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) The intersection of two sets in $$\mathfrak A \ $$ is in $$\mathfrak A \ $$.
 * 2) The difference of two sets in $$\mathfrak A \ $$ is in $$\mathfrak A \ $$.
 * 3) $$X \in \mathfrak A$$.
 * 4) 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, $$\varnothing \in \mathfrak A$$.