Definition
Let $X$ be a set and $\mathcal{E} \subset \mathcal{P}(X)$. The family $\mathcal{E}$ is elemental if:
- $\emptyset \in \mathcal{E}$;
- $E \cap F \in \mathcal{E}$ for each $E , F \in \mathcal{E}$;
- For all $E \in \mathcal{E}$, there is a finite family ${\{E_i\}}_{i = 1}^n \subset \mathcal{E}$, with $E_i \cap E_j = \emptyset$ if $i \neq j$, such that $$ E^c = \bigcup_{i = 1}^n E_i\mbox{.} $$
I want to prove that $$ \mathcal{A} = \left\{\bigcup_{i = 1}^n E_i : {\{E_i\}}_{i = 1}^n \mbox{ is a disjoint family in } \mathcal{E} \right\} $$ is an algebra on $X$. I have already proved that $X \in \mathcal{A}$ and $E^c \in \mathcal{A}$ if $E \in \mathcal{A}$. Fixed $E , F \in \mathcal{A}$, I want to show now that $E \cup F \in \mathcal{A}$. We have that $$ E = \bigcup_{i = 1}^n E_i \qquad \mbox{ and } \qquad F = \bigcup_{j = 1}^m F_i\mbox{,} $$ being ${\{E_i\}}_{i = 1}^n$ and ${\{F_i\}}_{j = 1}^m$ two disjoint families in $\mathcal{E}$, so we can write $$ E \cup F = \bigcup_{k = 1}^{n + m} A_k\mbox{,} $$ where $$ A_k = \left\{ \begin{array}{lcl} E_k & \mbox{ if } & k \leq n\mbox{;} \\ F_{k - n} & \mbox{ if } & k > n\mbox{.} \end{array} \right. $$ If we fix $k , l \in \{1 , \ldots , n + m\}$, then $A_k \cap A_l = \emptyset$ if $k \leq n$ and $l \leq n$ or $k > n$ and $l > n$, so only I need to show that $A_k \cap A_l = \emptyset$ if $k \leq n$ and $l > n$ or $k > n$ and $l \leq n$. The two cases have the same difficulty, so we suppose for example that $k \leq n$ and $l > n$. In this case, $A_k = E_k$ and $A_l = F_{l - n}$. How can I prove that $E_k \cap F_{l - n} = \emptyset$? My attemps have no result. Thank you in advance.
Given that you have proved that $\mathcal A$ is closed under complement, it's easier to prove first that if $E, F\in\mathcal{A}$ then $E\cap F\in\mathcal{A}$ and then use the fact that $$E\cup F=(E^c\cap F^c)^c.$$
To prove that $E\cap F\in\mathcal{A}$, simply observe that $$E\cap F=\bigcup_\limits{i,j} E_i\cap F_j$$ is a partition by elements of $\mathcal E$ as required.