A $\lambda$-system is a set $X$ along with $S$, a collection of subsets of $X$ with the following properties :
(i) $\varnothing \in S$
(ii) If $A \in S$ then $A^{c} = X \setminus A \in S$ ($S$ is closed under complementation)
(iii) If $A_{1}, A_{2}, … \in S$ and are pairwise-disjoint, then $\bigcup_{i \in \mathbb{N}}^{} A_{i} \in S $ ($S$ is closed under countable disjoint unions)
How can one prove that S is also closed under finite disjoint unions ? (1)
The closest I've been able to come is to prove that if $A_{1}, A_{2}, … \in S$ and are pairwise-disjoint then the union of any finite number of the $A_{i}$ is also in $S$, that is to say, $\bigcup_{i \in \mathbb{F}}^{} A_{i} \in S$ for all finite $F \subset \mathbb{N}$
Proof :
Let $\alpha = \bigcup_{i \in \mathbb{N}}^{} A_{i}$ and $F$ be a finite subset of $\mathbb{N}$
Then $\bigcup_{i \in \mathbb{N} \setminus F}^{} A_{i} \in S$ by (iii) and
$\alpha^{c} \in S$ by (ii)
Let $P = \alpha^{c} \cup \bigcup_{i \in \mathbb{N} \setminus F}^{} A_{i}$
Then $P \in S$ by (iii) and $P^{c} = \bigcup_{i \in F} A_{i} \in S$ by (ii)
Since $F$ can be chosen to be any finite subset of $\mathbb{N}$ then $\bigcup_{i \in F} A_{i} \in S$ for all finite $F \subset \mathbb{N}$
However I'm not sure how to prove the general case referred to in (1).
Best Answer
Per @peek-a-boo's comment, for pairwise-disjoint $A_{1}, A_{2}, ..., A_{n} \in S$, set $A_{k} = \varnothing$ for all $k > n$. The result follows from (iii).