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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.

Jr Antalan
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  • Strong induction is never necessary - anything that can be done with strong induction can be done with regular induction by moving to a stronger inductive hypothesis. – Carl Mummert Oct 01 '18 at 13:25
  • @Carl Mummert. Thank you for your kind comment Sir. I will be happy to recieve an illustration on "anything that can be done with strong induction can be done with regular induction by moving to a stronger inductive hypothesis." Thanks a lot Prof. – Jr Antalan Oct 01 '18 at 13:48
  • There is an answer at https://math.stackexchange.com/a/15002/630 about it. – Carl Mummert Oct 01 '18 at 14:19
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    @CarlMummert. Let $A$ be the set of primes not congruent to $3$ modulo $4$. Let $B$ be the set of $p\in A$ such that $p=x^2+y^2$ for some $x,y \in \Bbb N.$ A standard elementary proof that $B=A $ has a step where, if $2<p,$ we have some $q\in A$ with $q<p$, and we deploy the strong-inductive hypothesis ${q\in A:q<p}\subset B,$ as we do not know whether there are other members of $A$ between $q$ and $ p$ unless $p=5$. Of course with a few extra paragraphs you could turn it into a proof by regular induction, but you wouldn't enjoy it. – DanielWainfleet Oct 01 '18 at 23:59
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    @Daniel Wainfleet: it does not take any extra paragraphs. Using standard induction, just make the inductive hypothesis stronger to mimic "strong induction". There is no reason, in standard induction, that the inductive hypothesis cannot contain quantifiers. In particular, we could make the induction hypothesis "for every $q < p$, if $q \in A$ then $q \in B$" (or similar) using standard induction. – Carl Mummert Oct 02 '18 at 00:06

1 Answers1

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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.

Jr Antalan
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