A note on Maximality of ideal-independent sets

TL;DR

This paper explores properties of maximal ideal-independent sets in boolean algebras and discusses their implications for continuum cardinal invariants p and s_mm(P(omega)/fin).

Contribution

It introduces a new property of maximal ideal-independent sets and connects it to continuum cardinal characteristics.

Findings

Derived a property of maximal ideal-independent sets.

Established corollaries relating to continuum cardinals p and s_mm.

Provided insights into the structure of boolean algebras.

Abstract

In this note we derive a property of maximal ideal-independent subsets of boolean algebras which has corollaries regarding the continuum cardinals p and s_mm(P(omega)/fin).

Equations

$$s\left(A\right)=\sup \left\{\left|X\right|:X\text{mbox}isideal-independent.\right\}$$

$$s_{mm}\left(A\right)\stackrel{\text{def}}{=}\min \left\{\left|X\right|:X\text{mbox}isinfiniteandmaximalideal-independent\right\}.$$

$${\aleph }_0<s_{mm}\left(\mathcal{P}\left(\omega \right)/\mathrm{fin}\right)<{\beth }_1.$$

$$x\cdot \left(-b+x_2+x_3+\dots +x_n\right)=x\cdot -b+x\cdot \left(x_2+x_3+\dots +x_n\right)=0+y<x,$$

$$\mathfrak{p}\left(A\right)\stackrel{\text{def}}{=}\min \left\{\left|Y\right|:\sum Y=1\text{mbox}and\sum Y^{\prime }\ne 1\text{mbox}foreveryfinite{Y}^{\prime }\subseteq Y\right\}.$$

$$\left\{\left|X\right|:X\text{mbox}ismaximalideal-independent\right\}\subseteq$$

$$\left\{\left|Y\right|:\sum Y=1\text{mbox}and\sum Y^{\prime }\ne 1\text{mbox}foreveryfinite{Y}^{\prime }\subseteq Y\right\}$$

Full text

A note on maximality of ideal-independent sets

Corey T. Bruns, University of Wisconsin-Whitewater, brunsc@uww.edu

1 Introduction and Definitions

We give a solution to a problem originally posed in a draft of J.D. Monk [4]. This result is now listed as further fact 4 after theorem 2.16. In doing so, we will consider some subsets of boolean algebras.

We will follow S. Koppelberg [2] for notation. In particular $+$, $\cdot$, and $-$ will used as the boolean operators, and [math] and $1$ as the least and greatest element. By extension, the least upper bound of a set $M$ will be denoted $\sum M$.

Definition 1.1**.**

A subset $X$ of a boolean algebra is ideal-independent if $0,1\notin X$ and $\forall x\in X,x\not\in\left<X\smallsetminus\left\{x\right\}\right>^{id}$; equivalently, for distinct $x,x_{1},\ldots,x_{n}\in X$, $x\not\leq x_{1}+x_{2}+\ldots+x_{n}$.

By Zorn’s lemma, there are maximal ideal-independent sets.

This is used in one of several equivalent definitions in Monk [3] of the spread $s$ of a boolean algebra $A$:

[TABLE]

Related to this cardinal function is the minimaximal version:

[TABLE]

Among the results of Monk [4] is that it is consistent with ZFC that

[TABLE]

We will further show how $s_{mm}$ compares to the pseudo-intersection number $\mathfrak{p}$.

2 Maximality

Our main result is a necessary condition for maximality of an ideal-independent set.

Theorem 2.1**.**

Let $A$ be a boolean algebra. If $X\subseteq A$ is maximal for ideal-independence, then $\sum X=1$.

Proof.

Let $X\subseteq A$ be ideal-independent and let $b\in A\smallsetminus\left\{1\right\}$ be such that $\forall x\in Xx\leq b$. We claim that $X^{\prime}\stackrel{{\scriptstyle\text{\tiny def}}}{{=}}X\cup\left\{-b\right\}$ is ideal-independent (thus proving the theorem).

We need to show that for distinct $x,x_{1},x_{2},\ldots,x_{n}\in X^{\prime}$, $x\not\leq x_{1}+x_{2}+\ldots+x_{n}$. There are three cases to consider.

$-b\not\in\left\{x,x_{1},x_{2},\ldots,x_{n}\right\}$.

Then $x\not\leq x_{1}+x_{2}+\ldots+x_{n}$ by the ideal-independence of $X$. 2. 2.

$-b=x$.

Assume otherwise, that is, $-b\leq x_{1}+x_{2}+\ldots+x_{n}$. Since $x_{1}+x_{2}+\ldots+x_{n}\leq b$, we have $-b\leq b$, thus $b=1$, contradiction. 3. 3.

$-b=x_{1}$.

Then as $x\in X$, $x\leq b$, that is, $x\cdot-b=0$. By the ideal-independence of $X$, $x\not\leq x_{2}+x_{3}+\ldots+x_{n}$, so that $y\stackrel{{\scriptstyle\text{\tiny def}}}{{=}}x\cdot\left(x_{2}+x_{3}+\ldots+x_{n}\right)<x$.

Then

[TABLE]

thus $x\not\leq-b+x_{2}+x_{3}+\ldots+x_{n}$.

∎

It is worth noting that a converse of Theorem 2.1 does not hold for infinite sets. That is, there is an ideal-independent set $X$ with $\sum X=1$ that is not maximal.

Proposition 2.2**.**

*Let $\mathbb{P}=\left\{p\in\omega:p\mbox{ is prime}\right\}$. In the boolean algebra

${\mathscr{P}(\omega\smallsetminus\left\{0,1\right\})}$ where $+$ is union, $\cdot$ is intersection, and $-x$ is $\left(\omega\smallsetminus\left\{0,1\right\}\right)\smallsetminus x$, set $X=\left\{p\mathbb{Z}^{+}:p\in\mathbb{P}\right\}$ (where $n\mathbb{Z}^{+}$ is the set of all nonzero multiples of n). Then all of the following are true:*

$\sum X=1$** 2. 2.

$X$* is an ideal-independent set.* 3. 3.

$X\cup\left\{\mathbb{P}\right\}$* is also ideal-independent.*

Proof.

$\bigcup X=\omega\smallsetminus\left\{0,1\right\}$ as every integer other than [math] and $1$ is a nonzero multiple of a prime, so $\sum X=1$. 2. 2.

If $p$ is prime, then $p$ is not the multiple of any other prime, so $p\mathbb{Z}^{+}\not\leq p_{1}\mathbb{Z}^{+}+p_{2}\mathbb{Z}^{+}+\ldots+p_{n}\mathbb{Z}^{+}$ if $p,p_{1},\ldots,p_{n}$ are all distinct primes.

Thus $X$ is ideal-independent. 3. 3.

We have two parts to show that $X\cup\left\{\mathbb{P}\right\}$ is ideal-independent; Let $Y$ be a finite subset of $X$ and $q,r$ distinct primes such that $q\mathbb{Z}^{+}\notin Y$ and $r\mathbb{Z}^{+}\notin Y$. We must show that $\mathbb{P}\not\leq\sum Y$ and that $q\mathbb{Z}^{+}\not\leq\mathbb{P}+\sum Y$.

$q\in\mathbb{P}$ and $q\notin\sum Y$, so the first is true.

$rq\in q\mathbb{Z}^{+}$, but $rq\notin\mathbb{P}$ as it is composite and $rq\notin\sum Y$ since neither $r\mathbb{Z}^{+}\in Y$ nor $q\mathbb{Z}^{+}\in Y$.

Thus $X\cup\left\{\mathbb{P}\right\}$ is ideal-independent and so $X$ is not maximal.

∎

Corollary 2.3**.**

*For $A$ infinite, $\mathfrak{p}\left(A\right)\leq s_{mm}\left(A\right)$; in particular

$\mathfrak{p}=\mathfrak{p}\left(\mathscr{P}\left(\omega\right)/{\mathrm{fin}}\right)\leq s_{mm}\left(\mathscr{P}\left(\omega\right)/{\mathrm{fin}}\right)$.*

We recall from Monk [3] that the pseudo-intersection number $\mathfrak{p}$ is defined as:

[TABLE]

Proof.

Let $X$ be maximal ideal-independent (and thus infinite). By Theorem 2.1, $\sum X=1$.

If $X^{\prime}\subseteq X$ is finite, $\sum X^{\prime}\neq 1$, otherwise, take $x\in X\smallsetminus X^{\prime}$, then $x\leq 1=\sum X^{\prime}$, contradicting the ideal-independence of $X$. Thus

[TABLE]

[TABLE]

and so $\mathfrak{p}\left(A\right)\leq s_{mm}\left(A\right).$ ∎

Now, $\mathfrak{p}$ is one of the smaller continuum cardinals, so that $s_{mm}$ is larger is not particularly surprising. However, the method of proof does turn out to be useful with other similar properties. Other set properties involving sums of finite subsets, such as independence and n-independence, are amenable to these methods. In [1], we introduce some variations on $\mathfrak{i}$; their placement in order of continuum cardinals can be similarly determined.