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?

  • 1
    $\begingroup$ 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

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 $A_n$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.