Why is the dual of a filter an ideal?

Question

Jech's set theory, (3rd edition) says that if $F$ is a filter on $S$

Let $I = \left\{ {S - X: X \in F}\right\}$

then $I$ is an ideal of $S$ (dual to $F$).

However, let $X,Y \subset S$, $X \in I$ and $Y \subset X$.

I am having a hard time showing that $Y \in I$, to fulfill the requirements of being an ideal. Can someone please show how $Y \in I$ ?

Thanks!

Answer 1

$S \setminus X \subseteq S \setminus Y$ whenever $Y \subseteq X \subseteq S$. And for filters you know that supersets of filter elements are filter elements...

Answer 2

HINT: If $Y\subseteq X$, when can you say about the subset relation between $S\setminus X$ and $S\setminus Y$? Moreover, what do you know about $S\setminus X$?