We are asked to prove
Let $X$ be a set and $\mathscr{A}$ be an algebra of subsets of $X$. Show that $\mathscr{A}$ is closed under finite union. That is, if $A_1,A_2,\ldots, A_n\in \mathscr{A}$ then $\bigcup_{i=1}^n{A_i}\in \mathscr{A}$.
I plan to prove this problem via strong induction and it is given below.
My Proof:
Basis Step:
Note that $A_1\in \mathscr{A}$ by assumption.
If in addition, $A_2\in \mathscr{A}$ then $(A_1\cup A_2)\in\mathscr{A}$ since $\mathscr{A}$ is an algebra.
Induction Hypothesis:
Suppose that for $i\leq k$ we have
$\bigcup_{i=1}^k{A_i}\in \mathscr{A}$.
We must show that:
$\bigcup_{i=1}^{k+1}{A_i}\in\mathscr{A}$ where $A_{k+1}\in \mathscr{A}$.
This clearly follows from the Induction Hypothesis and the facts $A_{k+1}\in \mathscr{A}$ and $\mathscr{A}$ is an algebra.
I am doubtful on my proof. Is my proof correct? Is strong induction necessary?
For i can just consider at the very first place the union
$\bigcup_{i=1}^n{A_i}$
and get a union two at a time.
Thank you very much for your help.
At first I think that the answer is NO. For I am just asked to prove "closure" under finite union. Before, I think that using strong induction in this case means that the statement will be valid for an infinite union at least for countable one (since the set of positive integers is infinite).
However after some rethinking it seems that my first thinking was wrong. Since the fact that the statement is true for all positive integers $n$, then the intersection of $n$ number of elements in the collection is still in the collection. Again this is TRUE for all positive integers $n$ which is finite. Got it now.