1
$\begingroup$

$S$ be a set. A subset $I\subset \mathcal P(S)$ is called an ideal of $S$ if $I$ satisfies the following :

$$1)\Phi\in I.\\ 2)A\subset B, B\in I \implies A\in I.\\3) A,B\in I\implies A\cup B\in I.$$

The Dual filter $\mathcal F(I)$ is defined as $$\mathcal F(I)=\{A\in\mathcal P(S):S\backslash A \in I\}.$$

Is it possible that $I\cap \mathcal F(I)\neq \Phi$ unless we allow $\mathcal F(I)$ to contain $\Phi$ or $I$ to contain $X\ ?$ In that case though $I=\mathcal F(I).$ Or we take $I=\mathcal P(S)$

My guess is not. Except for the above three cases mentioned, $I\cap \mathcal F(I)=\Phi$ must hold.Because if not then say $A\in I\cap \mathcal F(I)$ s.t $A\neq \Phi,S.$ Then $S\backslash A\in I\cap \mathcal F(I)$ which in turn implies that $\Phi \in \mathcal F(I)$ and $S\in I$ which happen to be two of the cases above.

Correct me if I'm wrong.

  • 0
    If a filter contains both a set and its complement then it contains the empty set. It follows from that a filter is closed under finite intersections. – 2017-02-04

1 Answers 1

0

Yes you are absolutely right.An usual definition of filter says $\Phi\notin F$ but in those special cases, the other two criteria ensure that $\Phi\in F.$ No problem at all.