Properties of Algebras of Sets

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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.

So by Intersection with Relative Complement is Empty:

$\O \in \mathfrak A$

$\blacksquare$