I just realised two things.
$1)$ Ideals are closed under arbitrary intersection.
$2)$ Filters are closed under arbitrary union.
$X$ is a set.$I\subset \mathcal P(X)$ is called an Ideal if the following are satisfied.
$i)\Phi\in I$
$ii)A,B\in I \implies A\cup B\in I$
$iii)B\subseteq A, A\in I \implies B\in I$
Now take an arbitrary collection of members of $I$, say, $\mathcal A=\{A_i:i\in \Lambda , A_i\in I \forall i\}$ and let $C=\bigcap_{i\in \Lambda}A_i$ then $C\subset A_i\forall i\in \Lambda$ and hence $C\in I.$
$F\subset \mathcal P(X)$ is called a filter if the followings are satisfied
$i)\Phi \notin F$
$ii)A,B\in F \implies A\bigcap B\in F.$
$iii)A\in F, A\subseteq B\implies B\in F$
Now in the same manner as above take an arbitrary collection of members of $F$,say, $\mathcal B=\{B_i:i\in \Lambda,B_i\in F\forall i\in\Lambda \}$ and $D=\bigcup_{i\in \Lambda}B_i$ then $B_i\subseteq D\forall i\in \Lambda$ hence $D\in F.$
Thus my claims are proved.
And if $F=F(I)$,i.e. the dual filter of $I$ then the result about $F$ follows directly from the result about $I.$
$1$ and $2$ easily follow from the definitions but apparently the converse is not true (at least I don't see it being.)
Thoughts please. Thank you.