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!

  • $\begingroup$ 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 $N$ $\endgroup$ – saulspatz Sep 1 '20 at 2:25
  • $\begingroup$ 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. $\endgroup$ – Robert Shore Sep 1 '20 at 2:32
  • $\begingroup$ @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? $\endgroup$ – JoZ Sep 1 '20 at 23:33
  • $\begingroup$ 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 $A_m' \cap A_n' = \emptyset$. $\endgroup$ – Robert Shore Sep 2 '20 at 1:39

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.


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