Properties of Algebras of Sets

Theorem
Let $$X$$ be a [[Definition:Set|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 $$A, B \in \mathfrak{A}$$.

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


 * 1) $$A \cup B \in \mathfrak{A} \ $$;
 * 2) $$\mathcal{C} \left({A}\right) \in \mathfrak{A} \ $$.

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