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According to my lecture notes, with the help of a little "trick" it is possible to write an infinte union of sets as the infinite union of disjoint sets as follows:

$$A = \bigcup \limits_{k=1}^{\infty} A_k = \bigcup \limits_{k=1}^{\infty} \tilde{A_k}$$

with the disjoint sets $\tilde{A_k}$ whereas $\tilde{A_1} := A_1$ and $\tilde{A_k} := A_k \setminus A_{k-1}$ for $k \geq 2$.

My question: Given we have three sets $A_1,A_2$ and $A_3$ and none of them is disjoint. When I define $\tilde{A_3}$ as $A_3 \setminus A_2$ I don't see how $A_1$ and $A_3$ are supposed to be disjoint..so on what am I missing out?

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  • $\begingroup$ It seems like it should be $\tilde{A}_k = A_k \setminus \tilde{A}_{k-1}$, with a tilde on the right hand side. $\endgroup$ – Dan Shved Mar 6 '13 at 18:37
  • $\begingroup$ Hm. Doesn't seem to be correct neither. $\endgroup$ – TestGuest Mar 6 '13 at 18:44
  • $\begingroup$ you are right, I haven't thought this through. What I should have said is in coffeemath's answer. $\endgroup$ – Dan Shved Mar 6 '13 at 18:47
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    $\begingroup$ It could also be that $A_i$ in your lecture notes are an increasing chain, i.e. $A_i \subseteq A_{i+1}$. If so, the definition makes perfect sense as it is. $\endgroup$ – Dan Shved Mar 6 '13 at 18:48
  • $\begingroup$ !! Ah, it is true, it is a chain. Sorry then, the property of the sets bein ordered as chain was half a page above! $\endgroup$ – TestGuest Mar 6 '13 at 18:49
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The definition of $\bar{A^k}$ should be $A_k \setminus (A_1 \cup ... \cup A_{k-1}).$

One way to think of $\bar{A^k}$ is that it consists of all points (if any) which first appear in set $A_k$, not yet having appeared in any of the previous sets $A_1,...,A_{k-1}.$ To me this makes it clear the $\bar{A^k}$ are pairwise disjoint, with the same union as the original collection of $A_k$ sets.

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  • $\begingroup$ Yeah, I do understand the intuitive idea..but still am not sure about the definition. So according to you the definition is entirely wrong? That would be a bit odd since it is used throughout the whole lecture notes. $\endgroup$ – TestGuest Mar 6 '13 at 18:46
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    $\begingroup$ The definition in the question would have been correct if we were dealing with an increasing union, i.e., if $A_k\subseteq A_{k+1}$ for all $k$. Perhaps the author had this assumption in mind and neglected to state it. $\endgroup$ – Andreas Blass Mar 6 '13 at 18:49
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    $\begingroup$ It was stated, it was my fault that I missed out on it. Sorry! $\endgroup$ – TestGuest Mar 6 '13 at 18:54

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