(Edit: I'm splitting the question, leaving here only what is answered by Ashutosh, and moving the rest to another question.)


This question assumes familiarity with combinatorial cardinal characteristics of the continnum. It is a refined version of an [earlier question](https://mathoverflow.net/questions/238424/what-is-the-height-or-depth-of-mathbbn-infty).


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 $[\mathbb{N}]^\infty$ that
is $\subset^\*$-decreasing as the ordinal number $\alpha$ increases.


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.
Let us call this axiom BT (and ignore the coincidence).


It is [known](http://dx.doi.org/10.1090/S0002-9939-10-10407-9) or easy to see that:


1. If there is an unbounded tower of any cardinality, then BT holds. (The present proof is dichotomic.)
2. If $\mathfrak{t}=\mathfrak{b}$ or $\mathfrak{b}<\mathfrak{d}$, then BT holds.


**Question.** Is BT consistent with "$\aleph\_1=\mathfrak{t}<\mathfrak{b}=\mathfrak{c}=\aleph\_2$"?


Will Brian's answer for my [previous question](https://mathoverflow.net/questions/238424/what-is-the-height-or-depth-of-mathbbn-infty) implies that BT fails in the Hechler model. BT also fails in the Laver model, indirectly by the main result of the [linked paper](http://dx.doi.org/10.1090/S0002-9939-10-10407-9).



When is there an unbounded tower in $[\mathbb{N}]^\infty$?
A model for Question 1: Let $V \models$ MA + $2^{\aleph\_0} = \kappa \geq \aleph\_2$. Using MA, construct a $\supseteq^{\star}$-chain $\overline{A} = \langle A\_i : i < \kappa \rangle$ such that $\mathcal{A} = \{p\_{A\_i} : i < \kappa\}$ (where $p\_X(k)$ is the $k$th member of $X$) is a dominating family in $\omega^{\omega}$. Let $P$ add $\aleph\_1$ random reals. Since the random reals added constitute a non null set of size $\aleph\_1$, standard arguments show that $V^P \models \mathfrak{p} = \mathfrak{t} = \aleph\_1$. Since $P$ satisfies ccc and is $\omega^{\omega}$-bounding, $V^P \models \mathfrak{b} = \kappa = 2^{\aleph\_0}$. Finally, $\mathcal{A}$ remains dominating in $V^P$ so the principle BT continues to hold. So $V^p \models \aleph\_1 = \mathfrak{t} < \mathfrak{b} = \kappa = 2^{\aleph\_0} $ plus BT.


