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:
- $\O \in \mathfrak A$
$\blacksquare$