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 $\varnothing$ 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:

• $A \cup B \in \mathfrak A$
• $\complement_X \left({A}\right) \in \mathfrak A$.

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle A, B \in \mathfrak A\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \complement_X \left({A}\right) \cup \complement_X \left({B}\right) \in \mathfrak A\)	\(\displaystyle \)	Definition of algebra of sets
	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \complement_X \left({A \cap B}\right) \in \mathfrak A\)	\(\displaystyle \)	De Morgan's laws
	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle A \cap B \in \mathfrak A\)	\(\displaystyle \)	Definition of algebra of sets

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

Next:

	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle A, B \in \mathfrak A\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle A \cap \complement_X \left({B}\right) \in \mathfrak A\)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle A \setminus B \in \mathfrak A\)	\(\displaystyle \)	Set Difference Relative Complement

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

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

Then:

	\(\displaystyle \)	\(\displaystyle \complement_X \left({A}\right)\)	\(\in\)	\(\displaystyle \mathfrak A\)	\(\displaystyle \)	Definition of Algebra of Sets
	\(\displaystyle \implies\)	\(\displaystyle \complement_X \left({A}\right) \cup A\)	\(\in\)	\(\displaystyle \mathfrak A\)	\(\displaystyle \)	Definition of Algebra of Sets
	\(\displaystyle \implies\)	\(\displaystyle X\)	\(\in\)	\(\displaystyle \mathfrak A\)	\(\displaystyle \)	Union with Relative Complement

Also, $\complement_X \left({A}\right) \cap A \in \mathfrak A$ from above, and so by Intersection with Relative Complement, $\varnothing \in \mathfrak A$.

$\blacksquare$