This question assumes familiarity with combinatorial cardinal characteristics of the continnum. It continues an earlier question.

Let $[\mathbb{N}]^\infty$ be the family of infinite subsets of $\mathbb{N}$,
partially ordered by $\subseteq^*$, where $a\subseteq^* b$ means $a\setminus b$ is finite.

Let $\kappa$ be a cardinal number.
A *tower of height $\kappa$* is a $\kappa$-sequence 
$\langle\, s_\alpha : \alpha<\kappa\,\rangle$ in $[\mathbb{N}]^\infty$ such that

 1. This $\kappa$-sequence is $\subset^*$-decreasing as the ordinal number $\alpha$ increases.
 2. The set $\{\,s_\alpha : \alpha<\kappa\,\}$ has no pseudointersection.
(That is, there is no infinite set $s$ such that $s\subseteq^* s_\alpha$ for
all $\alpha<\kappa$).

An element $a\in [\mathbb{N}]^\infty$ is identified with its increasing enumeration. This way, the set $[\mathbb{N}]^\infty$ 
becomes the family of increasing functions in $\mathbb{N}^\mathbb{N}$,
and the standard relation $\le^*$ is defined on 
$[\mathbb{N}]^\infty$.
A set $X\subseteq [\mathbb{N}]^\infty$ is *bounded* if it is bounded (from above) with respect to $\le^*$.

The general goal is to understand when is there an *unbounded* tower of height $\mathfrak{b}$. Let us call this axiom BT (and ignore the coincidence).

It is [known][1] or easy to see that:

 1. An unbounded set has no pseudointersection. So we may remove the need for no pseudointersection from the definition of *tower* without altering BT.
 2. If there is an unbounded tower of any cardinality, then BT holds. (The present proof is dichotomic.)
 3. If $\mathfrak{t}=\mathfrak{b}$ or $\mathfrak{b}<\mathfrak{d}$, then BT holds.

**Open-ended question.** Can the axiom BT be expressed using (standard) 
cardinal characteristics of the continuum?

[Ashutosh][2] proved that BT is consistent with "$\aleph_1=\mathfrak{t}<\mathfrak{b}=\mathfrak{c}=\aleph_2$".

[Will Brian][4] points out that that BT fails in the Hechler model. BT also fails in the Laver model, indirectly by the main result of the [linked paper][1].

**Question 1.** Does any additional inequality or inequality among cardinals of the continuum (one not following from 
$\mathfrak{t}=\mathfrak{b}$ or $\mathfrak{b}<\mathfrak{d}$) imply BT? 

**Question 2.** Does BT imply any equality among cardinals of the continuum? 

**Question 3.** Does BT fail in the Mathias model?

Since BT follows from CH, it does not imply any inequality.

**Motivation.** BT implies that, in the realm of real sets, the selective covering property $\operatorname{S}_1(\Gamma,\Gamma)$ (which is [consistently trivial][1]) is nontrivial.


  [1]: http://dx.doi.org/10.1090/S0002-9939-10-10407-9
  [2]: http://mathoverflow.net/questions/238500/when-is-there-an-unbounded-tower-in-mathbbn-infty
  [3]: http://www.math.lsa.umich.edu/~ablass/hbk.pdf
  [4]: http://mathoverflow.net/questions/238424/what-is-the-height-or-depth-of-mathbbn-infty