Is $sat(I)$ always a regular cardinal?

Question

Let $I$ be an ideal and let $I^+$ denote its complement (the so-called $I$-positive sets). Now we say that $I$ is $\lambda$-saturated iff each antichain in $I^+$ has size less than $\lambda$. Further $sat(I)$ is the least cardinal $\kappa$ such that $I$ is $\kappa$-saturated.

It can be shown that if $sat(I)$ is infinite then it has to be uncountable. I think that a similar argument gives us that $sat(I)$ is a regular cardinal, provided that $I$ is a $\kappa$-complete ideal on $\kappa$.

My question now is: Do we need the $\kappa$-completeness of $I$ or is $sat(I)$ always a regular cardinal, no matter if $I$ is $\kappa$-complete or not.

Answer 1

You do not need the $\kappa$-completeness of $I$. In fact the following holds for an arbitrary partial order and is exercise F4 of chapter VII in Kunen's book Set Theory: an Introduction to Independence Proofs.

Theorem. (Tarski) Let $\mathbb{P}$ be a poset, and let $\kappa$ be the least cardinal for which $\mathbb{P}$ has no antichain of size $\kappa.$ Then $\kappa$ is a regular cardinal.

Here we consider elements $p,q$ of $\mathbb{P}$ incompatible if there is no $r\leq p,q$. So $sat(I)=\kappa$ for the partial order on $I^+$ ordered by $\subseteq$.

Answer 2

If $I$ is an ideal on a set $X$, the saturation of $I$ is the saturation of the Boolean algebra $\mathbb{B} = \mathcal{P}(X)/I$. Let $\mathbb{B}'$ be the completion of $\mathbb{B}$. The completion clearly has all the antichains of the original algebra, so its saturation can only be higher. But suppose $\mathbb{B}'$ has an antichain $W'$ bigger than any that $\mathbb{B}$ has. Since $\mathbb{B}$ is dense in $\mathbb{B}'$, we can easily obtain from $W'$ an antichain $W \subset \mathbb{B}$ with $|W'| = |W|$, contradiction. ($W$ is obtained by taking, for each element of $W'$, a non-zero element of $\mathbb{B}$ below that element - such elements exist by the density of $\mathbb{B}$ in $\mathbb{B}'$).