5
$\begingroup$

Update: Perhaps the question is too difficult. I would appreciate, thus, even just comments or related observations.

This question assumes familiarity with combinatorial cardinal characteristics of the continnum. It is the essence 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 $[\mathbb{N}]^\infty$ 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.

It is known 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.
  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 proved that BT is consistent with "$\aleph_1=\mathfrak{t}<\mathfrak{b}=\mathfrak{c}=\aleph_2$".

Will Brian 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. I suspect that BT also fails in the Mathias model.

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?

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

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.