If $\mathcal{E}$ is an elementary family of sets, then $\mathcal{A}$ whose elements are finite sums of disjoint sets of $\mathcal{E}$ is an algebra What is an elementary family of sets $\mathcal{E}$? It's such a family of sets of subsets of $X$, that

*

*$\varnothing \in \mathcal{E}$


*If $A, B \in \mathcal{E}$, then $A \cap B \in \mathcal{E}$


*If $A \in \mathcal{E}$, then $X \backslash A$ is a finite sum of dijoint elements of $\mathcal{E}$
My attempt
We have to prove these conditions:

*

*$\bigcup_{i=1}^n A_i \in \mathcal{A}$ for $\forall_i \: A_i \in \mathcal{A}$


*$X \backslash A \in \mathcal{A}$ for $A \in \mathcal{A}$


*$X \in \mathcal{A}$
That's how I'd do it:

*

*Because $A_i \in \mathcal{A}$ is a finite sum of disjoint elements of $\mathcal{E}$, then naturally, the sum of those $A_i$ will still be a finite sum of disjoint elements of $\mathcal{E}$, thus an element of the family $\mathcal{A}$


*Because it's true, that $A \in \mathcal{A} \subset \mathcal{E}$, then we know that $X \backslash A$ must be a finite sum of disjoint elements of $\mathcal{E}$, which again by the very nature of $\mathcal{A}$ makes it an element of the family $\mathcal{A}$


*$\mathcal{A} \ni X \backslash (X \backslash A) = X$, it's a consequence from the second point as we've shown that $X \backslash A \in \mathcal{A}$ if $A \in \mathcal{A}$.

My question is: Is this proof alright? Because I've seen a solution to that exercise which relied on some complicated calculations by creating another family of sets by which you could prove these points.
I can't imagine that such an exercise would require such a complicated solution, but I'm not sure if my attempt is not actually "too simple" and "too trivial" to be accepted.
 A: Your 1 needs some refining: show that a finite normal union of disjoint unions, is again a disjoint finite union. We'll need the intersection axiom on $\mathcal E$, I think. Make it explicit: look at a simple example and generalise. Introduce notation etc.
Your 2 is also too easy: yes, if $A \in \mathcal E$ then by definition $X\setminus A$ is a disjoint finite union of sets from $\mathcal{E}$ and $\mathcal{E} \subseteq \mathcal{A}$ trivially (a single element $A$ is the union of $\{A\}$ which is trivially a finite disjoint family).
But if $A \in \mathcal{A}$ we know $A = \bigcup_{i=1}^n E_i$ where all $E_i \in \mathcal{E}$.
In that case $X\setminus A = \bigcap_{i=1}^n (X\setminus E_i)$ by de Morgan and we can write $X\setminus E_i = \bigcup_{j=1}^{m_i} E^{(i)}_j$ where the right hand union is disjoint, $m_i \in \Bbb N$ and all $E^{(i)}_j \in \mathcal E$.
Now try to reason further from that to get $X\setminus A$ in $\mathcal A$ as well.
If 2 is done, we know that $\emptyset \in \mathcal E \subseteq \mathcal A$ and $X\setminus \emptyset = X$.
