A question about $cov(I)$ Given a set $X$ and an ideal $I \subseteq \mathcal P(X)$, 

$cov(I) := \min \{ |\mathcal A| | \mathcal A \subseteq I (\bigcup \mathcal A = X)\}$.

My question is: isn't it possible that for each $\mathcal A \subseteq I$, $\bigcup \mathcal A \neq X$? In this case, what will $cov(I)$ be?
Thank you!

Note that given any ideal $I$ on $X$, then $I$ is an ideal on any $X'\supseteq X$. But if $X'\neq X$ then $\bigcup I\neq X'$ since $\bigcup I\subseteq X$.
But at the same time, if $I$ is an ideal on $X$, it is also an ideal on $\bigcup I$. So we may assume without loss of generality that $X=\bigcup I$.
Formally, you're right. If $\bigcup I\subsetneqq X$ then the definition is problematic.
