TITLE: Cofinality of a $\sigma$-ideal of $\mathbb{R}$
QUESTION [7 upvotes]: 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 :)

REPLY [3 votes]: 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)