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Consider the sequence $\{A_n\}_{n=1}^\infty$, satisfying $A_1 \subseteq A_2 \subseteq \cdots$. Since this sequence is monotone nondecreasing, we know its limit exists: $$ \lim_{n \to \infty} A_n = \bigcup_{n=1}^\infty A_n $$ Suppose from $\{A_n\}_{n=1}^\infty$, I construct a new sequence of disjoint members $\{B_n\}_{n=1}^\infty$, defined \begin{align*} B_1&=A_1 \\ B_n&=A_n-A_{n-1} \end{align*} How would I prove $\bigcup_{n=1}^\infty B_n=\bigcup_{n=1}^\infty A_n$?


Proof attempt posted as a potential answer below. Apologies for all the confusion and the large number of times I edited this post. I had originally tried to flesh out step 3 of this proof about Dynkin systems. However it became clear to me that the proof was flawed because the sequence $\{B_n\}_{n=1}^\infty$ they constructed is not pairwise disjoint (e.g., $B_3 \cap B_1 \ne \emptyset$). I had the right idea changing their definition of $\{B_n\}_{n=1}^\infty$ to $B_n=A_n-A_{n-1}$, however got confused along the way between $\lim_{n\to\infty} B_n$ and $\bigcup_{n=1}^\infty B_n$ (since in the case of $\{A_n\}_{n=1}^\infty$, the limit and union were equal).

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  • $\begingroup$ I think you want to prove $\lim\limits_{n\to\infty} \bigcup_{n=1}^\infty B_n=\lim\limits_{n\to\infty} A_n$ $\endgroup$ – nicomezi Sep 12 '16 at 5:20
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    $\begingroup$ The sequence $\lbrace B_n \rbrace$ has not the same limit as $\lbrace A_n \rbrace$. But $\lbrace \bigcup B_n \rbrace$, yes. $\endgroup$ – nicomezi Sep 12 '16 at 5:28
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    $\begingroup$ But it doesn't. Any element of $A_1$ will only be in $B_1$. Generalise a little and you'll find the $B_n$'s to be mutually disjoint... $\endgroup$ – Henrik Sep 12 '16 at 5:28
  • $\begingroup$ At this point and after several contradictory edits, it is entirely unclear what you are asking. The original claim (edited out, then restored back again) is obviously false, as pointed out in several comments and an answer. Maybe you should rethink what you really mean to ask, then repost the question afterwards. $\endgroup$ – dxiv Sep 12 '16 at 5:58
  • $\begingroup$ Is there a sequence of pairwise disjoint members we could construct which has the same limit as the monotonic sequence? No. The monotonic sequence (and its limit) would always include elements that the other sequence would have necessarily discarded at some point due to the disjoint restriction. However, if you mean the union $\cup{B_n}$ (as hinted already) then the answer is of course yes. $\endgroup$ – dxiv Sep 12 '16 at 6:11
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I don't think that this is true.

For example, suppose $A_n = \{1, 2, ..., n\} $. Then $A_n \subset A_{n+1} $ and $B_n=A_n-A_{n-1} =\{n\} $ so $\lim_{n \to \infty} B_n = \emptyset $.

You may mean $\lim_{n \to \infty} \cup_{k=1}^n B_k =\lim_{n \to \infty} A_n $..

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To prove $\bigcup_{n=1}^\infty B_n=\bigcup_{n=1}^\infty A_n$:

First, we show the left hand side is a subset of the right hand side. Suppose $x \in \bigcup_{n=1}^\infty B_n$. Then $x \in B_n$ for at least one $n \in \mathbb{N}$. This means $x \in A_n$ for at least one $n \in \mathbb{N}$. Thus, by definition of arbitrary union, $x \in \bigcup_{n=1}^\infty A_n$.

Next, we show the right hand side is a subset of the left. Consider $y \in \bigcup_{n=1}^\infty A_n$. This means $y \in A_n$ for at least one $n \in \mathbb{N}$, say $A_k$. Since $B_k=A_k-A_{k-1}$, this implies $y \in B_k$. Again by definition of arbitrary union, this impliease $y \in \bigcup_{n=1}^\infty B_n$.

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    $\begingroup$ By a similar argument, you could easily prove the stronger statement that $\cup_{n=1}^{N} B_n = A_N$ for all $N$ i.e. not only are the limits the same, but the actual sequences are identical. $\endgroup$ – dxiv Sep 12 '16 at 6:40
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    $\begingroup$ Great suggestion. You're right - that argument is even stronger and more elegant. Thanks again for all your help and patience! $\endgroup$ – Mathemanic Sep 12 '16 at 6:54
  • $\begingroup$ The second part of the argument is not quite correct: what if $y$ is in both $A_k$ and $A_{k-1}$? You need to let $k=\min\{n\in\Bbb Z^+:y\in A_n\}$. Then either $k=1$, in which case $y\in A_1=B_1$, or $k>1$, in which case $y\in A_k\setminus A_{k-1}$. Alternatively, you can indeed prove by induction on $n$ that $\bigcup_{k=1}^nB_k=\bigcup_{k=1}^nA_k=A_n$. $\endgroup$ – Brian M. Scott Sep 12 '16 at 8:11

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