5
$\begingroup$

(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.

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 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 implies that BT fails in the Hechler model. BT also fails in the Laver model, indirectly by the main result of the linked paper.

$\endgroup$
2
  • 1
    $\begingroup$ I think adding $\aleph_1$ random reals to a model of MA plus not CH should work for Q1. This is because random forcing is $\omega^{\omega}$-bounding. $\endgroup$
    – Ashutosh
    May 10, 2016 at 20:03
  • 1
    $\begingroup$ @Ashutosh: Could you provide some more details? Intuitively, MA gives t=b=c so we have an unbounded tower. Then the randoms only add bounded reals, so the same tower remains unbounded in the extension. Would appreciate if you could post some more details as an answer (i.e., not just comment). $\endgroup$ May 10, 2016 at 21:46

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Does $\operatorname{cov}(\mathcal{N})$ remain $\mathfrak{c}$ in this extension? $\endgroup$ May 16, 2016 at 18:02
  • 1
    $\begingroup$ Yes, adding any number of random reals increases covering of null. This can be proved using Fubini's theorem. $\endgroup$
    – Ashutosh
    May 16, 2016 at 19:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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