A tower is a subset $T\subset [\omega]^\omega$ of the family $[\omega]^\omega$ of all infinite subsets of $\omega$ such that $T$ is well-ordered by the relation $\supset^*$ of almost inclusion and has no infinite pseudointersections. A tower is regular if its cardinality is a regular cardinal.

Consider two small uncountable cardinals:

$\mathfrak t=\min\{|T|:T\subset[\omega]^\omega$ is a tower$\}$;

$\hat{\mathfrak t}=\sup\{|T|:T\subset[\omega]^\omega$ is a regular tower$\}$.

It is clear that $$\mathfrak t\le\hat{\mathfrak t}\le\mathfrak c.$$ Martin's Axiom implies $\mathfrak t=\hat{\mathfrak t}=\mathfrak c$. On the other hand the strict inequality $\mathfrak t<\hat{\mathfrak t}$ is known to be consistent (but I do not know the reference).

The cardinal $\mathfrak t$, called the tower number, is well-studied in Set Theory.

I am interested in the cardinal $\hat{\mathfrak t}$, more precisely in its relation to other known cardinal characteristics of the continuum.

Problem 1. Is there any non-trivial upper or lower bound for the cardinal $\hat{\mathfrak t}$? In particular, is $\mathfrak b\le\hat{\mathfrak t}$?

Problem 2. In which known models of ZFC does the strict inequality $\mathfrak t<\hat{\mathfrak t}$ hold?

Maybe there is some fixed standard notation for $\hat{\mathfrak t}$?

  • $\begingroup$ Did you really mean $\sup$ and not $\min$ in $\hat t$? $\endgroup$ Dec 28, 2019 at 10:07
  • $\begingroup$ @JoelDavidHamkins Yes, I had in mind sup, not min (min yields the well-known cardinal $t=p$). $\endgroup$ Jan 1, 2020 at 11:18

2 Answers 2


Assuming I understand the definitions correctly, I can give you a couple of references.

(1) Dordal (see below) gives a model in which $\mathfrak{b}=\mathfrak{c}=\aleph_2$ and all towers have cardinality $\aleph_1$. Thus, in his model $\mathfrak{t}=\hat{\mathfrak{t}}=\aleph_1<\aleph_2=\mathfrak{b}=\mathfrak{c}$, and so the last question in your first problem has a negative answer.

Edit: In fact, $\mathfrak{h}=\aleph_2$ in the model since he is using Mathias forcing, so I'm not sure of any good candidate for a non-trivial lower bound for $\hat{\mathfrak{t}}.$

(2) On the other hand, the model of Blass and Shelah discussed here gives a model in which $\mathfrak{t}=\mathfrak{b}=\aleph_1$ and there is a tower of length $\aleph_2$ (namely, the generating tower for the simple $P_{\aleph_2}$-point). Thus, this gives a model where $\mathfrak{t}<\hat{\mathfrak{t}}$.

Edit 3/17/20

I tracked down another reference (due to Dordal [2]) with some more information on this question. Theorem 1.3 in the paper gives a connection between towers in the structures $^\omega\omega$ and $[\omega]^\omega$:

Theorem (Dordal, Theorem 1.3 of [2])

Let $\kappa$ be an uncountable regular cardinal, and suppose there is a $\kappa$-tower in $^\omega\omega$ but not $[\omega]^\omega$. Then there is a $\kappa$-scale in $^\omega\omega$ (that is, $\mathfrak{b}=\mathfrak{d}=\kappa$).

With regard to Problem 1, we may say that if $\mathfrak{b}<\mathfrak{d}$ then $\mathfrak{b}\leq\hat{\mathfrak{t}}$ (apply the above theorem, and use the fact that there is a tower of length $\mathfrak{b}$ in the structure $^\omega\omega$. On the other hand, it is consistent that $\mathfrak{b}=\mathfrak{d}=\aleph_2$ and $\hat{\mathfrak{t}}=\aleph_1$ (this occurs in the model from [1]).

The paper has many other forcing constructions showing that not much more can be said.

[1] Dordal, Peter Lars, A model in which the base-matrix tree cannot have cofinal branches, J. Symb. Log. 52, 651-664 (1987). ZBL0637.03049.

[2] Dordal, Peter Lars, Towers in $[\omega]^{\omega}$ and $^{\omega}\omega$, Ann. Pure Appl. Logic 45, No. 3, 247-276 (1989). ZBL0686.03024.

  • $\begingroup$ Thank you very much for the answer, which I noticed only today, when got interested in $\hat{\mathfrak t}$ once more. $\endgroup$ Oct 16, 2021 at 10:02

This is not a full answer, but I claim that the restriction to regular towers is not relevant.

Theorem. The cardinal $\hat t$ is also the supremum of all tower sizes, not just the regular towers. $$\hat t=\sup\{\ |T|\ \mid\ T\subset[\omega]^\omega\text{ is a regular tower }\}=\sup\{\ |T|\ \mid\ T\subset[\omega]^\omega\text{ is a tower }\}.$$

This will be a consequence of the following lemma.

Lemma. If $T$ is any tower and $S\subset[\omega]^\omega$ is any chain of sets well-ordered by $\supset^*$, then $T$ is a subtower of a tower $T^*$ with $|T^*|\geq|S|$.

Proof of lemma. We may assume $T$ is continuous, and so there are successive sets $A$ and $B$ appearing in $T$ with infinite difference $A-B$. We may build a copy of $S$ on this difference set, containing $B$, and place it between $A$ and $B$ in $T$, thereby constructing a new tower $T^*$ of size at least $|S|$. This new chain $T^*$ is a tower since it is still well-ordered by $\supset^*$ and since it agrees with $T$ on a final segment, it still has no infinite pseudointersection. $\Box$

Proof of theorem. Suppose that $S$ is any tower, and that $T$ is a regular tower. If $|S|$ is regular, then by a direct application of the lemma, we find a regular tower $T^*$ of size at least $|S|$. If $|S|$ is singular, then this is a limit of regular cardinals, and we may extend $T$ to various $T'$ of at least those regular sizes, by using only part of $S$ in the extending process. So the supremum of the sizes of the regular towers is the same as the supremum of the sizes of any tower. $\quad\Box$

I am not sure, however, how large the towers can be. (Without the well-ordered requirement, we can of course find towers of size continuum, using Dedekind cuts in the rationals.)

  • $\begingroup$ Unfortunately, your proof is not clear to me. If you create a tower on the base of a tower $T$ inserting new sets, then you do not change the cofinality of that tower. Consequently, the cofinality of the new tower $T^*$ is equal to the cofinality of the tower $T$ and this does not produce a regular tower. Or I do not see something important? $\endgroup$ Jan 2, 2020 at 9:05
  • $\begingroup$ I agree, I have changed only the cardinality and not the cofinality of the tower. But you defined regular tower in terms of its cardinality, not cofinality. I found that strange, and this is why I posted this answer. Of course, the cofinality of a tower will always be a regular cardinal. $\endgroup$ Jan 2, 2020 at 12:43
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    $\begingroup$ Thank you for your comment due to which I have understood that we (with Bardyla) introduced a wrong defintion of $\hat t$. In that definition the cardinality should be changed by cofinality. $\endgroup$ Jan 5, 2020 at 9:05

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