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Prove that intersection of principal filters on $I$ is still a principal filter.
Let S be a nonempty set and C be a nonempty subset. Then {C} is a filter base. The filter it generates (i.e., the collection of all subsets containing C) is called the principal filter generated by C. Sorry for posting in this way, I understand it's not polite, but could you give any tip-off.
Some preliminary definitions:
“D is a filter” = df.
i.D $\subseteq$S(I) for some nonempty set I, where S(I) is the powerset of I, &
ii. I$\in$D;
iii. Given X, Y$\in$D, then X$\cap$Y $\in$D;
iv. Given X $\in$ D & X$\subseteq$Z $\subseteq$I then Z$\subseteq$ D.
“D is principal” =df. D is a filter and D = {X$\subseteq$I:Y $\subseteq$ X}.
Left to Right of “iff”: Suppose that D is a principal filter. Then D = {X$\subseteq$I: Y$\subseteq$ X}. So take $\cap$D. Clearly $\cap$D = Y, and Y $\in$ D (Y$\subseteq$ Y, and Y is thus one of the subsets of I having Y as subset). Thus $\cap$D$\in$D. <END OF LEFT TO RIGHT>.
Right to Left of “iff”: Suppose that $\cap$D$\in$D, and that D is a filter. We show D is principal. Thus by assumption D $\subseteq$ S(I), and so $\cap$ D$\in$S(I) hence $\cap$ D$ $ $\subset$ I. Thus $\exists$Y: D$\subseteq$Y & Y$\subseteq$I. Now call {Y: D $\subseteq$ Y & Y $\subseteq$ I} = Z. Clearly $\cap$D $\in$ Z. But $\cap$D $\in$ D as well (by initial assumption); so by extensionality D = Z; and so we have the appropriate D, so that D must be principal. <END OF RIGHT TO LEFT>.
<END OF PROOF>.
Take any two non-empty subsets $A,B$ of $I$, and let $\mathcal E,\mathcal F$ the principal filters on $I$ that they generate. Then $\mathcal E\cap\mathcal F$ is the set of subsets $C$ of $I$ that contain both $A$ and $B$ (that is, having both $A$ and $B$ as subsets). If $C\subseteq I$ is such that $A,B\subseteq C,$ what can you say with regard to $A\cup B$ and $C$? If $A\cup B\subseteq C,$ what can you say with regard to $A,B$ and $C$?
