In the book I'm reading, the author defines a filter like this:
Given a non-empty set $X$ a filter over $X$ is a set $f \subseteq P(X)$ such that:
(i) $f \not= \emptyset$
(ii) If $S_{1}, S_{2} \in f$ then $S_{1} \cap S_{2} \in f$
(iii) If $S_{1} \subseteq S_{2} \subseteq X$ and $S_{1} \in f$ then $S_{2} \in f$
He then defines an ideal like this:
Given a non-empty set $X$ , an ideal over $X$ is a set $f \subseteq P(X)$ such that:
(i) $f \not= \emptyset$
(ii) If $S_{1}, S_{2} \in f$ then $S_{1} \cup S_{2} \in f$
(iii) If $S_{1} \subseteq S_{2} \subseteq X$ and $S_{2} \in f$ then $S_{1} \in f$
Then he asks the reader to prove that $f$ is a filter if and only if $P(X) \backslash f$ is an ideal. Is this correct? I've been trying to prove it but I think I got a counterexample instead.
That's false. What's true is that if $f$ is a filter over $X$, then $$\{X\setminus S: S\in f\}$$ is an ideal over $X$.