Cofinality of a $\sigma$-ideal of $\mathbb{R}$

Question

The cofinality of a partially ordered set $\left( P,\leq \right)$, written $cof(P)$, is the smallest cardinality of a subset $T$ of $P$ that is [EDIT: cofinal] in $P$, i.e. for every element $p\in P$ there is a larger element $q\in T$ such that $p\leq q$.

A $\sigma$-ideal $I$ of $\mathbb{R}$ is a collection of subsets of $\mathbb{R}$ that is closed under taking subsets and countable unions, i.e. $U\in I, V\subset U \Rightarrow V\in I$ and $U_{n}\in I, n<\omega \Rightarrow \cup _{n<\omega }U_{n} \in I$.

We can now define the cofinality $cof(I)$ of a $\sigma$-ideal I as the cofinality of the partial order $\left( I,\subseteq \right)$.

We can prove some neat results such that if the ideal I is nonprincipal (i.e. it is not the set of all subsets of some subset $U\subset \mathbb{R}$), then its cofinality is at least uncountable.

My question is: are there any $\sigma$-ideals of $\mathbb{R}$ whose cofinality is greater than continuum, the cardinality of $\mathbb{R}$?

EDIT: thanks for the terminology and markup tips :)

Answer 1

$\textbf{A counterexample when $2^{\aleph_{0}}$ is regular}$.

This holds if $2^{\aleph_{0}}$ is a regular cardinal. In fact, it holds for any regular cardinal. If $\kappa$ is a regular cardinal, then the ideal of non-stationary sets in $\kappa$ cannot be generated by $\kappa$ many elements. To prove this fact, suppose to the contrary that $(C_{\alpha})_{\alpha<\kappa}$ generates the filter generated by all club sets. Then let $D=\Delta_{\alpha<\kappa}C_{\alpha}$ be the diagonal intersection. Then $D$ is a club set such that for each club set $C$, we have $D\subseteq C\cup A$ for some bounded $A$. However, if $E$ is the collection of limit points of $D$, then $D\not\subseteq E\cup A$ for each bounded $A\subseteq\kappa$. This is a contradiction.

$\textbf{A counterexample that works regardless of the regularity of $2^{\aleph_{0}}$}$.

One can modify the above example to get such an ideal even when $2^{\aleph_{0}}$ is not regular. The idea is to take the filter of club sets on $P_{\kappa}(X)$ instead of $\kappa$ and generalize the above argument. Suppose that $\kappa$ is an uncountable cardinal such that $\mathfrak{c}^{<\kappa}=\mathfrak{c}$ (i.e. $\mathfrak{c}^{\lambda}=\mathfrak{c}$ for each $\lambda<\kappa$. For example, we could have $\kappa=\aleph_{1}$. Let $X$ be any set of cardinality continuum. Then define $P_{\kappa}(X)=\{R\subseteq X:|R|<\kappa\}$. Then $|P_{\kappa}(X)|=\mathfrak{c}$. A subset $\mathcal{P}\subseteq P_{\kappa}(X)$ is said to be an unbounded set if for each $P\in P_{\kappa}(X)$ there is some $Q\in X$ with $P\subseteq Q$. We say that $\mathcal{P}\subseteq P_{\kappa}(X)$ is closed if whenever $\lambda<\kappa$ and $P_{\alpha}\in\mathcal{P}$ for $\alpha<\lambda$, then $\bigcup_{\alpha<\lambda}P_{\alpha}\in\mathcal{P}$. As usual, a set is a club set if it is closed and unbounded. It is easy to show that the intersection of less than $\kappa$ club sets in $P_{\kappa}(X)$ is a club set. If $C_{x}\subseteq P_{\kappa}(X)$ for each $x\in X$, then define the diagonal intersection by letting $R\in\Delta_{x\in X}C_{x}$ iff $R\in P_{\kappa}(X)$ and $R\in C_{x}$ for each $x\in R$. It is easy to show that the diagonal intersection of club sets in $P_{\kappa}(X)$ is a club set in $P_{\kappa}(X)$. The filter generated by the club sets in $P_{\kappa}(X)$ is $\sigma$-complete and even $\kappa$-complete, but I claim that this filter is not generated by continuumly many elements.

Suppose for the sake of contradiction, that the filter generated by the club sets in $P_{\kappa}(X)$ is generated by continuumly many elements. Then let $(C_{x})_{x\in X}$ be a system of club sets that generates the filter of club sets. Then let $D=\Delta_{x\in X}C_{x}$. Then $D$ is a club set in $P_{\kappa}(X)$. Define $\uparrow x=\{R\in P_{\kappa}(X)|x\in R\}$. If $C$ is a club set, then $C_{x}\subseteq C$ for some $x\in X$. Therefore if $R\in D,x\in R$, then $R\in C_{x}\subseteq C$. In other words, for each club set $C$ there is an $x\in X$ where $D\cap\uparrow x\subseteq C$. Therefore $\{D\cap\uparrow x|x\in X\}$ generates the club filter.

Now assume that $A_{x}\in D\cap\uparrow x$ for each $x\in X$. Let $B_{x}\in P_{\kappa}(X)$ be a set such that $A_{x}\subseteq B_{x}$ but $A_{x}\neq B_{x}$. Let $E\subseteq P_{\kappa}(X)$ be the collection of all subsets $L\in P_{\kappa}(X)$ such that if $x\in L$, then $B_{x}\subseteq L$. Then $E$ is a club set. However, $A_{x}\in D\cap\uparrow x$, but $A_{x}\not\in E$. Therefore $D\cap\uparrow x\not\subseteq E$ for each $x\in X$. This is a contradiction. We conclude that the filter generated by the club sets in $P_{\kappa}(X)$ cannot be generated with continuumly many elements.

Answer 2

If the continuum hypothesis holds, then there is such an ideal. Indeed, we need only to assume that $2^\omega\lt 2^{\omega_1}$, a weakening of CH.

Consider the tree $T=2^{\lt\omega_1}$ of all binary sequences of some countable ordinal length. This tree has continuum many nodes, and so let us associate with every node of $T$ a distinct real number. If $s\in 2^{\omega_1}$ is an $\omega_1$-branch through $T$, let $A_s$ be the set of reals appearing on the branch $s$. Let $I$ be the $\sigma$-ideal on $\mathbb{R}$ generated by these sets $\{A_s\mid s\in 2^{\omega_1}\}$. Thus, a set is in $I$ if and only if it is contained in $\bigcup_n A_{s_n}$, for some countable collection of $\omega_1$-branches $s_n\in 2^{\omega_1}$ through $T$.

I claim that no continuum many sets can be cofinal in $I$. To see this, suppose toward contradiction that $J\subset I$ and $J$ has size continuum. Each $B\in J$ is contained in some countable union $\bigcup_n A_{s_n^B}$. In particular, there are only $2^\omega$ many branches $s_n^B$ mentioned for $B\in J$. But since there are $2^{\omega_1}$ many branches through $T$, which is strictly larger than the continuum by our assumption, there is some $\omega_1$-branch $s$ through $T$ that is not $s_n^B$ for any $B\in J$. But it now follows that $A_s$ is not contained in any element of $J$, and so $J$ is not cofinal.

Answer 3

A set $X \subseteq \mathbb R$ is of ``strong measure zero'' if the following holds:

• For every sequence $(\varepsilon_n:n\in \mathbb N)$ of positive numbers there is a sequence $(I_n:n\in \mathbb N)$ of intervals such that $I_n$ has length $\varepsilon_n$, and $X$ is covered by the union of these intervals.

The smz (strong measure zero) sets are a proper subideal of the ideal of Lebesgue null sets, as the classical Cantor set (and indeed any perfect set) is not smz.

It is consistent that the cofinality of the $\sigma$-ideal smz is greater than continuum (e.g. under CH). It is also consistent that this cofinality is equal to the continuum, or less than the continuum.

(Teruyuki Yorioka, The cofinality of the strong measure zero ideal. J. Symbolic Logic 67 (2002), no. 4, 1373–1384. MR1955243)