Ordered Pairs and Ordinals, is there an equivalence? Can anyone explain if there is a relationship between ordered pairs and the Von Neumann ordinals?  It's my understanding the generally accepted definition of an ordered pair is:
$\langle x,y \rangle = \{\{x\}, \{x, y\}\} $
It's also my understanding the successor ordinal is:
$S(\alpha) = \alpha \space \cup \{\alpha\}$
however, looking at the ordinals, they don't appear in the form of:
$\langle\langle x,y\rangle, z\rangle = \langle x,y,z\rangle$
which would imply a ordered sequence.  Is it that each ordinal is just composed of all the objects less then the limit and there is no order sequence involved unless the subset relation is used?
PS = I also tried using the short definition of the ordered pair $O_{short} = \{x, \{x, y\}\} $, however I do not see a relationship here either.
Appreciate any pointers to help understand this.
Thanks again!
 A: Unfortunately I don't understand the main point of your question. But let me answer the following 

[..] however, looking at the ordinals, they don't appear in the form of:
$\lt\lt x,y\gt, z\gt = \lt x,y,z\gt$

Claim. The only ordinal expressible as a (nested) pair via the Kuratowski pairing function is $2 = \{ \emptyset, \{\emptyset\}$ }$.
Proof. $2 = \{ \emptyset, \{ \emptyset \} \} = \langle \emptyset, \emptyset \rangle$.
For an ordinal $\alpha \neq 2$ just note that $\alpha$ has either at most one or at least three members (namely $0,1,2$) and that any (nested) Kuratowski pair has exactly two members. Q.E.D.
A: 
there is no order sequence involved unless the subset relation is used?

Yes, and I think this is a key part of your question. You can define an ordered set as a certain structure $(S,<)$ where $S$ is a set and $<$ is a set of ordered pairs of elements in $S$. That is, ${<}\subset S\times S$. Then of course you impose the usual axioms on $<$, like transitivity.
With this formulation, the relation $<$ need not have anything to do with $\in$ or $\subset$. And $<$ need not be a subset of $S$, either, or have any intersection with $S$. So it doesn't matter whether some of the elements of $S$ are ordered pairs themselves: whether Kuratowski ordered pairs, or any other definition.
Finally, you can say that the set $S$ itself doesn't have any ordering until you specify $(S,<)$. And whenever you speak of $S$ as an ordered set, we understand that it's a slight abuse of notation, and you really mean $(S,<)$.
