega_1}$ of subsets of $\omega$ such that

$A_0=A$;

$|A_{\alpha+1}\setminus A_\alpha|<\omega$; and

$|A_\alpha\setminus A_{\alpha+1}|=\omega$

for every $\alpha<\omega_1$?

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  • $\begingroup$ Do you want $\subseteq^*$ or just finite differences? Because the answer is yes in either case, but for different reasons. Please settle on a version before I put any more energy into writing an answer... $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 10, 2018 at 17:36
  • $\begingroup$ @AsafKaragila I would like for $A_{\alpha+1}$ to be contained in $A_\alpha$, up to some finite error. $\endgroup$
    – Forever Mozart
    Commented Dec 10, 2018 at 17:37
  • $\begingroup$ As regards the current question, the constant sequence $A_\alpha=A$ works... Second, your axioms on the sequence make no relation between $A_\alpha$ and $A_\beta$ except when $\alpha$ and $\beta$ differ by addition by an integer. Please write your question more carefully. (Your last edit doesn't address my second sentence. Currently it's obvious to arrange a sequence of length $2^{\aleph_0}$.) $\endgroup$
    – YCor
    Commented Dec 10, 2018 at 17:43
  • $\begingroup$ @YCor I would like the elements of the sequence to be distinct if possible $\endgroup$
    – Forever Mozart
    Commented Dec 10, 2018 at 17:45
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    $\begingroup$ Obligatory keyword reference: $\frak t$, by the way. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 10, 2018 at 17:48

1 Answer 1

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Yes, easily. Consider all well-ordered sequences $(A_\alpha)$, of any length, such that each $A_\alpha$ is infinite and satisfying your almost containment condition. Make one such sequence less than another if the first is an initial segment of the second. Zornicate to get a maximal such sequence, and it is easy to show that a maximal sequence cannot have countable length.

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    $\begingroup$ One should add, CH has nothing to do with it. Although it is consistent that a maximal one does have length $\omega_1$ which is strictly smaller than the continuum. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 10, 2018 at 17:44
  • $\begingroup$ @Nik Thank you, your assumption was correct too. I would also be interested in a more constructive (inductive) proof if it exists, because I am trying to apply this to construct a particular example. $\endgroup$
    – Forever Mozart
    Commented Dec 10, 2018 at 17:49
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    $\begingroup$ It's not what I mean. I mean that one should ask some relation (other than just being distinct) between $A_\alpha$ and $A_\beta$ for all $\alpha<\beta$, not only between $A_\alpha$ and $A_{\alpha+n}$ for $n<\omega$. $\endgroup$
    – YCor
    Commented Dec 10, 2018 at 17:51
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    $\begingroup$ So you just consider the poset $\omega_1$ as a disjoint union of plenty of copies of the poset $\omega$. $\endgroup$
    – YCor
    Commented Dec 10, 2018 at 17:56
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    $\begingroup$ YCor is right, you want to require $A_\beta\setminus A_\alpha$ to be finite for all $\beta > \alpha$. $\endgroup$
    – Nik Weaver
    Commented Dec 10, 2018 at 18:18

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