TITLE: Understanding Specker's disproof of the axiom of choice in New Foundations
QUESTION [17 upvotes]: Hi all! I am trying to understand Specker (1953)'s proof (found here) that the axiom of choice is false in New Foundations. I am stuck on the following point. At 3.5 Specker writes:
3.5. The cardinal numbers are well ordered by the relation "there are sets $a,b$ such that $a \in n, b \in m$ and $a \subseteq b$" (axiom of choice).
I am assuming that this is a consequence of the axiom of choice, which he is using to derive a contradiction. Is that true? If so, how is it a consequence of the axiom of choice?
Another, broader question: can anybody give an intuitive explanation of why AC fails in NF?
Thank you!

REPLY [3 votes]: See lemma 2.2.1 in my dissertation of NF in 1989 (G. Wagemakers, New Foundations: A survey of Quine’s Set Theory). 
More info can be found there, for example that there is no relation between $n$ and $T(n)$ at all. The one can be larger, equal or smaller than the other.
You're asking for an intuitive explanation why AC fails in NF. Let me give that a try (terminology as in my document; note that I skip parentheses that are not essential).
For cardinals $m$ it is easy to prove the following ($\Omega = |V|$):
\begin{cases}
2^{T\Omega}=\Omega\\
m \le T\Omega \implies 2^{Tm} = T2^m\\
\Omega > T\Omega\\
2^m \ne \Lambda \implies m < 2^m
\end{cases}
This leads to a descending series of cardinals:
$$\Omega > T\Omega > T^2\Omega > ...$$
If we would be able to form the set $Clos(\{\Omega\}, T)$ containing all these cardinals, then we would contradict AC as that set would not have a smallest element (the set of all cardinals, hence every subset of it, is well-ordered by $\le$).
But, alas, this set cannot be formed as $T$ is type-raising.
So we have to resort to a real set, intuitively containing at least the cardinal numbers in this list, and hope for the best. This set is $A = \{m \in NC |\Phi m \in Fin\}$, where
\begin{cases}
\Phi m = Clos(\{m\}, 2^{.}) - \{\Lambda\}\\
Fin = \bigcap Nn\\
Nn = Clos (\{0\}, . + 1)
\end{cases}
This set exists, as $2^.$ is type-preserving, and indeed contains $\Omega$ ($|\Phi \Omega| =1$), $T\Omega$ ($|\Phi T\Omega| =2$), etcetera (note that this fact doesn't helps us much, as this series of descending cardinals could be all larger than the minimum element of the set).
Assuming AC, A is well-ordered by $\le$ (NC is, and $A \subseteq NC$). Specker proceeds to derive an inconsistency: for the smallest element $n$ of this set, it can be derived that $n=Tn$. But then $|\Phi n| = |\Phi Tn| = T|\Phi n| + (1 \text{ or } 2)$, contradicting the fact that there can be no natural number $|\Phi n|$ having this property.
Obviously, a number of lemmata is required to derive these facts.
The conclusion is that the FULL axiom of choice AC does not hold in NF. Note that for cardinals m less than that of $A$, $AC(m)$ has not been disproved by Specker's argument.