I wonder if there are natural examples of sets whose cardinality is $\frak p$ and $\frak t$? Examples that could provide intuition for non- set-theory experts? I find the definitions in the Malliaris & Shelah paper that established $\frak p=t$ difficult to grok, so to speak. I am not asking to understand the proof, just to understand the cardinals themselves.

Malliaris, Maryanthe, and Saharon Shelah. "Cofinality spectrum theorems in model theory, set theory, and general topology." Journal of the American Mathematical Society 29.1 (2016): 237-297. (Journal link; Link to earlier arXiv version.)

Related MO question: Short proof of $\frak p=t$.

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    $\begingroup$ Now that they are equal? Apparently they were always equal. $\endgroup$
    – Asaf Karagila
    Sep 14 '17 at 5:10

Each element of p and each element of t is a countable family of infinite sets of natural numbers. Let's start by giving an example of such family.

A simple example of a countable family of infinite sets of natural numbers would be $$E = \bigotimes_{j\in \Bbb N} \{k \in \Bbb N : \exists m | k = jm \}$$ so that two elements of $E$ would be $\{ 2,4,6,8\cdots\}$ and $\{ 3,6,9,12\cdots\}$. We can say that a countable family of infinite sets of natural numbers is a subset of $[\Bbb N]^{\aleph_0}$. (We will see that this particular $E$ is not in p.)

Now let's define a relationship between sets $A$ and $B$ which I will call "is almost a subset of," written as $A \subseteq^* B$. In words, $A$ is almost a subset of $B$ iff only a finite set of elements of $A$ are not also in $B$. That is, $$A \subseteq^* B \Leftrightarrow |\{ x:x\in A \wedge x \not\in B\}| < \infty $$ For example if $A$ is the set of primes and $B$ is the set of odd natural numbers, then $\{ x:x\in A \wedge x \not\in B\} = \{2\}$ and $A \subseteq^* B$.

Note that it is quite possible for $A \subseteq^* B$ and $B \subseteq^* A$.

Now let's define p and give an example of one element of p.

p is the set of all countable families of infinite sets of natural numbers such that family $f \in \mathbf{p}$ if and only if two conditions are met:

  • Every non-empty finite subfamily of $f$ (every finite collection of sets, all of which are in $f$) has an infinite intersection.

  • But there is no infinite set of integers $A$ such that $A$ is almost a subset of each element of $f$. That is, $\left( \forall A \subseteq \Bbb N: B \in f \implies A \subseteq^* B \right) \implies |A| = \infty$.

You can see that these two conditions work against one another in some sense. A simple example of a family of sets $f_1$ meeting the first condition is $\left\{ \{ n+0 :n \in \Bbb N\},\{ n+1 :n \in \Bbb N\}, \cdots \right\}$ since the intersection of any finite number of elements of $f_1$ consists of all numbers $\geq$ the highest starting point of any of that finite set of elements.

But that family $f_1$ fails the second condition, because if $A$ is the (infinite) set of even integers, then $A$ is almost a subset of each member of $f_1$.

An example of an element of p is the family of sets (indexed by $k\in \Bbb N$) of the form $\{ m^k : k\in \Bbb N\}$. The first condition is met since if $\ell$ is the least common multiple of all the indices in the finite sub-family then the set $\{ m^\ell : m\in \Bbb N\}$ is in every set in the finite subfamily. The second condition is met since there are an infinite number of integers that are cubes but not squares and vice-versa, so any infinite set $A$ of integers is either not almost a subset of the squares, or not almost a subset of the cubes, unless all but a finite number of elements of $A$ are sixth powers. But then $A$ won't be almost a subset of the fifth powers, unless all but a finite number of elements of $A$ are $30$-th powers; and so forth.

Now let's define t and give an example of one element of t.

t is the set of all countable families of infinite sets of natural numbers that can be well-ordered by the $\subseteq^*$ relation. (Remember, well-ordering will mean that if family $T$ is in t then every (non-equivalent under $A \subseteq^* B$) pair of elements of $T$ can be related by $\subseteq^*$ in one direction or the other, and every subset of $T$ has a least element in that ordering.)

A simple example of an element of t is the family $f_1$ described above. This "tower" happens not to be in p.

The p = t question, then, is whether p can be mapped onto t, or t can be mapped onto p, or they are in $1:1$ correspondence.

In the paper by Malliaris and $(\mbox{Shelah})^2$ there is a minor wording error in definition 1.1 whch might be causing some confusion: The first bullet should read

$D$ has an infinite pseudo-intersection if there is an infinite $A\subseteq \Bbb N$ such that $\forall B\in D, A \subseteq^* B$.

  • $\begingroup$ Thanks. May I ask, what does "indexed by $k \in BbbN$" mean? $\endgroup$ Sep 13 '17 at 22:03
  • $\begingroup$ @JosephO'Rourke I guess it is meant to $\mathbb{N}$, and Mark accidentally used the command he is used to using in a personal setup. $\endgroup$ Sep 14 '17 at 2:38
  • $\begingroup$ @DavidRoberts I think he just forgot a backslash... \Bbb works with MathJax it seems and is equivalent to \mathbb. Fixed now. $\endgroup$ Sep 14 '17 at 2:49
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    $\begingroup$ The paper is by M. Malliaris and S. Shelah and S. Shelah; thus the $\mbox{(Shelah)}^2$. On a move serious note, I neglected an important condition to be a tower: It also needs to have no infinite pseudo-intersection. I think my example $f_1$ meets this condition but I'm not certain. $\endgroup$ Sep 14 '17 at 14:02
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    $\begingroup$ Not that it matters, but the AMS journal link says the authors are M.M. and S.Shelah, punkt: no second S.Shelah. $\endgroup$ Sep 14 '17 at 21:45

A couple of simple examples; they don't have cardinality $\mathfrak p$ or $\mathfrak t$ but depending on your knowledge they may or may not help.

They are families of sets of natural numbers that I'll call $\mathcal F$, $\mathcal L$, $\mathcal M$, and $\mathcal D$.

Let $\mathcal F=\{X\subseteq\mathbb N: X\text{ is co-finite}\}$ and $\mathcal L=\{[n,\infty)\cap\mathbb N: n\in\mathbb N\}$.

In terms of the definitions on page 2 of

M. Malliaris, S. Shelah, Cofinality spectrum theorems in model theory, set theory and general topology, arXiv:1208.5424 (or direct pdf link)

$\mathcal L$ is not a tower, because even though it is well-ordered by $\supseteq^*$, it has $\mathbb N$ itself as a pseudo-intersection (the elements of $\mathcal L$ are too big, so to speak).

Anyway, $\mathcal L$ is countable, so it's plausible that a tower would have to be uncountable, and indeed that's the cardinality $\mathfrak t$.

$\mathcal F$ is a bit bigger than $\mathcal L$, is not well-ordered by $\supseteq^*$, but still has the s.f.i.p. since the intersection of finitely many co-finite sets is still co-finite (since the union of finitely many finite sets is finite).

Like $\mathcal L$, the family $\mathcal F$ is countable and unfortunately has $\mathbb N$ as pseudo-intersection. So it's again plausible that the corresponding cardinal $\mathfrak p$ is uncountable.

So let's consider $\mathcal M$, the family of sets whose limiting (Banach) density is 1. It has the s.f.i.p. and no pseudo-intersection. Great! However, $\mathcal M$ has cardinality $2^{\aleph_0}$.

Now consider $\mathcal D=\{\{k: k\text{ is divisible by }2^m\}: m\in\mathbb N\}$. This is also countable, well-ordered by $\supseteq$, but again unfortunately has an infinite pseudo-intersection, namely $\{1,2,4,8,16,\dots\}$.

$\begin{eqnarray*} \text{Family }\quad & \text{Well-ordered?}\quad&\text{S.f.i.p.? }\quad & \text{pseudo-$\cap$? }&\text{Cardinality}\\ \mathcal F & \text{no}& \text{yes} &\text{yes}& \aleph_0\\ \mathcal L & \text{yes}& \text{yes} &\text{yes}& \aleph_0\\ \mathcal M & \text{no}& \text{yes} &\text{no}& 2^{\aleph_0}\\ \mathcal D & \text{yes}& \text{yes} &\text{yes}& \aleph_0\\ \end{eqnarray*}$


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