# Let $A_1,A_2,…$ be a sequence of sets. Define $A'_1=A_1,A'_n=A_n\backslash\cup^{n=1}_{k=1}A_k,n=2,3…$ prove that $A_1',A_2'…$is disjoint

Sorry I do not have enough space to state the whole question. It is stated below:

Let $$A_1,A_2,...$$ be a sequence of sets. Define $$A'_1=A_1,A'_n=A_n\backslash\cup^{n-1}_{k=1}A_k,n=2,3....$$ prove that $$A_1',A_2'...$$is disjoint and $$\cup^\infty_{n=1}A_n=\cup^\infty_{n=1}A_n'$$

For the first part of the question, my approach is to first show the case is true for $$A'_1,A'_2$$ by saying that:

let any $$x\in A'_1, x\notin A_2',A_3',...$$ by definition, therefore $$A_1'\cap A'_2= \emptyset, A_1'\cap A'_3= \emptyset,...$$

let any $$x\in A'_2, x\notin A_3',A_4',...$$ by definition, therefore $$A_2'\cap A'_3= \emptyset, A_2'\cap A'_4= \emptyset,...$$

then extend it to $$x\in A'_n, x\notin A_{n+1}'=A_{n+1}\backslash\cup^{n}_{k=1}A_k$$ by definition, nor will it belongs to any of $$A_k'$$ where $$n=1,2,...,k-1$$. Therefore all sets should be pairwise disjoint. However, I am not sure if I am doing the question correctly and fear some important part is missing... (I feel my proof is not very rigorous).

For the second part of the question, I found this statement rather too abvious and do not know how to approach this quesiton ... Can someone pls: 1) take a look at my proof for the first part and give me some advise; 2) explain to me how should I approach the second part of the problem?

Thank you so much!

• I think the first part is okay. For the second part, we have $A_n'\subset A_n$ so one inclusion is obvious. For the other way, show that if $x\in A_n$ then $x\in A_k'$ for some $k$. Or alternatively, argue that $\cup^N_{n=1}A_n=\cup^N_{n=1}A_n'$ for every $N$ – saulspatz Sep 1 '20 at 2:25
• You can make your proof of the first part more rigorous by framing it as a proof by induction on $n$. A common way to prove set equality is to prove that each side of the equation is contained in the other, and that works here. – Robert Shore Sep 1 '20 at 2:32
• @RobertShore Hi, I was attempting to do the proof by induction but I only know how to do induction in the setting where we first suppose an equation hold for n=1, then suppose n=k holds and show n=k+1 also hold. I don't know how to do induction in a question of such... if possible could you please give me some hints? – JoZ Sep 1 '20 at 23:33
• Actually, I think it's easier to prove that for any positive $k, A_n \cap A_{n+k}' = \emptyset$. Then prove $A_n' \subseteq A_n$. Combined, these prove that if $m \neq n$, then $A_m' \cap A_n' = \emptyset$. – Robert Shore Sep 2 '20 at 1:39

Note that $$\forall n \in \Bbb N^+~(A_n' \subseteq A_n).$$ Also, if $$x \in \cup A_n$$, then there is some least $$n$$ for which $$x \in A_n$$, so for that $$n, x \in A_n'$$. Thus, $$\cup A_n \subseteq \cup A_n'$$ and equality follows.
Finally, choose $$x \in \cup A_n$$ and let $$m$$ be smallest such that $$x \in A_m$$. Then by hypothesis, $$x \notin \cup_{n=1}^{m-1} A_n$$ so $$x \in A_m'$$. $$A_n' \subseteq A_n \Rightarrow x \notin A_n'$$ for $$n \lt m$$. If $$n \gt m$$, then $$x \in A_m \Rightarrow x \notin A_n'.$$ Thus, for each $$x \in \cup A_n'$$, there is a unique $$m$$ such that $$x \in A_m$$, so the $$A_m$$ are pairwise disjoint.