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:

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

Next:

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:

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