# A ridiculous combinatorial cardinal characteristic of the continuum?

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.

• While I agree that we sort of exhausted the single lettered one, I don't think we should support ridiculous naming conventions like this. – Asaf Karagila Nov 8 '18 at 18:10
• What does centered mean in this context? – Douglas Ulrich Nov 8 '18 at 21:11
• @AsafKaragila :) Yes, you are right! Prove that the answer is positive, and then the ridiculous name will not be needed anymore! – Boaz Tsaban Nov 9 '18 at 7:08
• 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. – Taras Banakh Nov 9 '18 at 17:53
• 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). – Taras Banakh Nov 9 '18 at 18:51

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.