# Properties of Algebras of Sets

## Theorem

Let $X$ be a set.

Let $\mathfrak A$ be an algebra of sets on $X$.

Then the following hold:

$(1): \quad$ The intersection of two sets in $\mathfrak A$ is in $\mathfrak A$.
$(2): \quad$ The difference of two sets in $\mathfrak A$ is in $\mathfrak A$.
$(3): \quad$ $X \in \mathfrak A$.
$(4): \quad$ The empty set $\O$ is in $\mathfrak A$.

## Proof

Let:

$X$ be a set
$\mathfrak A$ be an algebra of sets on $X$
$A, B \in \mathfrak A$

By the definition of algebra of sets, we have that:

$A \cup B \in \mathfrak A$
$\relcomp X A \in \mathfrak A$

Thus:

 $\ds$  $\ds A, B \in \mathfrak A$ $\ds$ $\leadsto$ $\ds \relcomp X A \cup \relcomp X B \in \mathfrak A$ Definition of Algebra of Sets $\ds$ $\leadsto$ $\ds \relcomp X {A \cap B} \in \mathfrak A$ De Morgan's Laws: Complement of Intersection $\ds$ $\leadsto$ $\ds A \cap B \in \mathfrak A$ Definition of Algebra of Sets

and so we have that the intersection of two sets in $\mathfrak A$ is in $\mathfrak A$.

Next:

 $\ds$  $\ds A, B \in \mathfrak A$ $\ds$ $\leadsto$ $\ds A \cap \relcomp X B \in \mathfrak A$ from above $\ds$ $\leadsto$ $\ds A \setminus B \in \mathfrak A$ Set Difference as Intersection with Relative Complement

and so we have that the difference of two sets in $\mathfrak A$ is in $\mathfrak A$.

We have that $\mathfrak A \ne \O$ and so $\exists A \subseteq X: A \in \mathfrak A$.

Then:

 $\ds \relcomp X A$ $\in$ $\ds \mathfrak A$ Definition of Algebra of Sets $\ds \leadsto \ \$ $\ds \relcomp X A \cup A$ $\in$ $\ds \mathfrak A$ Definition of Algebra of Sets $\ds \leadsto \ \$ $\ds X$ $\in$ $\ds \mathfrak A$ Union with Relative Complement

Also, $\relcomp X A \cap A \in \mathfrak A$ from above.

$\O \in \mathfrak A$

$\blacksquare$