Hi all! I am trying to understand Specker (1953)'s proof (found [here](http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1063889/pdf/pnas01594-0080.pdf)) 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!



Understanding Specker's disproof of the axiom of choice in New Foundations
First notice that one can carry out in NF Zermelo's proof that the axiom of choice implies that all sets can be well-ordered. It follows that the relation on cardinals defined in your quote from Specker satisfies the conditions for a linear ordering. (It's trivially reflexive and transitive. Antisymmetry is the Cantor-Schröder-Bernstein theorem. Linearity follows from well-ordering because, for any two well-ordered sets, one is in (order-preserving) bijection with an initial segment (possibly improper) of the other.) It remains to show that, in any nonempty collection $X$ of cardinals, there is a smallest one. I'll use the fact that there is a set $U$ so big that every cardinal in $X$ occurs as the cardinality of a subset of $U$. In ZFC-style set theories, I'd get $U$ by choosing a representative set for each cardinality in $X$ and taking the union of these representatives; in NF it's easier, since the universe $V$ is a set that can serve as $U$. Well-order $U$. By what I already said, each cardinal $x\in X$ is the cardinality of either $U$ or a proper initial segment of $U$. If all are the cardinality of $U$, then $X$ contains just one cardinal, which is obviously least. So suppose some $x\in X$ is represented by an initial segment of $U$. (The same $x$ may correspond to several initial segments; that's OK.) Among all those initial segments, for all $x\in X$, there is a shortest because $U$ is well-ordered. The correponding $x$ is the smallest cardinal in $X$.


