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I have worked the following problem from a book on Abstract Analysis, and I am not sure if my approach is correct. I have listed the entire exercise below followed by my solution. I would greatly appreciate some guidance.

The Exercise:

Let $\{A_n:\space n=1,2,...\}$ be an arbitrary sequence of sets. Define $A_0=\displaystyle{\bigcap_{n=1}^\infty A_n}$ and $A_\infty=\displaystyle{\bigcup_{n=1}^\infty A_n}$.

a) Construct a monotone non-increasing sequence of sets $\{B_n\}$ such that $\{B_n\}\downarrow A_0$.

b) Construct a monotone non-decreasing sequence of sets $\{C_n\}$ such that $\{C_n\}\uparrow A_\infty$.

c) Given $\{C_n\}\uparrow A_\infty$, construct a pairwise disjoint sequence $\{D_n\}$ such that $\displaystyle{\sum_{n=1}^\infty D_n}=A_\infty$ (when a sequence of sets are disjoint the author writes $\displaystyle{\sum_{n=1}^\infty D_n}$ in place of $\displaystyle{\bigcup_{n=1}^\infty D_n}$)

My Solution

a) If $A_0=\displaystyle{\bigcap_{n=1}^\infty A_n}$ and $\{B_n\}$ is monotone non-increasing then we want that $A_0\subset B_1,B_2,...$ Now, suppose that we make a new set $\mathcal{F}$ That contains every set in the sequence $\{A_n\}$ Then by the well ordering theorem, we can impose an order on $\mathcal{F}$ so that it becomes monotonically non-increasing we can call this set the sequence $\{B_n\}$, then for each $n\in \mathbb N $ we must have that $A_0 \subset B_n$, and therefore $\{B_n\}\downarrow A_0$. (The book has not mentioned anything about the well ordering theorem up to this point, so I don't know if this is acceptable).

b) $A_\infty=\displaystyle{\bigcup_{n=1}^\infty A_n}$ means that we have that $A_1,A_2,...\subset A_\infty$. Again we consider the set $\mathcal{F}$ and impose an order on it so that is monotonically non-decreasing we then call this $\{C_n\}$, since for all $n\in \mathbb{N}$ we have that $C_n\subset A_\infty$ we have that $\{C_n\}\uparrow A_\infty$.

c) Now considering the sequence $\{C_n\}$ if we let $D_n = \bigcup\{A_n\} \setminus \displaystyle{\bigcup_{i<n}_{i>n} C_i}$ Then the sequence $\{D_n\}$ should be pairwise disjoint and $\sum_{n=1}^\infty D_n=A_\infty$.

Are these answers acceptable? If so how might I make them better? Or, how might I answer these questions without invoking the well-ordered theorem? Thanks a lot.

This Question is related to The convergence of a sequence of sets

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In (a) it’s not true that you can necessarily reorder the family $\mathscr{F}=\{A_n:n\in\Bbb Z^+\}$ so that it’s monotonically non-increasing. What if there are $m,n\in\Bbb Z^+$ such that $A_m=[0,2]\subseteq\Bbb R$ and $A_n=[1,3]$; then $A_m\nsubseteq A_n$ and $A_n\nsubseteq A_m$, for instance? No matter which of $A_m$ and $A_n$ comes first in your new ordering, it won’t contain the other one as a subset. The sets $A_n$ may simply not be nested.

The trick is to replace them with related sets that are nested. Specifically, for $n\in\Bbb Z^+$ let $$B_n=\bigcap_{k=1}^nA_k=A_1\cap A_2\cap\ldots\cap A_n\;.$$ Then for each $n\in\Bbb Z^+$ we have $B_{n+1}=B_n\cap A_{n+1}\subseteq B_n$, so the sequence $\langle B_n:n\in\Bbb Z^+\rangle$ is non-increasing, and it’s not hard to prove that $$\bigcap_{n\ge 1}B_n=\bigcap_{n\ge 1}A_n=A_0\;.$$

The same difficulty can arise in (b), so here again you can’t count on simply rearranging the sequence $\langle A_n:n\in\Bbb Z^+\rangle$ to make it non-decreasing. However, you can use a trick very similar to the one that I used in (a); can you see what it should be?

In (c) you don’t need to use the sets $A_n$ at all; in fact, they just get in the way. All you need to know is that

$$C_1\subseteq C_2\subseteq C_3\subseteq\ldots$$

and $\bigcup_{n\ge 1}C_n=A_\infty$. Just let $D_1=C_1$, and for $n>1$ let $D_n=C_n\setminus C_{n-1}$. If you make a Venn diagram it should be intuitively clear that the sets $D_n$ are pairwise disjoint (though some may be empty) and that $\sum_{n\ge 1}D_n=A_\infty$, and proving it isn’t hard. For instance, to show that $\sum_{n\ge 1}D_n\supseteq A_\infty$, let $x\in A_\infty$. Then there is some $k\in\Bbb Z^+$ such that $x\in C_k$. (Why?) Let $n$ be the smallest such $k$, and show that $x\in D_n$.

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  • $\begingroup$ Thanks a lot for this! I reworked (a),(b), and (c) I feel better about this topic now. It was kind of confusing because the author is using this exercise make us derive, in a way, the definition of a limit of a sequence of sets. In fact, even after working out and proving everything you showed above I am still unsure how I deduce that $\bigcap_{n=1}^{\infty} B_n=A_0$ is logically equivalent to the statement $\{B_n\}\downarrow A_0$ if it were not for your explanation in my previous post. $\endgroup$ – JimmyJackson Oct 9 '13 at 2:15
  • $\begingroup$ @Jimmy: You’re very welcome. $\endgroup$ – Brian M. Scott Oct 9 '13 at 3:30

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