We consider the Boolean ring $\mathcal{B}X$ of subsets of $X$, with the operations of symmetric difference as addition and intersection as multiplication.
One direction of part iii of the exercise then reads as follows:
A nonempty subset $I\subseteq\mathcal{B}X$ is an ideal if $A\in I$ implies that every subset of $A$ also lies in $I$.
I cannot prove that such an $I$ must be closed under addition, and indeed I believe the following is a counterexample:
$X=\{x,y,z\}$, $I=\mathcal{P}X\setminus\{X,\{x,y\}\}$.
Is this a valid counterexample? Is the exercise incorrect?
Thank you.
Your example is correct. Clearly $I$ has the property stated above, but $I$ is not an ideal since $\{ x\} \triangle \{ y\} = \{ x,y \}$ which is not in $I$.
I think that the only thing we can say is that every ideal must satisfy the property.