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The book that I am reading says that the decomposition of a set $A$ is any representation of $A$ as the union of a disjoint family of set, $A=\sum_{i\in I}A_i$ (if a family of set is pairwise disjoint the author writes $\bigcup_{i\in I}A_i$ as $\sum_{i\in I}A_i$).

The family $\{A_i:i\in I\}$ is referred to as the partiton of the set $A$.

Does this imply that all of the elements of a partition on the set $A$ are pairwise disjoint? Also, is it true that the union over every element of $\{A_i:i\in I\}$ is equal to $A$?

Also, a quick question on notation, when we write a union in the form $\bigcup_{i\in I}A_i$ do we mean that we take the union for all $i$ in $I$? Or, we take the union for any $i$ in $I$ not necessarily every $i$?

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Yes: a partition of $A$ is a family $\mathscr{P}$ of pairwise disjoint non-empty subsets of $A$ whose union is $A$. (Sometimes the requirement that the members of the partition be non-empty is relaxed.) Thus, if $\mathscr{P}=\{A_i:i\in I\}$, then

$$\bigcup\mathscr{P}=\bigcup_{i\in I}A_i=\bigcup\{A_i:i\in I\}=A$$

And yes, all of the union notations above mean the union of all of the sets in the collection.

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  • $\begingroup$ Thanks! That makes things a lot more clear. $\endgroup$ – JimmyJackson Oct 7 '13 at 22:41
  • $\begingroup$ @Jimmy: You’re welcome! $\endgroup$ – Brian M. Scott Oct 7 '13 at 22:42
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$\cup_{i\in I} A_i$ means you take every value of the index $i$, within $I$. Same goes for $\sum_{i\in I}$.

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  • $\begingroup$ Thanks for clarifying this, I have come across many summations of that form and I was never quite sure on the meaning of this bound. $\endgroup$ – JimmyJackson Oct 7 '13 at 22:43

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