6
$\begingroup$

This question assumes familiarity with combinatorial cardinal characteristics of the continuum. It is abstracted out of a question in a joint research with Jialiang He. I hope we've got the abstraction right.

A family of subsets of $\mathbb{N}$ is centered if every finite subfamily has an infinite intersection. A pseudointersection of a family is an infinite set that is almost contained in every member of the family.

Let $\mathfrak{ridiculous}$ be the the minimal cardinality of a centered family of subsets of $\mathbb{N}$ with no 2 to 1 image that has a pseudointersection.

By 2 to 1 image of a family $\mathcal{A}$ we mean the family $\{f[A] : A\in\mathcal{A}\}$, for some 2 to 1 function $f\colon \mathbb{N}\to \mathbb{N}$. ($f[A]:=\{f(n):n\in A\}$).

We know that $\mathfrak{p}\le\mathfrak{ridiculous}\le \operatorname{add}(\mathcal{M})$. (We see this using selection principles; direct arguments of course must exist, too.)

Question. Is $\mathfrak{ridiculous}=\mathfrak{p}$?

A negative answer (i.e., consistently ``no'') would be ridiculous.

Improve this question
$\endgroup$
7
  • 12
    $\begingroup$ While I agree that we sort of exhausted the single lettered one, I don't think we should support ridiculous naming conventions like this. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 8, 2018 at 18:10
  • 1
    $\begingroup$ What does centered mean in this context? $\endgroup$
    – Danielle Ulrich
    Commented Nov 8, 2018 at 21:11
  • 1
    $\begingroup$ @AsafKaragila :) Yes, you are right! Prove that the answer is positive, and then the ridiculous name will not be needed anymore! $\endgroup$
    – Boaz Tsaban
    Commented Nov 9, 2018 at 7:08
  • 6
    $\begingroup$ By the way, looking at $\mathfrak{stupid}$ as the product of known cardinal characteristics, we can conclude that $\mathfrak{stupid}=\mathfrak{ui}=\max\{\mathfrak u,\mathfrak i\}$ and this seems to be different from any known (single) cardinal characteristics of the continuum. $\endgroup$
    – Taras Banakh
    Commented Nov 9, 2018 at 17:53
  • 6
    $\begingroup$ Just to smoothen the discussion, the cardinal $\mathfrak{ridiculous}$ has already disappeared, being equal to $\mathfrak p$, exactly I was planned by Boaz (I hope). $\endgroup$
    – Taras Banakh
    Commented Nov 9, 2018 at 18:51

1 Answer 1

Reset to default
13
$\begingroup$

The cardinal $\mathfrak{ridiculous}$ is equal to $\mathfrak p$ (which is equal to the smallest character of a free filter without infinite pseudointesection on $\omega$). It suffices to prove that a free filter $\mathcal F$ on $\omega$ has infinite pseudointersection if $\mathcal F$ has a base $\mathcal B$ of cardinality $|\mathcal B|<\mathfrak{ridiculous}$.

The latter inequality implies that there exists a sequence $(\{x_n,y_n\})_{n\in\omega}$ of pairwise disjoint doubletons such that each basic set $B\in\mathcal B$ (and consequently, each set in the filter $\mathcal F$) intersects all but finitely many doubletons $\{x_n,y_n\}$.

If $\{x_n\}_{n\in\omega}$ is not a pseudointersection of $\mathcal F$, then there exists a set $E\in\mathcal F$ such that the set $\Omega=\{n\in\omega:x_n\notin E\}$ is infinite. We claim that the set $\{y_n\}_{n\in\Omega}$ is a pseudointersection of $\mathcal F$. Indeed, for any $F\in\mathcal F$ the set $F\cap E$ intersects all but finitely many doubletons $\{x_n,y_n\}$, $n\in\Omega$, and contains no points $x_n\in\Omega$. Then $F\cap E$ contains all but finitely many point $y_n$, $n\in\Omega$, and so does the larger set $E$, which means that $\{y_n\}_{n\in\Omega}$ is an infinite pseudointersection of the filter $\mathcal F$.

So, in both cases, $\mathcal F$ has infinite pseudointersection: $\{x_n\}_{n\in\omega}$ or $\{y_n\}_{n\in\Omega}$ for some infinite set $\Omega\subset\omega$.

Remark. This method easily generalizes to prove that $\mathfrak p$ is equal to the smallest cardinality $\mathcal F$ of a centered family $\mathcal F$ of subsets of $\omega$ such that for any $n$-to-1 map $\varphi:\omega\to\omega$ the image $\varphi(\mathcal F)$ has no infinite pseudointersection.

On the other hand, the smallest cardinality $\mathcal F$ of a centered family $\mathcal F$ of subsets of $\omega$ such that for any finite-to-one map $\varphi:\omega\to\omega$ the image $\varphi(\mathcal F)$ has no infinite pseudointersection is equal to $\mathfrak b$, see Theorem 9.10 in the survey "Combinatorial Cardinal Characteristics of the Continuum" by Andreas Blass.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .