# Why can't the tower number be $\aleph_0$?

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?

• 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? – ItsJustVennDiagramsBro Mar 10 at 21:29
• Yes, that is it. It must not appear in the sequence previously. Will now edit – Uri George Peterzil Mar 10 at 21:31
• Wait, the set $\mathbb N$ doesn't work as it is reverse inclusion. So no, any set. – Uri George Peterzil Mar 10 at 21:35
• The way you have defined almost containment makes $\mathbb{N}$ almost contained in anything, as $B\setminus\mathbb{N}=\emptyset$. – ItsJustVennDiagramsBro Mar 10 at 21:36
• You're right, corrected my definition. – 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 $$A_n$$.
• 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. – ItsJustVennDiagramsBro Mar 10 at 21:44
• 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. – Eric Wofsey Mar 10 at 21:51