View source for Template:MT book:C1:full proofs:Proposition1.1.1

<math>(\Rightarrow)</math>

Property 1, it's direct.

For property 2 , note that <math>\forall A \in \mathcal{A} </math>:

:<math> A^c = X \setminus A \in \mathcal{A} </math>

Finally for property 3 , since the property 2 holds, <math>\forall B \in \mathcal{A}</math> :

:<math> B^c \in \mathcal{A} \Rightarrow A \setminus B^c = A \cap B \in \mathcal{A}</math>

<math>(\Leftarrow)</math>

Trivially <math> X \in \mathcal{A} </math>

and using properties 2 and 3 it's clear that <math> \forall A,B \in \mathcal{A} \Rightarrow A\setminus B \in \mathcal{A} </math>