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Suppose that $A=\{A_i:i \in I\}$ is a family of sets. Consider the case where we want to define a set $S$ which has a unique element from each set $A_i$ and has no more elements than that.

For example, if $A= \{1,2,3\} , \{ a,b,c \} , \{ !,\#,@ \} \}$ then $S$ may be $\{1,a,!\}$ or $\{1,b,@\}$ but $S$ can't be $\{1,a\}$ as $S$ didn't have an element from the set $\{!,\#,@\}$ and the set $\{1,2,a,@\}$ can't be $S$ as it has two elements from the set $\{1,2,3\}$.

So $S$ has an element from each set $A_i$ and only one element from this $A_i$.

My question is how to define this set formally.

my attempt is , $S \subset \bigcup_{i \in I} A_i$, where for every $i \in I$ , $S \cap A_i =\{a\}$ for some $a \in A_i$.

Does this definition work ?

Are there any better definitions you suggest ( in builder-notation for instance?) ?

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Please use \{ and \} to put braces $\{$ and $\}$ in your math. You can click the edit button to fix this. – dfeuer Oct 2 '13 at 0:56
@Element Without further restriction such a set may not exists. – azarel Oct 2 '13 at 1:24
@azarel ,Can you clarify that more please ? Why without more restriction such a set may not exists ? and what are those conditions which guarantee its existence ? – Element Oct 2 '13 at 16:40
@Element For example, take $A$ equal to all infinite subsets of the natural numbers, then there is no $S$ satisfying your definition. – azarel Oct 2 '13 at 17:16
@azarel , Why not ? notice that $S$ take a unique element from each A , but it may take $1$ from a set $A$ and another $1$ from a set $B$ ( but here we would allow repition of elements of the set ). – Element Oct 2 '13 at 17:24

Your definition looks fine to me.

I would probably let $A$ be the union of the $A_i$, and then write:

"Pick $S \subset A$ such that $|A_{i} \cap S| = 1$ for all $i \in I$."

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What is $|A_i| \cap S $ ? – Element Oct 2 '13 at 16:48
a typo! I have fixed it; should be: $|A_i \cap S|$, which means "the number of elements in both $A_i$ and $S$." – Benjamin Dickman Oct 2 '13 at 17:24

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