[Math] Proving an Algebra is a semi-algebra. Finite disjoint union of $1$ element

Question

$\mathcal{f}$ is a semi-algebra if

• $\mathcal{F}$ contains $\emptyset$ and $\Omega$
• $\mathcal{F}$ is closed under finite intersections
• The complement of any $A \in \mathcal{F}$ is equal to the finite disjoint union of elements in $\mathcal{F}$

and $\mathcal{F}$ is an algebra if

• $\mathcal{F}$ contains $\emptyset$ and $\Omega$
• $\mathcal{F}$ is closed under complements
• $\mathcal{F}$ is closed under finite unions and intersections

My question is, for proving "The complement of any $A \in \mathcal{F}$ is equal to the finite disjoint union of elements in $\mathcal{F}$" is true when $\mathcal{F}$ is an algebra…

is this satisfied by the fact that $A^c \in \mathcal{F}$. That is, can I say that the complement of any element $A$ of $\mathcal{F}$ is the finite disjoint union of $A^c \in \mathcal{F}$ ($A^c$ is in $\mathcal{F}$ because $\mathcal{F}$ is closed under complements)

I am simply unsure of whether the "union" of one element makes sense?

Answer 1

In case you desire a complete answer:

Let $\mathcal F$ be an algebra.

1. Then it contains $\phi$ and $\Omega$ by assumption.
2. It is closed under finite intersections by assumption.
3. If $A\in\mathcal F$ then $A^c=A^c\cup \phi$, a finite disjoint union of two elements from $\mathcal F$ since $A^c\in\mathcal F$.

Hence $\mathcal F$ is a semi-algebra.