32
$\begingroup$

This question assumes familiarity with combinatorial cardinal characteristics of the continuum.

Identify an infinite set $a\subseteq\mathbb{N}$ with its increasing enumeration. Thus, for each natural number $n$, $a(n)$ is the $n$-th element of $a$ in increasing order. This way, every infinite set of a natural number is a member of $\mathbb{N}^{\mathbb{N}}$.

For $f,g\in\mathbb{N}^{\mathbb{N}}$, write $f\le^{\infty}g$ if $f(n)\le g(n)$ for infinitely many $n$. That is, if $g\not<^* f$.

For an infinite co-infinite set $a\subseteq\mathbb{N}$, let $a^{c}$ be its (infinite) complement in $\mathbb{N}$. Thus, $a$ and $a^{c}$ are both in $\mathbb{N}^{\mathbb{N}}$, and the following notion is natural.

A set $Y\subseteq\mathbb{N}^{\mathbb{N}}$ is bi-nondominating if there is an infinite co-infinite set $a\subseteq\mathbb{N}$ such that $Y\le^{\infty}a,a^{c}$. That is, for each $f\in Y$ we have that $f\le^{\infty}a$ and $f\le^{\infty}a^{c}$.

Definition. $\mathfrak{bidi}$ is the minimal cardinality of a set $Y\subseteq\mathbb{N}^{\mathbb{N}}$ that is not bi-nondominating. That is, such that for each infinite co-infinite set $a\subseteq\mathbb{N}$, there is $f\in Y$ such that $a\le^{*}f$ or $a^{c}\le^{*}f$.

It follows immediately that $\mathfrak{b}\le\mathfrak{bidi}\le\mathfrak{d}$. It can be shown that if we close the sequences in $Y$ under left shifts, then a witness for $Y$ being bi-nondominating actually reaps $Y$. Thus, $\mathfrak{bidi}\le\mathfrak{r}$, the reaping (or unsplitting) number. Since bounded sets in the Baire space $\mathbb{N}^{\mathbb{N}}$ are meager, and the operator $a\mapsto a^{c}$ is a hoemomorphism from the Cantor space $P(\mathbb{N})$ onto itself, we also have that the covering number for meager (or Baire first category) sets, $\mbox{cov}(\mathcal{M})$, is not greater than $\mathfrak{bidi}$. In summary, we have: $$ \max\left\{ \mbox{cov}(\mathcal{M}),\mathfrak{b}\right\} \le\mathfrak{bidi}\le\min\left\{ \mathfrak{r},\mathfrak{d}\right\} . $$

Questions. 1. Is $\mathfrak{bidi}$ a new combinatorial cardinal characteristic of the continuum?

  1. More precisely, is it conistently not in the lattice generated by the known combinatorial cardinal characteristics of the continuum by maxima and minima?

  2. More concretely, are the inequalities $\max\left\{ \mbox{cov}(\mathcal{M}),\mathfrak{b}\right\} <\mathfrak{bidi}$ and $\mathfrak{bidi}<\min\left\{ \mathfrak{r},\mathfrak{d}\right\} $ each consistent?

Guess. My guess is that the answer (to all questions) is "Yes", but I would be happier if it turns out to be "No."

Motivation. A topological space is Menger if for each countable family of open covers of that space, there are finite subsets of the open covers that together cover the space. This property was introduced by Menger in 1924, and in this form by Hurewicz (1925). It found numerous connections and applications to various branches of mathematics, even to additive Ramsey theory.

A famous and notorious open problem asks whether, consistently, the square of every Menger set of real numbers is Menger. In a joint work with Piotr Szewczak, we proved that if $\mathfrak{bidi}=\mathfrak{d}$, then there is a Menger set of real numbers whose square is not Menger. Such a result was earlier known under the stronger hypothesis $\mbox{cov}(\mathcal{M})=(\mathfrak{d}=)\mbox{cof}(\mathcal{M})$.

A survey of the standard models. Following are comments about $\mathfrak{bidi}$ in models obtained by adding standard generic reals (usually, $\aleph_2$ many) to a model of CH.

The Cohen model. $\mbox{cov}(\mathcal{M})=\mathfrak{c}$, so $\mathfrak{bidi}=\mathfrak{c}$, too.

The Random reals model. $\mathfrak{d}=\aleph_1$, and thus $\mathfrak{bidi}=\aleph_1$.

The Sack model. Rarely a good model to separate cardinals of this kind, since all classic cardinals are $\aleph_1$ there.

The Dominating reals (Hechler) model, the Laver model, the Mathias model. $\mathfrak{b}=\mathfrak{c}$, so $\mathfrak{bidi}=\mathfrak{c}$.

Miller's model. $\mathfrak{r}=\aleph_1$, and thus $\mathfrak{bidi}=\aleph_1$.

Mildenberger suggests that adding Miller reals and then random reals may result in $\max\left\{ \mbox{cov}(\mathcal{M}),\mathfrak{b}\right\} < \min\left\{ \mathfrak{r},\mathfrak{d}\right\}$. What is $\mathfrak{bidi}$ in this model? (I guess that adding enough random reals to any model of $\mbox{cov}(\mathcal{M})<\mathfrak{d}$ wound render $\max\left\{ \mbox{cov}(\mathcal{M}),\mathfrak{b}\right\} < \min\left\{ \mathfrak{r},\mathfrak{d}\right\}$.)

$\endgroup$
  • $\begingroup$ Just a short question (probably trivial and/or irrelevant): Let $K$ be the collection of infinite co-infinite sequences and let ${\frak bidi}'= \text{min}\{|S|: S\subseteq K \land \forall a\in K \exists s\in S (a \leq s \lor a^c \leq s)\}$ where $\leq$ is the componentwise comparison of members of $K$ if we see them as members of $\omega^\omega$. Do we have ${\frak bidi}'= {\frak bidi}$? (The analogous consideration for ${\frak d}$ shows that we can work with either $\leq$ or $\leq^*$, see mathoverflow.net/questions/206898/… ) $\endgroup$ – Dominic van der Zypen May 19 '15 at 7:53
  • $\begingroup$ @DominicvanderZypen: I think that the same argument why $\mathfrak{d}=\mathfrak{d}'$ shows that $\mathfrak{bidi}=\mathfrak{bidi}'$: If $Y$ is not bi-nondominating, you can close $Y$ under finite modifications of its elements. The cardinality will not increase (if $Y$ is infinite), and now it becomes not bi-nondominating' in your sense. $\endgroup$ – Boaz Tsaban May 19 '15 at 8:41
  • $\begingroup$ Maybe we can use the theorem of $\mathfrak{b}<\mathfrak{d}$ using a Cohen real, but showing that it adds a "bi-unbounded" real i.e. real $x$ such that both $x$ and $x^c$ are unbounded by the ground model reals (pretty much the same as Lemma IV 7.35 in Kunen's Set Theory book). For that we can 1) add to the forcing conditions $p$ a set $A$ indicating natural numbers that will belong to $x^c$ and 2) Show that given $h$ the new sets $D_m = \{p: |range(p)|\ge m, |A| \ge m, \exists n_1 \ge m: p'(n_1) \ge h(n_1), A'(n_1) \ge h(n_1) \}$ (where $y'$ is the range of $y$ in increasing order) are dense. $\endgroup$ – Eran May 19 '15 at 14:21
  • 2
    $\begingroup$ @Eran: See survey of standard models added now to the question. $\endgroup$ – Boaz Tsaban May 19 '15 at 18:28
  • 1
    $\begingroup$ @GuidoJorg see the paper "proofs from the book" for an introduction to these properties (including history and references). Available at my webpage ( math.biu.ac.il/~tsaban/papers.html ) $\endgroup$ – Boaz Tsaban Oct 24 '15 at 17:01
18
$\begingroup$

Using a characterisation of $\min\{\mathfrak{d},\mathfrak{r}\}$ that comes from dualizing a result in Kamburelis' and Węglorz's paper called "Splittings":

A. Kamburelis, B. Węglorz, Splittings, Archive for Mathematical Logic 35, Issue 4 (1996) 263-277, doi:10.1007/s001530050044,

the cardinal invariant $\mathfrak{bidi}$ is equal to that minimum.

To introduce their result, consider the following notion of splitting: given $a\in[\omega]^\omega$ and $\bar{I}=\langle I_n\rangle_{n<\omega}$ an interval partition of $\omega$, $a$ splits $\bar{I}$ iff $a$ contains infinitely many $I_n$'s and also its complement $\omega\smallsetminus a$ contains infinitely many $I_n$'s.

A family $\mathcal{I}$ of interval partitions of $\omega$ is unreapable if no $a\in[\omega]^\omega$ splits all the members of $\mathcal{I}$. On the other hand, say that $A\subseteq[\omega]^\omega$ is interval-splitting if any interval partition of $\omega$ is splitted by some member of $A$.

Kamburelis and Weglorz proved that $\max\{\mathfrak{b},\mathfrak{s}\}$ is equal to the smallest size of an interval-splitting family. On the other hand, the dual proof leads to see that $\min\{\mathfrak{d},\mathfrak{r}\}$ is the smallest size of an unreapable family of interval partitions of $\omega$.

We use this result to obtain $\min\{\mathfrak{d},\mathfrak{r}\}\leq\mathfrak{bidi}$ (the other inequality was pointed in Tsaban's question). Let $Y\subseteq\omega^\omega$ of size $<\min\{\mathfrak{d},\mathfrak{r}\}$, wlog, we may assume that, for any $h\in Y$, $h$ is increasing and $h(0)>0$. Now, consider $h^*\in\omega^\omega$ defined as $h^*(0)=0$ and $h^*(k+1)=h(h^*(k))$ and let $\bar{I}^h$ be the interval partition where $I^h_k=[h^*(2k),h^*(2k+2))$. Therefore (by Kamburelis and Weglorz), there exists an $a\in[\omega]^\omega$ that splits $\bar{I}^h$ for all $h\in Y$. It is very easy to conclude that $Y$ is bi-nondominating (as defined by Tsaban) by $a$, to be more specific, if $I^h_k\subseteq a$ then the $h^*(2k)$-th element of $\omega\smallsetminus a$ is bigger than $h^*(2k+1)=h(h^*(2k))$ and, likewise, if $I^h_k\subseteq \omega\smallsetminus a$ then the $h^*(2k)$-th element of $a$ is bigger than $h(h^*(2k))$.

Naturally, the previous argument would lead to a Tukey embedding that also gives $\max\{\mathfrak{b},\mathfrak{s}\}$ bigger than or equal (actually, just equal) to the dual of $\mathfrak{bidi}$, i.e., the minimal size of a family $A\subseteq[\omega]^\omega$ with the property that, for any $y\in\omega^\omega$ there is an $a\in A$ such that $y$ doesn't dominate both the increasing enumerations of $a$ and $\omega\smallsetminus a$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

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