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A set $A$ is said to be almost contained in a set $B$ if $A\setminus B$ is finite. A sequence $(A_\alpha)_{\alpha<\lambda}$ of infinite subsets of $\mathbb N$ will be called a tower if for every $\alpha<\beta<\lambda$, $A_\beta$ is almost contained in $A_\alpha$. A tower is said to have a continuation if exists some infinite subset of $\mathbb N$ that is almost contained in every element of the tower. The tower number is defined to be the minimal cardinality of the set of towers that don't have a continuation.

My question is: why can't the tower number be $\aleph_0$? Or equivalently, why does every tower of countable length necessarily have a continuation?

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    $\begingroup$ I think you need to be a little more specific with your definitions here; for instance, every tower has a continuation because I can just take the continuing set to be $\mathbb{N}$; presumably you want the continuing set to not show up in the sequence under consideration? $\endgroup$ – ItsJustVennDiagramsBro Mar 10 at 21:29
  • $\begingroup$ Yes, that is it. It must not appear in the sequence previously. Will now edit $\endgroup$ – Uri George Peterzil Mar 10 at 21:31
  • $\begingroup$ Wait, the set $\mathbb N$ doesn't work as it is reverse inclusion. So no, any set. $\endgroup$ – Uri George Peterzil Mar 10 at 21:35
  • $\begingroup$ The way you have defined almost containment makes $\mathbb{N}$ almost contained in anything, as $B\setminus\mathbb{N}=\emptyset$. $\endgroup$ – ItsJustVennDiagramsBro Mar 10 at 21:36
  • $\begingroup$ You're right, corrected my definition. $\endgroup$ – Uri George Peterzil Mar 10 at 21:38
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This is a simple diagonalization argument. Suppose we have a countable tower. Taking a cofinal subsequence, we may assume it has length $\omega$ and write it as $(A_n)_{n<\omega}$. We may moreover assume that the $A_n$ are actually literally contained in each other instead of almost contained in each other, by replacing $A_n$ with $A_0\cap\dots\cap A_n$.

It's now easy to construct a set $B$ which is almost contained in each $A_n$: just pick one element from each $A_n$ to be in $B$. Picking these elements one by one, we can arrange that they are all distinct (since each $A_n$ is infinite), so that the resulting set $B$ is infinite. For each $n$, all but possibly the first $n$ elements we put in $B$ must be in $A_n$ (since they are in $A_m$ for some $m\geq n$), so $B$ is almost contained in $A_n$.

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  • $\begingroup$ Does the edit to the definition of almost containment change your construction of $B$ at all? To be perfectly frank, I have a hard time following your answer due to the lack of justification for reducing to a sequence of length $\omega$ of literally contained sets. $\endgroup$ – ItsJustVennDiagramsBro Mar 10 at 21:44
  • $\begingroup$ I wrote this answer using the correct definition; it's a very commonly used definition in set theory. $\endgroup$ – Eric Wofsey Mar 10 at 21:45
  • $\begingroup$ Cool. I am not familiar with this at all, so I was just curious. Can you still provide more context on why your reduction is okay? $\endgroup$ – ItsJustVennDiagramsBro Mar 10 at 21:46
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    $\begingroup$ It is irrelevant whether $B$ is in the original sequence; what matters is that $B$ is almost contained in every term of the original sequence. That will be true because $B$ is almost contained in every term of a cofinal subsequence of the original sequence. $\endgroup$ – Eric Wofsey Mar 10 at 21:51
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    $\begingroup$ The asker added that remark because they were confused by your questions and didn't realize the issue was just that they had miswritten the definition of "almost contain". $\endgroup$ – Eric Wofsey Mar 10 at 21:55

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