Template:MT book:full proofs:Proposition1.1.1

$(\Rightarrow )$

Property 1, it's direct.

For property 2 , note that $\mathrm{\forall }A\in 𝒜$:

$A^c=X\setminus A\in 𝒜$

Finally for property 3 , since the property 2 holds, $\mathrm{\forall }B\in 𝒜$ :

$B^c\in 𝒜\Rightarrow A\setminus B^c=A\cap B\in 𝒜$

$(\Leftarrow )$

Trivially $X\in 𝒜$

and using properties 2 and 3 it's clear that $\mathrm{\forall }A,B\in 𝒜\Rightarrow A\setminus B\in 𝒜$