I need to disprove an alternate definition of an ordered pair. Why is $\langle a,b\rangle = \{a,\{b\}\}$ incorrect? So we know that the an ordered pair $(a,b) = (c,d)$ if and only if $a = c$ and $b = d$.  And we know the Kuratowski definition of an ordered pair is: $(a,b) = \{\{a\},\{a,b\}\}$
http://en.wikipedia.org/wiki/Ordered_pair#Kuratowski_definition
The proof for the Kuratowski definition is in the wikipedia link.
Now, why is this alternate definition, $(a,b) = \{a,\{b\}\}$ incorrect?
I'm trying to follow the proof for $(a,b) = (c,d)$ iff $a = c$ and $b = d$ as given in the wikipedia link, only for this alternate definition for an ordered pair, in order to search for a contradiction.  But I don't think I'm dong it right.
I started with...


*

*$(a,b) = (c,d)$

*Then $\{a,\{b\}\} = \{c,\{d\}\}$ based on the alternate definition


Now...


*

*Suppose $a \neq b$

*

*$\{a,\{b\}\} = \{c,\{d\}\}$

*But since it's an ordered pair, either of the following can be true?

*

*$a = c$ and $\{b\} = \{d\}$ ?

*OR $a = \{d\}$ and $\{b\} = c$ ?




Yeah I have no idea where to reallly go from here.   Is that a contradiction in itself? I can't tell.
Thank you for the help. 
Edit: Alright, I have developed a counter example based mostly off of Asaf Karagila answer (Thanks Asaf!).  Essentially what I needed to do was prove that, by this definition, a != c or b != d, even when (a,b) = (c,d).
So using what Asaf told me, I set a = {x} and b = y. Which by the incorrect definition gives...
(a,b) = {{x},{y}}
Then I set c = {y} and d = x, which gives
(c,d) = {{y},{x}} which is equivalent to {{x},{y}}
So, (a,b) = (c,d) even though a != c and b != d, which is a contradiction. I cleared this method with my professor.
Thanks for the help everyone! 
 A: The decoding of the first and second elements is not unique. Suppose that $c\neq d$, now we have:
$$(\{c\},d)=\{\{c\},\{d\}\}=(\{d\},c)$$
Recall that the definition of ordered pairs should not only hold for pairs of numbers. It should allow set theory be adequate for expressing a lot. Using this definition, the above shows that the two functions:


*

*$f(x)=\{x\}$

*$g(\{x\})=x$


Are both represented by the same set. Which is usually a sign of an inadequate representation of an ordered pair.
A: The ordered pair definition is meant to work for any kind of set, so assuming you work with numbers is really not the right way to go. 
There is a reason why the "second coordinate" must contain two elements. Here's why.
Say you define ordered pairs $(a,b) = \{a, \{b\}\}$. Is there anyway you can decide "which one is the first coordinate"? a priori both elements of the ordered pair are sets, and you have no reason to believe that you can distinguish between one of the two by the fact that they belong to some other set. For instance if you take an ordered pair $(a,b) \in \mathcal P(X) \times X$, then for some element $x \in X$ you can consider the ordered pair $(\{x\},x)$, which under your definition gives the set $\{\{x\},\{x\}\}$. Therefore there is no way to distinguish between both coordinates.
Kuratowski's definition ensures that we can know which coordinate is the first /second one because there are well-defined operators that output the right coordinate. Your website linked explains it well.
Hope that helps,
