Ordered Pairs and Set Theory So, first of all I'm gonna start with stating that I'm a complete noob in mathematics that started doing this as a hobby. I encounter problems that I can't figure out on a regular basis, like a few times per page but I usually find someone to help me. This time those same people were of no help so I came here.
We have an ordered pair $(a,b)$, okay?
And if we wanted to show the world that ordered pair using set theory it would look like this: $\{\{a\},\{a,b\}\}$.
Yeah, I don't get that at all. As I've came to understand in the ordered pair order does matter. So how come it doesn't look like this when using set theory: $\{\{a\},\{b\}\}$? I know my question is probably retarded but this seems awfully confusing to me. How come the second coordinate can be either a or b? Logic makes me believe that my previous way of thinking can't be true and that I'm understanding it completely wrong. Can someone help me with this using simple language?
Improvements: Now I understand why $\{\{a\},\{b\}\}$ doesn't apply. Thanks for the first step helpers.
 A: Two sets are equal if and only if they share the same elements. Thus there is no distinction between the sets $\{\{a\},\{b\}\}$ and $\{\{b\},\{a\}\}.$ That's why we need a different trick to create a mathematical object involving $a$ and $b$ in some particular order so that $(a,b)\neq(b,a)$ unless $a=b.$ The model $\{\{a\},\{a,b\}\}$ that you cite is the most common one but not the only possibility.
In fact the full requirement for our model is: $(a,b)=(c,d)$ if and only if $a=c$ AND $b=d.$ It is not difficult to verify that the model in your book satisfies that requirement, using the criterion that sets are equal if and only if they contain the same elements: let us show that equality of the ordered pairs implies equality of the corresponding elements.
If the sets $\{\{a\},\{a,b\}\}$ and $\{c\},\{c,d\}\}$ are identical then we must have
$$(\{a\}=\{c\}\hbox{ OR }\{a\}=\{c,d\})\hbox{ AND }(\{a,b\}=\{c\}\hbox{ OR }\{a,b\}=\{c,d\})$$
and
$$(\{c\}=\{a\}\hbox{ OR }\{c\}=\{a,b\})\hbox{ AND }(\{c,d\}=\{a\}\hbox{ OR }\{c,d\}=\{a,b\}).$$
The first two equalities imply that $c$ is an element of the singleton $\{a\},$ so their OR implies $a=c.$
The last two equalities on the first line imply that $b=c$ or $b=d.$
The last two equalities on the second line imply that $d=a$ or $b=d.$
So we either have $b=d$ or $d=a=c=b.$ In either case, $a=c$ and $b=d.$
A: A short answer... but hopefully useful!
Indeed, the aim of an ordered pair, is that the order matters. Then the target is to define the ordered pair using classical "set constructions": union, intersection...
The issue with $\{\{a\},\{b\}\}$, is that you cannot differentiate the two elements of the set. You don't create "an order".
With $\{\{a\},\{a,b\}\}$ it is much better. One of the elements of the set $\{a\}$ has only ONE element, while the second $\{a,b\}$ has TWO elements. A good way to differentiate both of them... and create the "order" which was desired.
A: suppose $a$ and $b$ belong to a set $S$. then
$$
(a,b) =\{ \{a\}, \{a,b\}\} \in \mathfrak{P}^2(S)
$$
this implies the identification 
$$
(a,a) = \{\{a\}\}
$$
let $T=\{1,2\}$ and define the collection $P_S$ of p-maps on S  as the maps $t:T \to  \mathfrak{P}(S)$ satisfying:
$$
1 \le |t(k)| \le k 
$$
and
$$
j \lt k \Rightarrow t(j) \subseteq t(k)
$$
then every ordered pair is the image of a p-map and vice versa
A: It's because a set isn't concerned by order so the set $\{a, b\}$ is the same as the set $\{b, a\}$, but that's not all that's needs to be addressed. A set is not concerned with multiplicity either so the sets $\{a, a\}$ and $\{a\}$ are also the same. This is the reason one need to use something more clever than just putting the elements into a set to construct an ordered pair.
What we need is both a way to construct a set from an ordered pair, but also a way to from the set be able to extract the first and the second element.
Now there's multiple ways that would do to define an ordered set with the required properties. The way it's chosen here has a subtle "problem", if the elements are the same you would end up with $(a,a) = \{\{a\}, \{a, a\}\} = \{\{a\}\}$, but that's not really a problem. 
The recipe for extracting the first element from the pair is to in the set find (precicely) a set containing only one element. This element it contains is the first element.
The recipe for extracting the second element is that if there's only one set in the set, it contains the second element, otherwise there's a set containing two elements one of them which is the first - the other element in that set is the second element.
