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 :)



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


