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Ok, I think I got it.. There is no order among the elements of a set, hence, since an ordered pair (or n-touple in general) is a set, there has to be a way to imply order in set notation, as well as to keep the fact that (a, b) = (a', b') <=> a=a' & b=b' true. So, from a set {{a}, {a, b}} we can 'read': the set with only one element is {a}, which makes a the first element in the ordered pair (a, b). Assuming a does not equal b, we 'jump' to the next set {a, b}, and select the element b as the second element of (a, b).
Analogically, if we have a set { {a}, { {a}, {{b}, {b, c}} } }, we see that the set with one element is {a}, which makes a the first element in (a, b, c). Let's assume a, b and b, c are different. So, we 'jump' to the next set { {a}, {{b}, {b, c}} }. Since, a and b are different, we directly jump to the set {{b}, {b, c}} and select b for the second element of (a, b, c), since {b} is a singleton. And, finally, since b and c are different, we select c for the third element of (a, b, c)... Is this a correct way of thinking?
Analogically, if we have a set { {a}, { {a}, {{b}, {b, c}} } }, we see that the set with one element is {a}, which makes a the first element in (a, b, c). Let's assume a, b and b, c are different. So, we 'jump' to the next set { {a}, {{b}, {b, c}} }. Since, a and b are different, we directly jump to the set {{b}, {b, c}} and select b for the second element of (a, b, c), since {b} is a singleton. And, finally, since b and c are different, we select c for the third element of (a, b, c)... Is this a correct way of thinking?
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The whole point of "ordered pair" is that, unlike the set {a, b}, we distinguish between the two members. Writing (a,b)= {{a},{a,b}} just means that there are two members, a and b, and we distinguish between the two. Hurkyl's point about "the set {a, {a, b}} doesn't always have two elements" is that the "pair" (a,b) corresponds to the set {{a},{a,a}}. But since {a, a} is a set where we don't "double list" the same thing, {a,a} is the same as {a}. That means that {{a}, {a,a}}= {{a},{a}} which is exactly the same as {{a}}.
When talking about "ordered triples", we can think of (a,b,c) as the "ordered pair" ((a,b),c) where the first member is the ordered pair (a,b). That is the same as the set {{(a,b)}, {(a,b),c}}. But (a,b) is {{a},{a,b}} so {{(a,b)},{(a,b),c}}= {{{{a},{a,b}}},{{{a},{a,b}},c}}. Or we could write it as (a, (b,c))= {{a},{a,(b,c)}= {{a},{a,{b,{b,c}}}.
(That reminds me of the computer language "LISP"- "Lots of Insane, Silly Parentheses"!
When talking about "ordered triples", we can think of (a,b,c) as the "ordered pair" ((a,b),c) where the first member is the ordered pair (a,b). That is the same as the set {{(a,b)}, {(a,b),c}}. But (a,b) is {{a},{a,b}} so {{(a,b)},{(a,b),c}}= {{{{a},{a,b}}},{{{a},{a,b}},c}}. Or we could write it as (a, (b,c))= {{a},{a,(b,c)}= {{a},{a,{b,{b,c}}}.
(That reminds me of the computer language "LISP"- "Lots of Insane, Silly Parentheses"!
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