# 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): $V=\bigcup_{\alpha} V_{\alpha}$)

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.

• I wonder if instead of $V=\bigcup_{\alpha} V_{\alpha}$ we define $V_{Ord}=\bigcup_{\alpha} V_{\alpha}^{Ord}$ where $V_{\alpha}^{Ord}$ is the ordinal of the same cardinality of $V_{\alpha}$. In this case $V_{Ord}$ is a set, right? – Wolphram jonny Aug 18 '14 at 4:25
• No. You are taking a union over a proper class. Here's a nice analogy which might be helpful. Consider only the finite cases (i.e. work in $\sf ZFC-Inf+\lnot Inf$, where all the ordinals are just natural numbers). Then you are taking an infinite union over finite ordinals. Is the result likely to be a finite ordinal? – Asaf Karagila Aug 18 '14 at 4:46
• I want to understand what is wrong in the original argument. Is it then that"... 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)" is wrong? is it true only for a finite union?. – Wolphram jonny Aug 18 '14 at 5:06
• Ugh. No. Let me start over. Consider the axioms of $\sf ZFC$ but replace the axiom of infinity by its negation. Now every set is finite. Now you have a proper class (read: an infinite collection) of ordinals, and you take a union of that class. So you are asking whether or not a union of infinitely finite ordinals is a finite ordinal. Is it? Do you expect it to be? Now back in the usual $\sf ZFC$ axioms, we replace "finite" by "set" and "infinite" by "proper class". You are taking a union over a collection which is not a set, why do you expect it to be a set? – Asaf Karagila Aug 18 '14 at 5:33
• because wikipedia says so? quote: "... every set of ordinals has a supremum, ** bold 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)". – Wolphram jonny Aug 18 '14 at 6:56

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.

• I understand the argument that V cannot be a set. I am asking why it does not contradict the Von Neumann definition. Or at least what is wrong in the quoted text. – Wolphram jonny Aug 18 '14 at 2:09
• Asaf's answer was more direct, the wrong assumption was that the sets V_alpha were ordinals. I am not sure if you stated that, perhaps you did but I didnt get it. Thanks anyways. – Wolphram jonny Aug 18 '14 at 2:19
• I certainly didn't have in mind that $V_\alpha$ is an ordinal. Are you somehow getting that from my answer? – Michael Hardy Aug 18 '14 at 2:28
• No, I did! but that was wrong and I am not sure if your answer was pointing at that or not. Sorry for the confusion. – Wolphram jonny Aug 18 '14 at 2:32