Descending almost-contained subsets of $\omega$ [closed]

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 $$\alpha<\omega_1$$?

closed as off-topic by YCor, Monroe Eskew, Andreas Blass, Ramiro de la Vega, Mark WildonDec 14 '18 at 20:33

This question appears to be off-topic. The users who voted to close gave these specific reasons:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – YCor, Andreas Blass, Mark Wildon
• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Monroe Eskew, Ramiro de la Vega
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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... – Asaf Karagila Dec 10 '18 at 17:36
• @AsafKaragila I would like for $A_{\alpha+1}$ to be contained in $A_\alpha$, up to some finite error. – Forever Mozart Dec 10 '18 at 17:37
• 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 $2^{\aleph_0}$.) – YCor Dec 10 '18 at 17:43
• @YCor I would like the elements of the sequence to be distinct if possible – Forever Mozart Dec 10 '18 at 17:45
• Obligatory keyword reference: $\frak t$, by the way. – Asaf Karagila Dec 10 '18 at 17:48

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.
• 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. – Asaf Karagila Dec 10 '18 at 17:44
• 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 $n<\omega$. – YCor Dec 10 '18 at 17:51
• So you just consider the poset $\omega_1$ as a disjoint union of plenty of copies of the poset $\omega$. – YCor Dec 10 '18 at 17:56
• YCor is right, you want to require $A_\beta\setminus A_\alpha$ to be finite for all $\beta > \alpha$. – Nik Weaver Dec 10 '18 at 18:18