TITLE: Please Explain Kuratowski Definition of Ordered Pairs
QUESTION [25 upvotes]: I've seen this Kuratowski definition for ordered pairs, but can't fathom why it implies an order to $x$ and $y$
$(x,y):=\{\{x\}, \{x,y\}\}$
As I understand sets, $\{\{x\}, \{x,y\}\}$ is also $\{\{x,y\}, \{x\}\}$. Only when I think about the Axiom of Union does $\{\{x\}, \{x,y\}\}$ "collapse down" to $S = \{x, y\}$, but that doesn't give me much either. All I can see is some as yet hidden message in the set saying "I am the set $\{x,y\}$ and my order of $x$ first is indicated by having $\{x\}$ along for the ride."

REPLY [3 votes]: The point of the ordered is for it to satisfy the following property:
$$(a,b) = (c,d) \text{ if and only if }a=c\text{ and }b=d.$$
The definition given satisfies that condition; it doesn’t matter how it satisfies it, just that it does.
It is true that $(a,b)$ is both $\{ \{a\},\{a,b\}\}$ and $\{\{a\}, \{b,a\}\}$ and $\{\{a,b\},\{a\}\}$ and $\{\{b,a\},\{a\}\}$. It doesn’t matter: they are all the same set and they all “are” $(a,b)$. The point is that $(a,b)=(c,d)$ if and only if $a=c$ and $b=d$.
To verify that this happens, note that if $a=c$ and $b=d$, then
$$(a,b) = \{\{a\},\{a,b\}\} = \{\{c\},\{c,d\}\} = (c,d).$$
Conversely, suppose that $(a,b)=(c,d)$. That means that the sets
$\{\{a\},\{a,b\}\}$ and $\{\{c\},\{c,d\}\}$ are equal as sets.
To verify that this implies that $a=c$ and $b=d$, consider two cases:
Case 1. $a=b$. Then $(a,b) = \{ \{a\},\{a,b\}\} = \{\{a\},\{a,a\}\} = \{ \{a\}\}$. 
As this is equal to $\{\{c\},\{c,d\}\}$, then $\{c\}\in\{\{a\}\}$, which means $\{c\}=\{a\}$, which means $c=a$. And $\{c,d\}\in\{\{a\}\}$, so $\{c,d\} = \{a\}$, hence $d=a=b$. Thus, $a=c$, and $b=d$ in this case.
Case 2. $a\neq b$. 
Recall that is $X$ is a set, whose elements are sets, then
$$\begin{align*}
\cup X &= \bigcup_{S\in X}S\\
\cap X &= \{s\in\cup X\mid s\in S\text{ for all }S\in X\}.
\end{align*}$$
If $(a,b) = (c,d)$, then, as sets, $\cap (a,b) = \cap (c,d)$. We have
$$\begin{align*}
\cap (a,b) &= \cap\{ \{a\},\{a,b\}\}\\
&= \{a\}\cap\{a,b\} = \{a\}\\
\cap (c,d) &= \cap \{ \{c\},\{c,d\}\}\\
&= \{c\}\cap\{c,d\} = \{c\}.
\end{align*}$$
Therefore, $\{a\}=\{c\}$, so $a=c$.
And $\cup (a,b) - \cap (a,b)$ must be equal to $\cup(c,d)-\cap(c,d)$, so
$$\begin{align*}
\cup(a,b)-\cap(a,b) &= \cup\{ \{a\},\{a,b\}\} - \{a\}\\
&= (\{a\}\cup\{a,b\})- \{a\}\\
&= \{a,b\} - \{a\} = \{b\};\\
\cup(c,d)-\cap(c,d) &= \cup\{ \{c\}, \{c,d\}\} - \{c\}\\
&= (\{c\}\cup\{c,d\})-\{c\} = \{c,d\}-\{c\} = \{d\},
\end{align*}$$
so we must have $\{b\}=\{d\}$, and therefore $b=d$.
In both cases, if $(a,b)=(c,d)$, as sets, then $a=c$ and $b=d$.
Which is what we want from an ordered pair. So now that we know this definition will guarantee this, we can safely forget about it in general and just use the “defining property” of the ordered pair: $(a,b) = (c,d)$ if and only if $a=c$ and $b=d$.