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!
Best Answer
Your approach will work as noted in the comments. Here's another approach that might be a little simpler.
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.