Following the Von Neumann definition of ordinal, why $V$ is not a set? According to wikipedia (http://en.wikipedia.org/wiki/Ordinal_number#Closed_unbounded_sets_and_classes)
(section "Von Neumann definition of ordinals"):
"... every set of ordinals has a supremum, the ordinal obtained by taking the union of all the ordinals in the set (this union exists regardless of the set's size, by the axiom of union)". 
If this is correct, then why V is not a set? (assuming the Von Neumann definition of V (http://en.wikipedia.org/wiki/Von_Neumann_universe): $V=\bigcup_{\alpha} V_{\alpha}$)
 A: Your question has two parts which are really unrelated. Every set of sets has a union, that is the axiom of union. From this follows that every set of ordinals has a supremum, its union.
Then you ask on something which is very much not a set of ordinals $\{V_\alpha\mid\alpha\in\rm Ord\}$. Standard arguments show that $V$ is not a set, therefore this collection of sets is not a set either.
Similarly, when you have a regular cardinal, like $\omega_1$, the union of every countable set of countable ordinals is not $\omega_1$; or in general the countable union of countable sets is countable. When the collection you're taking union over is not a set, then there is no reason to expect that the resulting union is a set.
A: If $V$ were a set, then it would be a member of $V$, since all sets are members of $V$, but then it would have as its rank some ordinal, and there would be greater ordinals $\alpha$, hence sets $V_\alpha$ of which $V$ is a subset, and which do not belong to $V$.
Remember that the class of all ordinals is not a set.  If it were a set, then it would be an ordinal, and it would have a successor ordinal, and the ordinals would then go on from there, beyond the extent of how far ordinals go.
